NCERT Solutions-Redox Reactions for Class 11 Chemistry Chapter 8

February 14, 2025December 18, 2024 by Neeraj Anand, Param Anand

NCERT Solutions for Class 11 Chemistry Chapter Redox Reactions

1. Assign oxidation number to the underlined elements in each of the following species:

(a)

\(\begin{array}{l}NaH_2\underline{P}O_4\end{array} \)

(b)

\(\begin{array}{l}NaH\underline{S}O_{4}\end{array} \)

(c)

\(\begin{array}{l}H_{4}\underline{P}_{2}O_{7}\end{array} \)

(d)

\(\begin{array}{l}K_{2}\underline{Mn}O_{4}\end{array} \)

(e)

\(\begin{array}{l}Ca\underline{O}_{2}\end{array} \)

(f)

\(\begin{array}{l}Na\underline{B}H_{4}\end{array} \)

(g)

\(\begin{array}{l}H_{2}\underline{S}_{2}O_{7}\end{array} \)

(h)

\(\begin{array}{l}KAl(\underline{S}O_{4})_{2}.12H_{2}O\end{array} \)

Answer:

(a)

\(\begin{array}{l}NaH_2\underline{P}O_4\end{array} \)

Let x be the oxidation no. of P.

Oxidation no. of Na = +1

Oxidation no. of H = +1

Oxidation no. of O = -2

1. Assign oxidation number to the underlined elements in each of the following species:

Then,

1(+1) + 2(+1) + 1(x) + 4(-2) = 0

1 + 2 + x -8 = 0

x = +5

Therefore, oxidation no. of P is +5.

(b)

\(\begin{array}{l}NaH\underline{S}O_{4}\end{array} \)

Let x be the oxidation no. of S.

Oxidation no. of Na = +1

Oxidation no. of H = +1

Oxidation no. of O = -2

Then,

1(+1) + 1(+1) + 1(x) + 4(-2) = 0

1 + 1 + x -8 = 0

x = +6

Therefore, oxidation no. of S is +6.

(c)

\(\begin{array}{l}H_{4}\underline{P}_{2}O_{7}\end{array} \)

Let x be the oxidation no. of P.

Oxidation no. of H = +1

Oxidation no. of O = -2

Then,

4(+1) + 2(x) + 7(-2) = 0

4 + 2x – 14 = 0

2x = +10

x = +5

Therefore, oxidation no. of P is +5.

(d)

\(\begin{array}{l}K_{2}\underline{Mn}O_{4}\end{array} \)

Let x be the oxidation no. of Mn.

Oxidation no. of K = +1

Oxidation no. of O = -2

Then,

2(+1) + x + 4(-2) = 0

2 + x – 8 = 0

x = +6

Therefore, oxidation no. of Mn is +6.

(e)

\(\begin{array}{l}Ca\underline{O}_{2}\end{array} \)

Let x be the oxidation no. of O.

Oxidation no. of Ca = +2

Then,

(+2) + 2(x) = 0

2 + 2x = 0

2x = -2

x = -1

Therefore, oxidation no. of O is -1.

(f)

\(\begin{array}{l}Na\underline{B}H_{4}\end{array} \)

Let x be the oxidation no. of B.

Oxidation no. of Na = +1

Oxidation no. of H = -1

Then,

1(+1) + 1(x) + 4(-1) = 0

1 + x -4 = 0

x = +3

Therefore, oxidation no. of B is +3.

(g)

\(\begin{array}{l}H_{2}\underline{S}_{2}O_{7}\end{array} \)

Let x be the oxidation no. of S.

Oxidation no. of H = +1

Oxidation no. of O = -2

Then,

2(+1) + 2(x) + 7(-2) = 0

2 + 2x – 14 = 0

2x = +12

x = +6

Therefore, oxidation no. of S is +6.

(h)

\(\begin{array}{l}KAl(\underline{S}O_{4})_{2}.12H_{2}O\end{array} \)

Let x be the oxidation no. of S.

Oxidation no. of K = +1

Oxidation no. of Al = +3

Oxidation no. of O= -2

Oxidation no. of H = +1

Then,

1(+1) + 1(+3) + 2(x) + 8(-2) + 24(+1) + 12(-2) = 0

1 + 3 + 2x – 16 + 24 – 24 = 0

2x = +12

x = +6

Therefore, oxidation no. of S is +6.

Ignore the water molecules because it is neutral. Then, the summation of the oxidation no. of all atoms of water molecules can be taken as 0. Hence, ignore the water molecule.

1(+1) + 1(+3) + 2(x) + 8(-2) = 0

1 + 3 + 2x -16 = 0

2x = 12

x = +6

Therefore, oxidation no. of S is +6.

2.What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results?

(a)

\(\begin{array}{l}K\underline{I}_{3}\end{array} \)

(b)

\(\begin{array}{l}H_{2}\underline{S}_{4}O_{6}\end{array} \)

(c)

\(\begin{array}{l}\underline{Fe}_{3}O_{4}\end{array} \)

(d)

\(\begin{array}{l}\underline{C}H_{3}\underline{C}H_{2}OH\end{array} \)

(e)

\(\begin{array}{l}\underline{C}H_{3}\underline{C}OOH\end{array} \)

Answer:

(a)

\(\begin{array}{l}K\underline{I}_{3}\end{array} \)

Let x be the oxidation no. of I.

Oxidation no. of K = +1

Then,

1(+1) + 3(x) = 0

1 + 3x = 0

x =

\(\begin{array}{l}-\frac{1}{3}\end{array} \)

Oxidation no. cannot be fractional. Hence, consider the structure of

\(\begin{array}{l}KI_{3}\end{array} \)

molecule, an iodine atom forms a coordinate covalent bond with an iodine molecule.

Therefore, in

\(\begin{array}{l}KI_{3}\end{array} \)

molecule, the oxidation no. of I atoms forming the molecule

\(\begin{array}{l}I_{2}\end{array} \)

is 0, while the oxidation no. of I atom, which is forming coordinate bond is -1.

(b)

\(\begin{array}{l}H_{2}\underline{S}_{4}O_{6}\end{array} \)

Let x be the oxidation no. of S.

Oxidation no. of H = +1

Oxidation no. of O = -2

Then,

2(+1) + 4(x) + 6(-2) = 0

2 + 4x -12 = 0

4x = 10

x =

\(\begin{array}{l}+2\frac{1}{2}\end{array} \)

Oxidation no. cannot be fractional. Therefore, S would be present with different oxidation states in the molecule.

The oxidation no. of two out of the four S atoms is +5 while that of the other two atoms is 0.

(c)

\(\begin{array}{l}\underline{Fe}_{3}O_{4}\end{array} \)

Let x be the oxidation no. of Fe.

Oxidation no. of O = -2

Then,

3(x) + 4(-2) = 0

3x -8 = 0

x =

\(\begin{array}{l}\frac{8}{3}\end{array} \)

Oxidation no. cannot be fractional.

One of the three atoms of Fe has oxidation no. +2 and the other two atoms of Fe have oxidation no. +3.

(d)

\(\begin{array}{l}\underline{C}H_{3}\underline{C}H_{2}OH\end{array} \)

Let x be the oxidation no. of C.

Oxidation no. of O= -2

Oxidation no. of H = +1

Then,

2(x) + 4(+1) + 1(-2) = 0

2x + 4 -2 = 0

x = -2

Therefore, oxidation no. of C is -2.

(e)

\(\begin{array}{l}\underline{C}H_{3}\underline{C}OOH\end{array} \)

Let x be the oxidation no. of C.

Oxidation no. of O= -2

Oxidation no. of H = +1

Then,

2(x) + 4(+1) + 2(-2) = 0

2x + 4 -4 = 0

x = 0

Therefore, the average oxidation no. of C is 0. Both the carbon atoms are present in different environments, so they cannot have the same oxidation no. Therefore, carbon has oxidation no. of +2 and _2 in

\(\begin{array}{l}CH_{3}COOH\end{array} \)

3. Justify that the following reactions are redox reactions:

(a)

\(\begin{array}{l}CuO_{(s)} \; + \; H_{2 \; (g)} \; \rightarrow \; Cu_{(s)} \; + \; H_{2}O_{(g)}\end{array} \)

(b)

\(\begin{array}{l}Fe_{2}O_{3 \; (s)} \; + \; 3 \; CO_{(g)} \; \rightarrow \; 2 \; Fe_{(s)} \; + \; 3 \; CO_{2 \; (g)}\end{array} \)

(c)

\(\begin{array}{l}4 \; BCl_{3 \; (g)} \; + \; 3 \; LiAlH_{4 \; (s)} \; \rightarrow \; 2 \; B_{2}H_{6 \; (g)} \; + \; 3 \; LiCl_{(s)} \; + \; 3 \; AlCl_{3 \; (s)}\end{array} \)

(d)

\(\begin{array}{l}2\;K_{(s)}\;+\;F_{2\;(g)}\;\rightarrow\;2\;K\;+\;F_{(s)}\end{array} \)

(e)

\(\begin{array}{l}4\;NH_{3\;(g)}\;+\;5\;O_{2\;(g)}\;\rightarrow \; 4 \;NO_{(g)}\;+\;6\;H_{2}O_{(g)}\end{array} \)

Answer:

(a)

\(\begin{array}{l}CuO_{(s)} \; + \; H_{2 \; (g)} \; \rightarrow \; Cu_{(s)} \; + \; H_{2}O_{(g)}\end{array} \)

Oxidation no. of Cu and O in

\(\begin{array}{l}CuO\end{array} \)

is +2 and -2, respectively.

Oxidation no. of

\(\begin{array}{l}H_{2}\end{array} \)

is 0.

Oxidation no. of Cu is 0.

Oxidation no. of H and O in

\(\begin{array}{l}H_{2}O\end{array} \)

is +1 and -2, respectively.

The oxidation no. of Cu decreased from +2 in

\(\begin{array}{l}CuO\end{array} \)

to 0 in Cu. That is

\(\begin{array}{l}CuO\end{array} \)

is reduced to Cu.

The oxidation no. of H increased from 0 to +1 in

\(\begin{array}{l}H_{2}\end{array} \)

. That is

\(\begin{array}{l}H_{2}\end{array} \)

is oxidized to

\(\begin{array}{l}H_{2}O\end{array} \)

Therefore, the reaction is a redox reaction.

(b)

\(\begin{array}{l}Fe_{2}O_{3 \; (s)} \; + \; 3 \; CO_{(g)} \; \rightarrow \; 2 \; Fe_{(s)} \; + \; 3 \; CO_{2 \; (g)}\end{array} \)

In the above reaction,

Oxidation no. of Fe and O in

\(\begin{array}{l}Fe_{2}O_{3}\end{array} \)

is +3 and -2, respectively.

Oxidation no. of C and O in CO is +2 and -2, respectively.

Oxidation no. of Fe is 0.

Oxidation no. of C and O in

\(\begin{array}{l}CO_{2}\end{array} \)

is +4 and -2, respectively.

The oxidation no. of Fe decreased from +3 in

\(\begin{array}{l}Fe_{2}O_{3}\end{array} \)

to 0 in Fe. That is

\(\begin{array}{l}Fe_{2}O_{3}\end{array} \)

is reduced to Fe.

The oxidation no. of C increased from 0 to +2 in CO to +4 in

\(\begin{array}{l}CO_{ 2 }\end{array} \)

. That is, CO is oxidized to

\(\begin{array}{l}CO_{ 2 }\end{array} \)

Therefore, the reaction is a redox reaction.

(c)

\(\begin{array}{l}4 \; BCl_{3 \; (g)} \; + \; 3 \; LiAlH_{4 \; (s)} \; \rightarrow \; 2 \; B_{2}H_{6 \; (g)} \; + \; 3 \; LiCl_{(s)} \; + \; 3 \; AlCl_{3 \; (s)}\end{array} \)

the above reaction,

Oxidation no. of B and Cl in

\(\begin{array}{l}BCl_{3}\end{array} \)

is +3 and -1, respectively.

Oxidation no. of Li, Al and H in

\(\begin{array}{l}LiAlH_{4}\end{array} \)

is +1, +3 and -1, respectively.

Oxidation no. of B and H in

\(\begin{array}{l}B_{2}H_{6}\end{array} \)

is -3 and +1, respectively.

Oxidation no. of Li and Cl in LiCl is +1 and -1, respectively.

Oxidation no. of Al and Cl in

\(\begin{array}{l}AlCl_{3}\end{array} \)

is +3 and -1, respectively.

The oxidation no. of B decreased from +3 in

\(\begin{array}{l}BCl_{3}\end{array} \)

to -3 in

\(\begin{array}{l}B_{2}H_{6}\end{array} \)

. That is

\(\begin{array}{l}BCl_{3}\end{array} \)

is reduced to

\(\begin{array}{l}B_{2}H_{6}\end{array} \)

The oxidation no. of H increased from -1 in

\(\begin{array}{l}LiAlH_{4}\end{array} \)

to +1 in

\(\begin{array}{l}B_{2}H_{6}\end{array} \)

. That is

\(\begin{array}{l}LiAlH_{4}\end{array} \)

is oxidized to

\(\begin{array}{l}B_{2}H_{6}\end{array} \)

Therefore, the reaction is a redox reaction.

(d)

\(\begin{array}{l}2 \; K_{(s)} \; + \; F_{2 \; (g)} \; \rightarrow \; 2 \; K \; + \; F_{(s)}\end{array} \)

In the above reaction,

Oxidation no. of K is 0.

Oxidation no. of F is 0.

Oxidation no. of K and F in KF is +1 and -1, respectively.

The oxidation no. of K increased from 0 in K to +1 in KF. That is K is oxidized to KF.

The oxidation no. of F decreased from 0 in

\(\begin{array}{l}F_{2}\end{array} \)

to -1 in KF. That is

\(\begin{array}{l}F_{2}\end{array} \)

is reduced to KF.

Therefore, the reaction is a redox reaction.

(e)

\(\begin{array}{l}4 \; NH_{3 \; (g)} \; + \; 5 \; O_{2 \; (g)} \; \rightarrow \; 4 \; NO_{(g)} \; + \; 6 \; H_{2}O_{(g)}\end{array} \)

In the above reaction,

Oxidation no. of N and H in

\(\begin{array}{l}NH_{3}\end{array} \)

is -3 and +1, respectively.

Oxidation no. of

\(\begin{array}{l}O_{2}\end{array} \)

is 0.

Oxidation no. of N and O in NO is +2 and -2, respectively.

Oxidation no. of H and O in

\(\begin{array}{l}H_{2}O\end{array} \)

is +1 and -2, respectively.

The oxidation no. of N increased from -3 in

\(\begin{array}{l}NH_{ 3 }\end{array} \)

to +2 in NO.

The oxidation no. of

\(\begin{array}{l}O_{2}\end{array} \)

decreased from 0 in

\(\begin{array}{l}O_{ 2 }\end{array} \)

to -2 in NO and

\(\begin{array}{l}H_{ 2 }O\end{array} \)

. That is

\(\begin{array}{l}O_{ 2 }\end{array} \)

is reduced.

Therefore, the reaction is a redox reaction.

4. Fluorine reacts with ice and results in the change:

\(\begin{array}{l}H_{ 2 }O_{ (s) } \; + \; F_{ 2 \; (g) } \; \rightarrow \; HF_{ (g) } \; + \; HOF_{ (g) }\end{array} \)

Justify that this reaction is a redox reaction

Answer:

\(\begin{array}{l}H_{ 2 }O_{ (s) } \; + \; F_{ 2 \; (g) } \; \rightarrow \; HF_{ (g) } \; + \; HOF_{ (g) }\end{array} \)

In the above reaction,

Oxidation no. of H and O in

\(\begin{array}{l}H_{ 2 }O\end{array} \)

is +1 and -2, respectively.

Oxidation no. of

\(\begin{array}{l}F_{ 2 }\end{array} \)

is 0.

Oxidation no. of H and F in HF is +1 and -1, respectively.

Oxidation no. of H, O and F in HOF is +1, -2 and +1, respectively.

The oxidation no. of F increased from 0 in

\(\begin{array}{l}F_{ 2 }\end{array} \)

to +1 in HOF.

The oxidation no. of F decreased from 0 in

\(\begin{array}{l}O_{ 2 }\end{array} \)

to -1 in HF.

Therefore, F is both reduced as well as oxidized. So, it is a redox reaction.

5. Calculate the oxidation no. of sulphur, chromium and nitrogen in

\(\begin{array}{l}H_{ 2 }SO_{ 5 }\end{array} \)

\(\begin{array}{l}Cr_{ 2 }O_{ 7 }^{ 2- }\end{array} \)

and

\(\begin{array}{l}NO_{ 3 }^{ – }\end{array} \)

. Suggest structure of these compounds. Count for the fallacy.

Answer:

For

\(\begin{array}{l}H_{ 2 }SO_{ 5 }\end{array} \)

Let x be the oxidation no. of S.

Oxidation no. of O= -2

Oxidation no. of H = +1

Then,

2(+1) + 1(x) + 5(-2) = 0

2 + x – 10 = 0

x = +8

But the oxidation no. of S cannot be +8 as S has 6 valence electrons. Therefore, the oxidation no. of S cannot be more than +6.

The structure of

\(\begin{array}{l}H_{ 2 }SO_{ 5 }\end{array} \)

is as given below:

Now,

2(+1) + 1(x) + 3(-2) + 2(-1) = 0

2 + x – 6 – 2 = 0

x = +6

Therefore, the oxidation no. of S is +6.

For

\(\begin{array}{l}Cr_{ 2 }O_{ 7 }^{ 2- }\end{array} \)

Let x be the oxidation no. of Cr.

Oxidation no. of O= -2

Then,

2(x) + 7(-2) = -2

2x -14 = -2

x = +6

There is no fallacy about the oxidation no. of Cr in

\(\begin{array}{l}Cr_{ 2 }O_{ 7 }^{ 2- }\end{array} \)

The structure of

\(\begin{array}{l}Cr_{ 2 }O_{ 7 }^{ 2- }\end{array} \)

is as given below.

Each of the two Cr atoms has the oxidation no. of +6.

For

\(\begin{array}{l}NO_{ 3 }^{ – }\end{array} \)

Let x be the oxidation no. of N.

Oxidation no. of O= -2

Then,

1(x) + 3(-2) = -1

x – 6 = -1

x = +5

There is no fallacy about the oxidation no. of N in

\(\begin{array}{l}NO_{ 3 }^{ – }\end{array} \)

The structure of

\(\begin{array}{l}NO_{ 3 }^{ – }\end{array} \)

is as given below.

Nitrogen atom has the oxidation no. of +5.

6. Write formulas for the following compounds:

(a) Mercury (II) chloride (b) Nickel (II) sulphate

(c) Tin (IV) oxide (d) Thallium (I) sulphate

(e) Iron (III) sulphate (f) Chromium (III) oxide

Answer:

(a) Mercury (II) chloride

\(\begin{array}{l}HgCl_{ 2 }\end{array} \)

(b) Nickel (II) sulphate

\(\begin{array}{l}NiSO_{ 4 }\end{array} \)

\(\begin{array}{l}SnO_{ 2 }\end{array} \)

(d) Thallium (I) sulphate

\(\begin{array}{l}Tl_{ 2 }SO_{ 4 }\end{array} \)

(e) Iron (III) sulphate

\(\begin{array}{l}Fe_{ 2 }(SO_{ 4 })_{ 3 }\end{array} \)

(f) Chromium (III) oxide

\(\begin{array}{l}Cr_{ 2 }O_{ 3 }\end{array} \)

7. Suggest a list of the substances where carbon can exhibit oxidation states from –4 to +4 and nitrogen from –3 to +5.

Answer:

The compound where carbon has oxidation no. from -4 to +4 is as given below in the table:

Compounds	Oxidation no. of carbon
\(\begin{array}{l}CH_{ 2 }Cl_{ 2 }\end{array} \)	0
\(\begin{array}{l}HC\equiv CH\end{array} \)	-1
\(\begin{array}{l}ClC\equiv CCl\end{array} \)	+1
\(\begin{array}{l}CH_{ 3 }Cl\end{array} \)	-2
\(\begin{array}{l}CHCl_{ 3 }\end{array} \) , CO	+2
\(\begin{array}{l}H_{ 3 }C-CH_{ 3 }\end{array} \)	-3
\(\begin{array}{l}Cl_{ 3 }C-CCl_{ 3 }\end{array} \)	+3
\(\begin{array}{l}CH_{ 4 }\end{array} \)	-4
\(\begin{array}{l}CCl_{ 4 }\end{array} \) , \(\begin{array}{l}CO_{ 2 }\end{array} \)	+4

Compounds	Oxidation no. of nitrogen
\(\begin{array}{l}N_{ 2 }\end{array} \)	0
\(\begin{array}{l}N_{ 2 }H_{ 2 }\end{array} \)	-1
\(\begin{array}{l}N_{ 2 }O\end{array} \)	+1
\(\begin{array}{l}N_{ 2 }H_{ 4 }\end{array} \)	-2
\(\begin{array}{l}NO\end{array} \)	+2
\(\begin{array}{l}NH_{ 3 }\end{array} \)	-3
\(\begin{array}{l}N_{ 2 }O_{ 3 }\end{array} \)	+3
\(\begin{array}{l}NO_{ 2 }\end{array} \)	+4
\(\begin{array}{l}N_{ 2 }O_{ 5 }\end{array} \)	+5

8. While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?

Answer:

In sulphur dioxide (

\(\begin{array}{l}SO_{ 2 }\end{array} \)

), the oxidation no. of S is +4, and the range of oxidation no. of sulphur is from +6 to -2.

Hence,

\(\begin{array}{l}SO_{ 2 }\end{array} \)

can act as a reducing and oxidising agent.

In hydrogen peroxide (

\(\begin{array}{l}H_{ 2 }O_{ 2 }\end{array} \)

), the oxidation no. of O is -1, and the range of the oxidation no. of oxygen is from 0 to -2. Oxygen can sometimes attain oxidation no. +1 and +2.

Therefore,

\(\begin{array}{l}H_{ 2 }O_{ 2 }\end{array} \)

can act as a reducing and oxidising agent.

In ozone (

\(\begin{array}{l}O_{ 3 }\end{array} \)

), the oxidation no. of O is 0, and the range of the oxidation no. of oxygen is from 0 to –2. Hence, the oxidation no. of oxygen only decreases in this case.

Therefore,

\(\begin{array}{l}O_{ 3 }\end{array} \)

acts only as an oxidant.

In nitric acid (

\(\begin{array}{l}HNO_{ 3 }\end{array} \)

), the oxidation no. of nitrogen is +5, and the range of the oxidation no. that nitrogen can have is from +5 to -3. Hence, the oxidation no. of nitrogen can only decrease in this case.

Therefore,

\(\begin{array}{l}HNO_{ 3 }\end{array} \)

acts only as an oxidant.

9.Consider the reactions:

(a)

\(\begin{array}{l}6 \; CO_{ 2 \; (g) } \; + \; 6 \; H_{ 2 }O_{ (l) } \; \rightarrow \; C_{ 6 }H_{ 12 }O_{ 6 \; (aq) } \; + \; 6 \; O_{ 2 \; (g) }\end{array} \)

(b)

\(\begin{array}{l}O_{ 3 \; (g) } \; + H_{ 2 }O_{ 2 \; (l) } \; \rightarrow \; H_{ 2 }O_{ (l) } \; + \; 2 \; O_{ 2 \; (g) }\end{array} \)

Why it is more appropriate to write these reactions as :

(a)

\(\begin{array}{l}6 \; CO_{ 2 \; (g) } \; + \; 12 \; H_{ 2 }O_{ (l) } \; \rightarrow \; C_{ 6 }H_{ 12 }O_{ 6 \; (aq) } \; + \; 6 \; H_{ 2 }O_{ (l) } \; + \; 6 \; O_{ 2 \; (g) }\end{array} \)

(b)

\(\begin{array}{l}O_{ 3 \; (g) } \; + H_{ 2 }O_{ 2 \; (l) } \; \rightarrow \; H_{ 2 }O_{ (l) } \; + \; O_{ 2 \; (g) } \; + \; O_{ 2 \; (g) }\end{array} \)

Also suggest a technique to investigate the path of the above (a) and (b) redox reactions

Answer:

(a)

Step 1 :

\(\begin{array}{l}H_{ 2 }O\end{array} \)

breaks to give

\(\begin{array}{l}H_{ 2 }\end{array} \)

and

\(\begin{array}{l}O_{ 2 }\end{array} \)

\(\begin{array}{l}2 \; H_{ 2 }O_{ (l) } \; \rightarrow \; 2 \; H_{ 2 \; (g) } \; + \; O_{ 2 \; (g) }\end{array} \)

Step 2 :

The

\(\begin{array}{l}H_{ 2 }\end{array} \)

produced in earlier step reduces

\(\begin{array}{l}CO_{ 2 }\end{array} \)

, thus produce glucose and water.

\(\begin{array}{l}6 \; CO_{ 2 \; (g) } \; + \; 12 \; H_{ 2 \; (g) } \; \rightarrow \; C_{ 6 }H_{ 12 }O_{ 6 \; (s) } \; + \; 6 \; H_{ 2 }O_{ (l) }\end{array} \)

The net reaction is as given below:

[

\(\begin{array}{l}2 \; H_{ 2 }O_{ (l) } \; \rightarrow \; 2 \; H_{ 2 \; (g) } \; + \; O_{ 2 \; (g) }\end{array} \)

] × 6

\(\begin{array}{l}6 \; CO_{ 2 \; (g) } \; + \; 12 \; H_{ 2 \; (g) } \; \rightarrow \; C_{ 6 }H_{ 12 }O_{ 6 \; (s) } \; + \; 6 \; H_{ 2 }O_{ (l) }\end{array} \)

————————————————————————————————————————–

\(\begin{array}{l}6 \; CO_{ 2 \; (g) } \; + \; 12 \; H_{ 2 }O_{ (l) } \; \rightarrow \; C_{ 6 }H_{ 12 }O_{ 6 \; (g) } \; + \; 6 \; H_{ 2 }O_{ (l) } \; + \; 6 \; O_{ 2 \; (g) }\end{array} \)

This is the suitable way to write the reaction as the reaction also produces water molecules in the photosynthesis process.

The path can be found with the help of radioactive

\(\begin{array}{l}H_{ 2 }O^{ 18 }\end{array} \)

instead of

\(\begin{array}{l}H_{ 2 }O\end{array} \)

(b)

Step 1 :

\(\begin{array}{l}O_{ 2 }\end{array} \)

is produced from each of the reactants

\(\begin{array}{l}O_{ 3 }\end{array} \)

and

\(\begin{array}{l}H_{ 2 }O_{ 2 }\end{array} \)

. That is the reason

\(\begin{array}{l}O_{ 2 }\end{array} \)

is written two times.

\(\begin{array}{l}O_{ 3 }\end{array} \)

breaks to form

\(\begin{array}{l}O_{ 2 }\end{array} \)

and O.

Step 2 :

\(\begin{array}{l}H_{ 2 }O_{ 2 }\end{array} \)

reacts with O produced in the earlier step, thus producing

\(\begin{array}{l}H_{ 2 }O\end{array} \)

and

\(\begin{array}{l}O_{ 2 }\end{array} \)

\(\begin{array}{l}O_{ 3 \; (g) } \; \rightarrow \; O_{ 2 \; (g) } \; + \; O_{ (g) }\;\end{array} \)

\(\begin{array}{l}\;H_{ 2 }O_{ 2 \; (l) } \; + \; O_{ (g) } \; \rightarrow \; H_{ 2 }O_{ (l) } \; + \; O_{ 2 \; (g) }\end{array} \)

———————————————————————————————-

\(\begin{array}{l}H_{ 2 }O_{ 2 \; (l) } \; + \; O_{ 3 \; (g) } \; \rightarrow \; H_{ 2 }O_{ (l) } \; + \; O_{ 2 \; (g) } \; + \; O_{ 2 \; (g) }\end{array} \)

The path can be found with the help of

\(\begin{array}{l}H_{ 2 }O_{ 2 }^{ 18 }\end{array} \)

\(\begin{array}{l}O_{ 3 }^{ 18 }\end{array} \)

10. The compound AgF₂ is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why?

Answer:

The oxidation no. of Ag in

\(\begin{array}{l}AgF_{ 2 }\end{array} \)

is +2. But, +2 is very unstable oxidation no. of Ag. Hence, when

\(\begin{array}{l}AgF_{ 2 }\end{array} \)

is formed, silver accepts an electron and forms

\(\begin{array}{l}Ag^{ + }\end{array} \)

. This decreases the oxidation no. of Ag from +2 to +1. +1 state is more stable. Therefore,

\(\begin{array}{l}AgF_{ 2 }\end{array} \)

acts as a very strong oxidizing agent.

11. Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.

Justify the above statement with three examples.

Answer:

When there is a reaction between a reducing agent and an oxidizing agent, a compound is formed, which has a lower oxidation number if the reducing agent is in excess. A compound is formed which has a higher oxidation number if the oxidizing agent is in excess.

(i)

\(\begin{array}{l}P_{ 4 }\end{array} \)

and

\(\begin{array}{l}F_{ 2 }\end{array} \)

are reducing and oxidizing agent respectively.

In an excess amount of

\(\begin{array}{l}P_{ 4 }\end{array} \)

is reacted with

\(\begin{array}{l}F_{ 2 }\end{array} \)

, then

\(\begin{array}{l}PF_{ 3 }\end{array} \)

would be produced, where the oxidation no. of P is +3.

\(\begin{array}{l}P_{ 4 } \; _{(excess)}\; + \; F_{ 2 } \; \rightarrow \; PF_{ 3 }\end{array} \)

\(\begin{array}{l}P_{ 4 }\end{array} \)

is reacted with excess of

\(\begin{array}{l}F_{ 2 }\end{array} \)

, then

\(\begin{array}{l}PF_{ 5 }\end{array} \)

would be produced, where the oxidation no. of P is +5.

\(\begin{array}{l}P_{ 4 } \; + \; F_{ 2 } \; _{(excess)} \; \rightarrow \; PF_{ 5 }\end{array} \)

(ii) K and

\(\begin{array}{l}O_{ 2 }\end{array} \)

acts as a reducing agent and oxidizing agent respectively.

If an excess of K reacts with

\(\begin{array}{l}O_{ 2 }\end{array} \)

, it produces

\(\begin{array}{l}K_{ 2 }O\end{array} \)

. Here, the oxidation number of O is -2.

\(\begin{array}{l}4 \; K \; _{(excess)} \; + \; O_{ 2 } \; \rightarrow \; 2 \; K_{ 2 }O^{ -2 }\end{array} \)

If K reacts with an excess of

\(\begin{array}{l}O_{ 2 }\end{array} \)

, it produces

\(\begin{array}{l}K_{ 2 }O_{ 2 }\end{array} \)

, where the oxidation number of O is –1.

\(\begin{array}{l}2 \; K \; + \; O_{ 2 } \; _{(excess)} \; \rightarrow \; K_{ 2 }O_{ 2 }^{ -1 }\end{array} \)

(iii) C and

\(\begin{array}{l}O_{ 2 }\end{array} \)

acts as a reducing agent and oxidizing agent respectively.

If an excess amount of C is reacted with an insufficient amount of

\(\begin{array}{l}O_{ 2 }\end{array} \)

, then it produces CO, where the oxidation number of C is +2.

\(\begin{array}{l}C \; _{(excess)} \; + \; O_{ 2 } \; \rightarrow \; CO\end{array} \)

If C is burnt in excess amount of

\(\begin{array}{l}O_{ 2 }\end{array} \)

, then

\(\begin{array}{l}CO_{ 2 }\end{array} \)

is produced, where the oxidation number of C is +4.

\(\begin{array}{l}C \; + \; O_{ 2 } \; _{(excess)} \; \rightarrow \; CO_{ 2 }\end{array} \)

12. How do you count for the following observations?

(a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write a balanced redox equation for the reaction.

(b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why?

Answer:

(a) While manufacturing benzoic acid from toluene, alcoholic potassium permanganate is used as an oxidant due to the given reasons.

(i) In a neutral medium,

\(\begin{array}{l}OH^{ – }\end{array} \)

ions are produced in the reaction. Due to that, the cost of adding an acid or a base can be reduced.

(ii)

\(\begin{array}{l}KMnO_{ 4 }\end{array} \)

and alcohol are homogeneous to each other as they are polar. Alcohol and toluene are homogeneous to each other because both are organic compounds. Reactions can proceed at a faster rate in a homogeneous medium compared to a heterogeneous medium. Therefore, in alcohol,

\(\begin{array}{l}KMnO_{ 4 }\end{array} \)

and toluene can react at a faster rate.

The redox reaction is as given below:

(b) When concentrated

\(\begin{array}{l}H_{ 2 }SO_{ 4 }\end{array} \)

is added to an inorganic mixture containing bromide, firstly HBr is produced. HBr, a strong reducing agent, reduces

\(\begin{array}{l}H_{ 2 }SO_{ 4 }\end{array} \)

\(\begin{array}{l}SO_{ 2 }\end{array} \)

with the evolution of bromine’s red vapour.

\(\begin{array}{l}2 \; NaBr \; + \; 2 \; H_{ 2 }SO_{ 4 } \; \rightarrow \; 2 \; NaHSO_{ 4 } \; + \; 2 \; HBr\end{array} \)

\(\begin{array}{l}2 \; HBr \; + \; H_{ 2 }SO_{ 4 } \; \rightarrow \; Br_{ 2 } \; + \; SO_{ 2 } \; + \; 2 \; H_{ 2 }O\end{array} \)

When concentrated

\(\begin{array}{l}H_{ 2 }SO_{ 4 }\end{array} \)

I added to an inorganic mixture containing chloride, a pungent smelling gas (HCl) is evolved. HCl, a weak reducing agent, cannot reduce

\(\begin{array}{l}H_{ 2 }SO_{ 4 }\end{array} \)

\(\begin{array}{l}SO_{ 2 }\end{array} \)

\(\begin{array}{l}2 \; NaCl \; + \; 2 \; H_{ 2 }SO_{ 4 } \; \rightarrow \; 2 \; NaHSO_{ 4 } \; + \; 2 \; HCl\end{array} \)

13. Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions:

(a)

\(\begin{array}{l}2 \; AgBr_{ (s) } \; + \; C_{ 6 }H_{ 6 }O_{ 2 \; (aq) } \; \rightarrow \; 2 \; Ag_{ (s) } \; + \; 2 \; HBr_{ (aq)} \; + \; C_{ 6 }H_{ 4 }O_{ 2 \; (aq) }\end{array} \)

(b)

\(\begin{array}{l}HCHO_{ (l) } \; + \; 2 \; [Ag(NH_{ 3 })_{ 2 }]^{ + }_{ (aq) } \; + \; 3 \; OH^{ – }_{ (aq) } \; \rightarrow \; 2 \; Ag_{ (s) } \; + \; HCOO^{ – }_{ (aq) } \; + \; 4 \; NH_{ 3 \; (aq) } \; + \; 2 \; H_{ 2 }O_{ (l) }\end{array} \)

(c)

\(\begin{array}{l}HCHO_{ (l) } \; + \; 2 \; Cu^{ 2+ }_{ (aq) } \; + \; 5 \; OH^{ – }_{ (aq) } \; \rightarrow \; Cu_{ 2 }O_{ (s) } \; + \; HCOO^{ – }_{ (aq) } \; + \; 3 \; H_{ 2 }O_{ (l) }\end{array} \)

(d)

\(\begin{array}{l}N_{ 2 }H_{ 4 \; (l) } \; + \; 2 \; H_{ 2 }O_{ 2 \; (l) } \; \rightarrow \; N_{ 2 \; (g) } \; + \; 4 \; H_{ 2 }O_{ (l) }\end{array} \)

(e)

\(\begin{array}{l}Pb_{ (s) } \; + \; PbO_{ 2 \; (s) } \; + \; 2 \; H_{ 2 }SO_{ 4 \; (aq) } \; \rightarrow \; 2 \; PbSO_{ 4 \; (aq) } \; + \; 2 \; H_{ 2 }O_{ (l) }\end{array} \)

Answer:

(a)

\(\begin{array}{l}2 \; AgBr_{ (s) } \; + \; C_{ 6 }H_{ 6 }O_{ 2 \; (aq) } \; \rightarrow \; 2 \; Ag_{ (s) } \; + \; 2 \; HBr_{ (aq)} \; + \; C_{ 6 }H_{ 4 }O_{ 2 \; (aq) }\end{array} \)

\(\begin{array}{l}C_{ 6 }H_{ 6 }O_{ 2 }\end{array} \)

=> Oxidized substance

AgBr => Reduced substance

AgBr =>Oxidizing agent

\(\begin{array}{l}C_{ 6 }H_{ 6 }O_{ 2 }\end{array} \)

=> Reducing agent

(b)

HCHO => Oxidized substance

\(\begin{array}{l}[Ag(NH_{ 3 })_{ 2 }]^{ + }\end{array} \)

=> Reduced substance

\(\begin{array}{l}[Ag(NH_{ 3 })_{ 2 }]^{ + }\end{array} \)

=> Oxidizing agent

HCHO=> Reducing agent

(c)

HCHO => Oxidized substance

\(\begin{array}{l}Cu^{ 2+ }\end{array} \)

=> Reduced substance

\(\begin{array}{l}Cu^{ 2+ }\end{array} \)

=> Oxidizing agent

HCHO => Reducing agent

(d)

\(\begin{array}{l}N_{ 2 }H_{ 4 \; (l) } \; + \; 2 \; H_{ 2 }O_{ 2 \; (l) } \; \rightarrow \; N_{ 2 \; (g) } \; + \; 4 \; H_{ 2 }O_{ (l) }\end{array} \)

\(\begin{array}{l}N_{ 2 }H_{ 4 }\end{array} \)

=> Oxidized substance

\(\begin{array}{l}H_{ 2 }O_{ 2 }\end{array} \)

=> Reduced substance

\(\begin{array}{l}H_{ 2 }O_{ 2 }\end{array} \)

=> Oxidizing agent

\(\begin{array}{l}N_{ 2 }H_{ 4 }\end{array} \)

=> Reducing agent

(e)

\(\begin{array}{l}Pb_{ (s) } \; + \; PbO_{ 2 \; (s) } \; + \; 2 \; H_{ 2 }SO_{ 4 \; (aq) } \; \rightarrow \; 2 \; PbSO_{ 4 \; (aq) } \; + \; 2 \; H_{ 2 }O_{ (l) }\end{array} \)

Pb=> Oxidized substance

\(\begin{array}{l}PbO_{ 2 }\end{array} \)

=> Reduced substance

\(\begin{array}{l}PbO_{ 2 }\end{array} \)

=> Oxidizing agent

Pb => Reducing agent

14. Consider the reactions :

\(\begin{array}{l}2 \; S_{ 2 }O^{ 2- }_{ 3 \; (aq) } \; + \; I_{ 2 \; (s) } \; \rightarrow \; S_{ 4 }O^{ 2- }_{ 6 \; (aq) } \; + \; 2 \; I^{ – }_{ (aq) }\end{array} \)

\(\begin{array}{l}S_{ 2 }O^{ 2- }_{ 3 \; (aq) } \; + \; 2 \; Br_{ 2 \; (l) } \; + \; 5 \; H_{ 2 }O_{ (l) } \; \rightarrow \; 2 \; SO^{ 2- }_{ 4 \; (aq) } \; + \; 4 \; Br^{ – }_{ (aq) } \; + \; 10 \; H^{ + }_{ (aq) }\end{array} \)

Why does the same reductant, thiosulphate react differently with iodine and bromine ?

Answer:

The average oxidation no. of S in

\(\begin{array}{l}S_{ 2 }O_{ 3 }^{ 2- }\end{array} \)

is +2.

The average oxidation no. of S in

\(\begin{array}{l}S_{ 4 }O_{ 6 }^{ 2- }\end{array} \)

is +2.5.

The oxidation no. of S in

\(\begin{array}{l}S_{ 2 }O_{ 3 }^{ 2- }\end{array} \)

is +2.

The oxidation no. of S in

\(\begin{array}{l}SO_{ 4 }^{ 2- }\end{array} \)

is +6.

\(\begin{array}{l}Br_{ 2 }\end{array} \)

is a stronger oxidizing agent than

\(\begin{array}{l}I_{ 2 }\end{array} \)

, it oxidizes S of

\(\begin{array}{l}S_{ 2 }O_{ 3 }^{ 2- }\end{array} \)

to a higher oxidation no. of +6 in

\(\begin{array}{l}SO_{ 4 }^{ 2- }\end{array} \)

\(\begin{array}{l}I_{ 2 }\end{array} \)

is a weaker oxidizing agent, it oxidizes S of

\(\begin{array}{l}S_{ 2 }O_{ 3 }^{ 2- }\end{array} \)

ion to a lower oxidation no. that is 2.5 in

\(\begin{array}{l}S_{ 4 }O_{ 6 }^{ 2- }\end{array} \)

ions.

Thus, thiosulphate reacts differently with

\(\begin{array}{l}I_{ 2 }\end{array} \)

and

\(\begin{array}{l}Br_{ 2 }\end{array} \)

15. Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.

Answer:

\(\begin{array}{l}F_{ 2 }\end{array} \)

can oxidize

\(\begin{array}{l}Cl^{ – }\end{array} \)

\(\begin{array}{l}Cl_{ 2 }\end{array} \)

\(\begin{array}{l}Br^{ – }\end{array} \)

\(\begin{array}{l}Br_{ 2 }\end{array} \)

, and

\(\begin{array}{l}I^{ – }\end{array} \)

\(\begin{array}{l}I_{ 2 }\end{array} \)

as:

\(\begin{array}{l}F_{ 2 \; (aq) } \; + \; 2 \; Cl^{ – }_{ (s) } \; \rightarrow \; 2 \; F^{ – }_{ (aq) } \; + \; Cl_{ 2 \; (g) }\end{array} \)

\(\begin{array}{l}F_{ 2 \; (aq) } \; + \; 2 \; Br^{ – }_{ (aq) } \; \rightarrow \; 2 \; F^{ – }_{ (aq) } \; + \; Br_{ 2 \; (l) }\end{array} \)

\(\begin{array}{l}F_{ 2 \; (aq) } \; + \; 2 \; I^{ – }_{ (aq) } \; \rightarrow \; 2 \; F^{ – }_{ (aq) } \; + \; I_{ 2 \; (s) }\end{array} \)

But,

\(\begin{array}{l}Cl_{ 2 }\end{array} \)

\(\begin{array}{l}Br_{ 2 }\end{array} \)

, and

\(\begin{array}{l}I_{ 2 }\end{array} \)

cannot oxidize

\(\begin{array}{l}F^{ – }\end{array} \)

\(\begin{array}{l}F_{ 2 }\end{array} \)

. The oxidizing power of halogens increases in the order as given below:

\(\begin{array}{l}I_{ 2 }\end{array} \)

\(\begin{array}{l}Br_{ 2 }\end{array} \)

\(\begin{array}{l}Cl_{ 2 }\end{array} \)

\(\begin{array}{l}F_{ 2 }\end{array} \)

Therefore, fluorine is the best oxidant among halogens.

\(\begin{array}{l}HI\end{array} \)

and

\(\begin{array}{l}HBr\end{array} \)

can reduce

\(\begin{array}{l}H_{ 2 }SO_{ 4 }\end{array} \)

\(\begin{array}{l}SO_{ 2 }\end{array} \)

, but

\(\begin{array}{l}HCl\end{array} \)

and

\(\begin{array}{l}HF\end{array} \)

cannot. Hence,

\(\begin{array}{l}HI\end{array} \)

and

\(\begin{array}{l}HBr\end{array} \)

are stronger reductants compared to

\(\begin{array}{l}HCl\end{array} \)

and

\(\begin{array}{l}HF\end{array} \)

\(\begin{array}{l}2 \; HI \; + \; H_{ 2 }SO_{ 4 } \; \rightarrow \; I_{ 2 } \; + \; SO_{ 2 } \; + \; 2 \; H_{ 2 }O\end{array} \)

\(\begin{array}{l}2 \; HBr \; + \; H_{ 2 }SO_{ 4 } \; \rightarrow \; Br_{ 2 } \; + \; SO_{ 2 } \; + \; 2 \; H_{ 2 }O\end{array} \)

\(\begin{array}{l}I^{ – }\end{array} \)

can reduce

\(\begin{array}{l}Cu^{ 2+ }\end{array} \)

\(\begin{array}{l}Cu^{ + }\end{array} \)

, but

\(\begin{array}{l}Br^{ – }\end{array} \)

cannot.

\(\begin{array}{l}4 \; I^{ – }_{ (aq) } \; + \; 2 \; Cu^{ 2+ }_{ (aq) } \; \rightarrow \; Cu_{ 2 }I_{ 2 \; (s) } \; + \; I_{ 2 \; (aq) }\end{array} \)

Therefore, hydrochloric acid is the best reductant among hydrohalic compounds.

Hence, the reducing power of hydrohalic acids increases as given below:

\(\begin{array}{l}HF\end{array} \)

\(\begin{array}{l}HCl\end{array} \)

\(\begin{array}{l}HBr\end{array} \)

\(\begin{array}{l}HI\end{array} \)

16. Why does the following reaction occur?

\(\begin{array}{l}XeO^{ 4- }_{ 6 \; (aq) } \; + \; 2 \; F^{ – }_{ (aq) } \; + \; 6 \; H^{ + }_{ (aq) } \; \rightarrow \; XeO_{ 3 \; (g) } \; + \; F_{ 2 \; (g) } \; + \; 3 \; H_{ 2 }O_{ (l) }\end{array} \)

What conclusion about the compound

\(\begin{array}{l}Na_{ 4 }XeO_{ 6 }\end{array} \)

( of which

\(\begin{array}{l}XeO_{ 6 }^{ 4- }\end{array} \)

is a part) can be drawn from the reaction?

Answer:

The oxidation no. of Xe reduces from +8 in

\(\begin{array}{l}XeO_{ 6 }^{ 4- }\end{array} \)

to +6 in

\(\begin{array}{l}XeO_{ 3 }\end{array} \)

The oxidation no. of F increases from -1 in

\(\begin{array}{l}F^{ – }\end{array} \)

to 0 in

\(\begin{array}{l}F_{ 2 }\end{array} \)

Hence,

\(\begin{array}{l}XeO_{ 6 }^{ 4- }\end{array} \)

is reduced on the other hand

\(\begin{array}{l}F^{ – }\end{array} \)

is oxidized. As

\(\begin{array}{l}Na_{ 2 }XeO_{ 6 }^{ 4- }\end{array} \)

(or

\(\begin{array}{l}XeO_{ 6 }^{ 4- }\end{array} \)

) is a stronger oxidizing agent compared to

\(\begin{array}{l}F_{ 2 }\end{array} \)

, this reaction occurs.

17. Consider the reactions:

(a)

\(\begin{array}{l}H_{ 3 }PO_{ 2 \; (aq) } \; + \; 4 \; AgNO_{ 3 \; (aq) } \; + \; 2 \; H_{ 2 }O_{ (l) } \; \rightarrow \; H_{ 3 }PO_{ 4 \; (aq) } \; + \; 4 \; Ag_{ (s) } \; + \; 4 \; HNO_{ 3 \; (aq) }\end{array} \)

(b)

\(\begin{array}{l}H_{ 3 }PO_{ 2 \; (aq) } \; + \; 2 \; CuSO_{ 4 \; (aq) } \; + \; 2 \; H_{ 2 }O_{ (l) } \; \rightarrow \; H_{ 3 }PO_{ 4 \; (aq) } \; + \; 2 \; Cu_{ (s) } \; + \; H_{ 2 }SO_{ 4 \; (aq) }\end{array} \)

(c)

\(\begin{array}{l}C_{ 6 }H_{ 5 }CHO_{ (l) } \; + \; 2 \; [Ag(NH_{ 3 })_{ 2 }]^{ + }_{ (aq) } \; + \; 3 \; OH^{ – }_{ (aq) } \; \rightarrow \; C_{ 6 }H_{ 5 }COO^{ – }_{(aq)} \; + \; 2\; Ag_{ (s) } \; + \; 4 \; NH_{ 3 \; (aq)} \; + \; 2 \; H_{ 2 }O_{ (l) }\end{array} \)

(d)

\(\begin{array}{l}C_{ 6 }H_{ 5 }CHO_{ (l) } \; + \; 2 \; Cu^{ 2+ }_{ (aq) } \; + \; 5 \; OH^{ – }_{ (aq) } \; \rightarrow\end{array} \)

No change is observed

What inference do you draw about the behavior of

\(\begin{array}{l}Ag^{ + }\end{array} \)

and

\(\begin{array}{l}Cu^{ 2+ }\end{array} \)

from these reactions?

Answer:

\(\begin{array}{l}Ag^{ + }\end{array} \)

and

\(\begin{array}{l}Cu^{ 2+ }\end{array} \)

behave as oxidizing agents in reactions (i) and (ii), respectively.

In reaction (iii),

\(\begin{array}{l}Ag^{ + }\end{array} \)

oxidizes

\(\begin{array}{l}C_{ 6 }H_{ 5 }CHO\end{array} \)

\(\begin{array}{l}C_{ 6 }H_{ 5 }COO^{ – }\end{array} \)

In reaction (iv),

\(\begin{array}{l}Cu^{ 2+ }\end{array} \)

cannot oxidize

\(\begin{array}{l}C_{ 6 }H_{ 5 }CHO\end{array} \)

Therefore,

\(\begin{array}{l}Ag^{ + }\end{array} \)

is a stronger oxidizing agent compared to

\(\begin{array}{l}Cu^{ 2+ }\end{array} \)

18. Balance the following redox reactions by ion – electron method :

(a)

\(\begin{array}{l}MnO^{ – }_{ 4 \; (aq) } \; + \; I_{ (aq) }^{ – } \; \rightarrow \; MnO_{ 2 \; (s) } \; + \; I_{ 2 \; (s) }\end{array} \)

(Basic medium)

(b)

\(\begin{array}{l}MnO^{ – }_{ 4 \; (aq) } \; + \; SO_{ 2 \; (g) } \; \rightarrow \; Mn^{ 2+ }_{ (aq) } \; + \; H_{ 2 }SO_{ 4 }^{ – }\end{array} \)

(Acidic medium)

(c)

\(\begin{array}{l}H_{ 2 }O_{ 2 \; (aq) } \; + \; Fe^{ 2+ }_{ (aq) } \; \rightarrow \; Fe_{ (aq) }^{ 3+ } \; + \; H_{ 2 }O_{ (l) }\end{array} \)

(Acidic medium)

(d)

\(\begin{array}{l}Cr_{ 2 }^{ 2- }O_{ 7 \; (aq) } \; + \; SO_{ 2 \; (g) } \; \rightarrow \; Cr^{ 3+ }_{ (aq) } \; + \; SO^{ 2- }_{ (aq) }\end{array} \)

(Acidic medium)

Answer:

(a)

\(\begin{array}{l}MnO^{ – }_{ 4 \; (aq) } \; + \; I_{ (aq) }^{ – } \; \rightarrow \; MnO_{ 2 \; (s) } \; + \; I_{ 2 \; (s) }\end{array} \)

Step 1

The two half-reactions are given below:

Oxidation half-reaction:

\(\begin{array}{l}I_{ (aq) } \; \rightarrow \; I_{ 2 \; (s) }\end{array} \)

Reduction half-reaction:

\(\begin{array}{l}MnO_{ 4 }^{ – } \; \rightarrow \; MnO_{ 2 }\end{array} \)

Step 2

Balance I in oxidation half-reaction:

\(\begin{array}{l}2 \; I^{ – }_{(aq)} \; \rightarrow \; I_{ 2 \; (s) }\end{array} \)

Add 2

\(\begin{array}{l}e^{ – }\end{array} \)

to the right-hand side of the reaction to balance the charge:

\(\begin{array}{l}2I^{ – }_{ (aq) } \; \rightarrow \; I_{ 2 \; (s) } \; + \; 2 \; e^{ – }\end{array} \)

Step 3

The oxidation no. of Mn has decreased from +7 to +4 in the reduction half-reaction. Therefore, 3 electrons are added to the left-hand side of the reaction.

\(\begin{array}{l}MnO^{ – }_{ 4 \; (aq) } \; + \; 3 \; e^{ – } \; \rightarrow \; MnO_{ 2 \; (aq)}\end{array} \)

Add 4

\(\begin{array}{l}OH^{ – }\end{array} \)

ions to the right-hand side of the reaction to balance the charge.

\(\begin{array}{l}MnO^{ – }_{ 4 \; (aq) } \; + \; 3 \; e^{ – } \; \rightarrow \; MnO_{ 2 \;(aq)} \; + \; 4 \; OH^{ – }\end{array} \)

Step 4

There are 6 oxygen atoms on the right-hand side and 4 oxygen atoms on the left-hand side. Hence, 2 water molecules are added to the left-hand side.

\(\begin{array}{l}MnO^{ – }_{ 4 \; (aq) } \; + \; 2 \; H_{ 2 }O \; + \; 3\; e^{ – }\; \rightarrow \; MnO_{ 2 \; (aq) } \; + \; 4 OH^{ – }\end{array} \)

Step 5

Equal the no. of electrons on both sides by multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 2:

\(\begin{array}{l}6 \; I_{ (aq) }^{ – } \; \rightarrow \; 3 \; I_{ 2 \; (s)} \; + \; 6 \; e^{ – }\end{array} \)

\(\begin{array}{l}2 \; MnO^{ – }_{ 4 \; (aq) } \; + \; 4 \; H_{ 2 }O \; + \; 6\; e^{ – } \; \rightarrow 2 \; MnO_{ 2 \; (s)} \; + \; 8\; OH^{ – }_{ (aq) }\end{array} \)

Step 6

After adding both the half-reactions, we get the balanced reaction as given below:

\(\begin{array}{l}6 \; I_{ (aq) }^{ – } \; + \; 2 \; MnO^{ – }_{ 4 \; (aq) } \;+ \; 4 \; H_{ 2 }O_{ (l) } \; \rightarrow \; 3 \; I_{ 2 \; (s)} \; + \; 2 \; MnO_{ 2 \; (s)} \; + \; 8\; OH^{ – }_{ (aq) }\end{array} \)

(b)

\(\begin{array}{l}MnO^{ – }_{ 4 \; (aq) } \; + \; SO_{ 2 \; (g) } \; \rightarrow \; Mn^{ 2+ }_{ (aq) } \; + \; H_{ 2 }SO_{ 4 }^{ – }\end{array} \)

Step 1

Similar to (i), the oxidation half-reaction is:

\(\begin{array}{l}SO_{ 2 \; (g) } \; + \; 2\; H_{ 2 }O_{ (l) } \; \rightarrow \; HSO^{ – }_{ 4 \; (aq) } \; + \; 3 \; H^{ + }_{ (aq) } \; + \; 2 \; e^{ – }_{ (aq) }\end{array} \)

Step 2

Reduction half-reaction is:

\(\begin{array}{l}MnO^{ – }_{ 4 \; (aq) } \; + \; 8\; H^{ + }_{ (aq) } \; + \; 5 \; e^{ – } \; \rightarrow \; Mn^{ 2+ }_{ (aq) } \; + \; 4 \; H_{ 2 }O_{ (l) }\end{array} \)

Step 3

Multiply the oxidation half-reaction with 5 and the reduction half-reaction with 2, then add them. We get the balanced reaction as given below:

\(\begin{array}{l}2 \; MnO^{ – }_{ 4 \; (aq) } \; + \; 5 \; SO_{ 2 \; (g) } \; + \; 2\; H_{ 2 }O_{ (l) } \; +\; H^{ + }_{ (aq) } \; \rightarrow \; 2 \; Mn^{ 2+ }_{ (aq) } \; + \; 5 \; HSO^{ – }_{ 4 \; (aq) }\end{array} \)

(c)

\(\begin{array}{l}H_{ 2 }O_{ 2 \; (aq) } \; + \; Fe^{ 2+ }_{ (aq) } \; \rightarrow \; Fe_{ (aq) }^{ 3+ } \; + \; H_{ 2 }O_{ (l) }\end{array} \)

Step 1

Similar to (i), oxidation half-reaction is:

\(\begin{array}{l}Fe^{ 2+ }_{ (aq) } \; \rightarrow \; Fe_{ (aq) }^{ 3+ } \; + \; e^{ – }\end{array} \)

Step 2

Reduction half-reaction is:

\(\begin{array}{l}H_{ 2 }O_{ 2 \; (aq) } \; + \; 2\; H_{ (aq) }^{ + } \; + \; 2\; e^{ – } \; \rightarrow \; 2\; H_{ 2 }O_{ (l) }\end{array} \)

Step 3

Multiply the oxidation half-reaction with 2 then add it to the reduction half-reaction. We get the balanced reaction as given below:

\(\begin{array}{l}H_{ 2 }O_{ 2 \; (aq) } \; + \; 2 \;Fe^{ 2+ }_{ (aq) } \; 2\; H_{ (aq) }^{ + } \; \rightarrow \; 2 \;Fe_{ (aq) }^{ 3+ }2\; H_{ 2 }O_{ (l) }\end{array} \)

(d)

\(\begin{array}{l}Cr_{ 2 }^{ 2- }O_{ 7 \; (aq) } \; + \; SO_{ 2 \; (g) } \; \rightarrow \; Cr^{ 3+ }_{ (aq) } \; + \; SO^{ 2- }_{ (aq) }\end{array} \)

Step 1

Similar to (i), oxidation half-reaction is:

\(\begin{array}{l}SO_{ 2 \; (g) } \; + \; 2 \; H_{ 2 }O_{ (l) } \; \rightarrow \; SO^{ 2- }_{ 4 \; (aq) } \; + \; 4 \; H_{ (aq) }^{ + } \; + \; 2 \; e^{ – }\end{array} \)

Step 2

Reduction half-reaction is:

\(\begin{array}{l}Cr_{ 2 }O^{ 2- }_{ 7 \; (aq) } \; + \; 14 \; H^{ + }_{ (aq) } \; + \; 6 \; e^{ – } \; \rightarrow \; 2 \; Cr_{ (aq) }^{ 3+ } \; + \; 7 \; H_{ 2 }O_{(l)}\end{array} \)

Step 3

Multiply the oxidation half-reaction with 2, then add it to the reduction half-reaction. We get the balanced reaction as given below:

\(\begin{array}{l}Cr_{ 2 }^{ 2- }O_{ 7 \; (aq) } \; + \;3 \; SO_{ 2 \; (g) } \; + \; 2 \; H^{ + }_{ (aq) } \; \rightarrow \; 2 \; Cr_{ (aq) }^{ 3+ } \; + \;3 \; SO^{ 2- }_{ 4 \; (aq) } \;+ \; H_{ 2 }O_{(l)}\end{array} \)

19. Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.

(a)

\(\begin{array}{l}P_{ 4 \; (s) } \; + \; OH^{ – }_{ (aq) } \; \rightarrow \; PH_{ 3 \; (g) } \; + \; HPO^{ – }_{ 2 \; (aq) }\end{array} \)

(b)

\(\begin{array}{l}N_{ 2 }H_{ 4 \; (l) } \; + \; ClO^{ – }_{ 3 \; (aq) } \; \rightarrow \; NO_{ (g) } \; + \; Cl_{ (g) }^{ – }\end{array} \)

(c)

\(\begin{array}{l}Cl_{ 2 }O_{ 7 \; (g) } \; + \; H_{ 2 }O_{ 2 \; (aq) } \; \rightarrow \; ClO^{ – }_{ 2 \; (aq) } \; + \;O_{ 2 \; (g) } \; + \; H^{ + }_{ (aq) }\end{array} \)

Answer:

(a) The Oxidation no. of P reduces from 0 in

\(\begin{array}{l}P_{ 4 }\end{array} \)

to – 3 in

\(\begin{array}{l}PH_{ 3 }\end{array} \)

The oxidation no. of P increases from 0 in

\(\begin{array}{l}P_{ 4 }\end{array} \)

to + 2 in

\(\begin{array}{l}HPO_{ 2 }^{ – }\end{array} \)

. Therefore,

\(\begin{array}{l}P_{ 4 }\end{array} \)

behaves both as a reducing agent as well as an oxidizing agent in the reaction.

Ion–electron method:

– The oxidation half-reaction:

\(\begin{array}{l}P_{ 4 \; (s) } \; \rightarrow \; HPO^{ – }_{ 2 \; (aq) }\end{array} \)

– Balance atom P:

\(\begin{array}{l}P_{ 4 \; (s) } \; \rightarrow \;4 \; HPO^{ – }_{ 2 \; (aq) }\end{array} \)

– Add 8 electrons to balance oxidation no.

\(\begin{array}{l}P_{4\;(s)}\;\rightarrow \;4 \; HPO^{ – }_{ 2 \; (aq) } \; + \; 8 \; e^{ – }\end{array} \)

– Add

\(\begin{array}{l}12 \; OH^{ – }\end{array} \)

to balance the charge:

\(\begin{array}{l}P_{4\;(s)}\;+\;12\;OH^{-}_{(aq)}\;\rightarrow\;4\;HPO^{-}_{2\;(aq)}\;+\;8\;e^{-}\end{array} \)

– Add 4

\(\begin{array}{l}H_{ 2 }O\end{array} \)

to balance H and O atoms:

\(\begin{array}{l}P_{ 4 \; (s) } \; + \; 12 \; OH^{-}_{ (aq) } \; \rightarrow \;4 \; HPO^{ – }_{ 2 \; (aq) } \; + \; 4 \; H_{ 2 }O_{ (l) } \; + \; 8 \; e^{ – }\end{array} \)

———-(1)

– The reduction half-reaction:

\(\begin{array}{l}P_{ 4 \; (s) } \; \rightarrow \; PH_{ 3 \; (g) }\end{array} \)

– Balance atom P:

\(\begin{array}{l}P^{0}_{ 4 \; (s) } \; \rightarrow \;4 \; P^{-3} H_{ 3 \; (g) }\end{array} \)

– Add 12 electrons to balance oxidation no.

\(\begin{array}{l}P_{4\;(s)}\;+\;12\;e^{-}\;\rightarrow\;4\;PH_{3\;(g)}\end{array} \)

– Add

\(\begin{array}{l}12 \; OH^{ – }\end{array} \)

to balance the charge:

\(\begin{array}{l}P_{4\;(s)}\;+\;12\;e^{-}\;\rightarrow\;4\;PH_{3\;(g)}\;+\;12\;OH^{-}_{(aq)}\end{array} \)

– Add 12

\(\begin{array}{l}H_{ 2 }O\end{array} \)

to balance H and O atoms:

\(\begin{array}{l}P_{4\;(s)}\;+\;12\;H_{2}O_{(l)}\;+\;12\;e^{-}\;\rightarrow\;4\;PH_{3\;(g)} \;+\;12\;OH^{-}_{(aq)}\end{array} \)

———- (2)

– Now, multiply the equation (1) by 3 and equation (2) by 2. Then, after adding them, we get the balanced redox reaction as given below:

\(\begin{array}{l}5\;P_{4\;(s)}\;+\; 12 \; H_{ 2 }O_{ (l) } \; + \; 12 \; HO^{ – }_{ (aq) } \; \rightarrow \;8 \; PH_{ 3 \; (g) } \; + \; 12 \; HPO^{ – }_{ 2 \; (aq ) }\end{array} \)

(b)

The Oxidation no. of N increases from -2 in

\(\begin{array}{l}N_{ 2 }H_{ 4 }\end{array} \)

to –+2 in NO.

The oxidation no. of Cl reduces from +5 in

\(\begin{array}{l}ClO_{ 3 }^{ – }\end{array} \)

to +-1 in

\(\begin{array}{l}Cl^{ – }\end{array} \)

Therefore,

\(\begin{array}{l}N_{ 2 }H_{ 4 }\end{array} \)

behaves as a reducing agent while

\(\begin{array}{l}ClO_{ 3 }^{ – }\end{array} \)

behaves as an oxidizing agent in the reaction.

Ion–electron method:

– The oxidation half-reaction:

\(\begin{array}{l}N_{2}H_{4\; (l) } \; \rightarrow \; NO_{ (g) }\end{array} \)

– Balance atom N:

\(\begin{array}{l}N_{2}H_{4\;(l)}\; \rightarrow\;2\;NO_{(g)}\end{array} \)

– Add 8 electrons to balance oxidation no:

\(\begin{array}{l}N_{2}H_{4 \; (l) } \; \rightarrow \; 2 \; NO_{ (g) } \; + \; 8 \; e^{-}\end{array} \)

– Add

\(\begin{array}{l}8 \; OH^{ – }\end{array} \)

to balance the charge:

\(\begin{array}{l}N_{2}H_{ 4 \; (l) } \; + \; 8 \; OH^{ – }_{ (aq) } \; \rightarrow \; 2\;NO_{(g)} \; + \; 8 \; e^{-}\end{array} \)

– Add 6

\(\begin{array}{l}H_{ 2 }O\end{array} \)

to balance O atoms:

\(\begin{array}{l}N_{2}H_{4\;(l)} \; + \; 8\; OH^{-}_{(aq)} \; \rightarrow \; 2\;NO_{(g)} \; + \; 6 \; H_{2}O_{(l)} \; + \; 8 \; e^{-}\end{array} \)

——— (1)

– The reduction half-reaction:

\(\begin{array}{l}ClO_{ 3 \; (aq) }^{ – } \; \rightarrow \; Cl_{ (aq) }^{ – }\end{array} \)

– Add 6 electrons to balance oxidation no.

\(\begin{array}{l}ClO_{3\;(aq)}^{-} \; + \; 6 \; e^{-} \; \rightarrow \; Cl_{(aq)}^{-}\end{array} \)

– Add

\(\begin{array}{l}6 \; OH^{ – }\end{array} \)

ions to balance the charge:

\(\begin{array}{l}ClO_{3\;(aq)}^{-} \; + \; 6 \; e^{-} \; \rightarrow \; Cl_{(aq)}^{-} \; + \; 6 \; OH^{-}_{(aq)}\end{array} \)

– Add 3

\(\begin{array}{l}H_{ 2 }O\end{array} \)

to balance O atoms:

\(\begin{array}{l}ClO_{3\;(aq)}^{-} \; + \; 3\; H_{2}O_{(l)} \; + \; 6 \; e^{-} \; \rightarrow \; Cl_{(aq)}^{-} \; + \; 6 \; OH^{-}_{(aq)}\end{array} \)

——— (2)

Now, multiply equation (1) by 3 and equation (2) by 4. Then, after adding them, we get the balanced redox reaction as given below:

\(\begin{array}{l}3 \; N_{ 2 }H_{ 4 \; (l) } \; + \; 4 \; ClO_{ 3 \; (aq) }^{ – } \; \rightarrow \;6 \; NO_{ (g) }\; + \; 4 \; Cl_{ (aq) }^{ – } \; + \; 6 \; H_{ 2 }O_{ (l) }\end{array} \)

Oxidation number method:

– Reduction in the oxidation no. of N = 2 × 4 = 8

– Increment in the oxidation no. of Cl = 1 × 6 = 6

Multiply

\(\begin{array}{l}N_{ 2 }H_{ 4 }\end{array} \)

by 3 and

\(\begin{array}{l}ClO_{ 3 }^{ – }\end{array} \)

by 4 to balance the reduction and increment of the oxidation no. :

\(\begin{array}{l}3 \; N_{ 2 }H_{ 4 \; (l) } \; + \; 4 \; ClO^{ – }_{ 3 \; (aq) } \; \rightarrow \; NO_{ (g) } \; + \; Cl^{ – }_{ (aq) }\end{array} \)

– Balance Cl and n atoms:

\(\begin{array}{l}3 \; N_{ 2 }H_{ 4 \; (l) } \; + \; 4 \; ClO^{ – }_{ 3 \; (aq) } \; \rightarrow \; 6 \; NO_{ (g) } \; + \; 4 \; Cl^{ – }_{ (aq) }\end{array} \)

– Add 6

\(\begin{array}{l}H_{ 2 }O\end{array} \)

to balance O atoms:

\(\begin{array}{l}3 \; N_{ 2 }H_{ 4 \; (l) } \; + \; 4 \; ClO^{ – }_{ 3 \; (aq) } \; \rightarrow \; 6 \; NO_{ (g) } \; + \; 4 \; Cl^{ – }_{ (aq) } \; + \; 6 \; H_{ 2 }O_{ (l) }\end{array} \)

This is the required reaction equation.

(c)

The Oxidation no. of Cldecreases from +7 in

\(\begin{array}{l}Cl_{ 2 }O_{ 7 }\end{array} \)

to +3 in

\(\begin{array}{l}ClO_{ 2 }^{ – }\end{array} \)

The oxidation no. of increases from -1 in

\(\begin{array}{l}H_{ 2 }O_{ 2 }\end{array} \)

to 0 in

\(\begin{array}{l}O_{ 2 }\end{array} \)

Therefore,

\(\begin{array}{l}H_{ 2 }O_{ 2 }\end{array} \)

behaves as a reducing agent while

\(\begin{array}{l}Cl_{ 2 }O_{ 7 }\end{array} \)

behaves as an oxidizing agent in the reaction.

Ion–electron method:

– The oxidation half-reaction:

\(\begin{array}{l}H_{ 2 }O_{ 2 \; (aq) } \; \rightarrow \; O_{ 2 \; (g) }\end{array} \)

– Add 2 electrons to balance oxidation no:

\(\begin{array}{l}H_{ 2 }O_{ 2 \; (aq) } \; \rightarrow \; O_{ 2 \; (g) } \; + \; 2 \; e^{ – }\end{array} \)

– Add

\(\begin{array}{l}2 \; OH^{ – }\end{array} \)

to balance the charge:

\(\begin{array}{l}H_{ 2 }O_{ 2 \; (aq) } \; + \; 2 \; OH^{ – }_{ (aq) } \rightarrow \; O_{ 2 \; (g) } \; + \; 2 \; e^{ – }\end{array} \)

– Add 2

\(\begin{array}{l}H_{ 2 }O\end{array} \)

to balance O atoms:

\(\begin{array}{l}H_{ 2 }O_{ 2 \; (aq) } \; + \; 2 \; OH^{ – }_{ (aq) } \rightarrow \; O_{ 2 \; (g) } \; + \; 2 \; H_{ 2 }O_{ (l) } \; + \; 2 \; e^{ – }\end{array} \)

——– (1)

– The reduction half-reaction:

\(\begin{array}{l}Cl_{ 2 }O_{ 7 \; (g) } \; \rightarrow \; ClO^{ – }_{ 2 \; (aq) }\end{array} \)

– Balance Cl atoms:

\(\begin{array}{l}Cl_{ 2 }O_{ 7 \; (g) } \; \rightarrow \; 2 \; ClO^{ – }_{ 2 \; (aq) }\end{array} \)

– Add 8 electrons to balance oxidation no.

\(\begin{array}{l}Cl_{ 2 }O_{ 7 \; (g) } \; + \; 8\; e^{ – } \; \rightarrow \; 2 \; ClO^{ – }_{ 2 \; (aq) }\end{array} \)

– Add

\(\begin{array}{l}6 \; OH^{ – }\end{array} \)

ions to balance the charge:

\(\begin{array}{l}Cl_{ 2 }O_{ 7 \; (g) } \; + \; 8\; e^{ – } \; \rightarrow \; 2 \; ClO^{ – }_{ 2 \; (aq) } \; + \;6 \; OH^{ – }_{ (aq) }\end{array} \)

– Add 3

\(\begin{array}{l}H_{ 2 }O\end{array} \)

to balance O atoms:

\(\begin{array}{l}Cl_{ 2 }O_{ 7 \; (g) } \; + \; 3 \; H_{ 2 }O_{ (l) } \; + 8\; e^{ – } \; \rightarrow \; 2 \; ClO^{ – }_{ 2 \; (aq) } \; + \;6 \; OH^{ – }_{ (aq) }\end{array} \)

Now, multiply the equation (1) by 4. Then, adding equation (1) and (2), we get the balanced redox reaction as given below:

\(\begin{array}{l}Cl_{ 2 }O_{ 7 \; (g) } \; + \; 4 \; H_{ 2 }O_{ 2 \; (aq) } \; + 2\; OH^{ – }_{ (aq) } \; \rightarrow \; 2 \; ClO^{ – }_{ 2 (aq) } \; + \; 4 \; O_{ 2 \; (g) } \; + \; 5 \; H_{ 2 }O_{ (l) }\end{array} \)

Oxidation number method:

– Reduction in the oxidation no. of

\(\begin{array}{l}Cl_{ 2 }O_{ 7 }\end{array} \)

= 4× 2 = 8

– Increment in the oxidation no. of

\(\begin{array}{l}H_{ 2 }O_{ 2 }\end{array} \)

= 2× 1 = 2

Multiply

\(\begin{array}{l}H_{ 2 }O_{ 2 }\end{array} \)

by 4 and

\(\begin{array}{l}O_{ 2 }\end{array} \)

by 4 to balance the reduction and increment of the oxidation no. :

\(\begin{array}{l}3 \; N_{ 2 }H_{ 4 \; (l) } \; + \; 4 \; ClO^{ – }_{ 3 \; (aq) } \; \rightarrow \; NO_{ (g) } \; + \; Cl^{ – }_{ (aq) }\end{array} \)

– Balance Cl and n atoms:

\(\begin{array}{l}Cl_{ 2 }O_{ 7 \; (g) } \; + \; 4 \; H_{ 2 }O_{ 2 \; (aq) } \; \rightarrow \; 2\; ClO^{ – }_{ 2 \;(aq) } \; + \; 4 \; O_{ 2 \; (g) }\end{array} \)

– Add 3

\(\begin{array}{l}H_{ 2 }O\end{array} \)

to balance O atoms:

\(\begin{array}{l}Cl_{ 2 }O_{ 7 \; (g) } \; + \; 4 \; H_{ 2 }O_{ 2 \; (aq) } \; \rightarrow \; 2 \; ClO^{ – }_{ 2 (aq) } \; + \; 4 \; O_{ 2 (g) } \; + \; 3 \; H_{ 2 }O_{ (l) }\end{array} \)

– Add

\(\begin{array}{l}2 \; OH^{ – }\end{array} \)

and

\(\begin{array}{l}2 \; H_{ 2 }O\end{array} \)

to balance H atoms:

\(\begin{array}{l}Cl_{ 2 }O_{ 7 \; (g) } \; + \; 4 \; H_{ 2 }O_{ 2 \; (aq) } \; 2 \; OH^{ – }_{ (aq) } \; \rightarrow \; 2 \; ClO^{ – }_{ 2 \; (aq) } \; + \; 4 \; O_{ 2 \; (g) } \; + \; 5 \; H_{ 2 }O_{ (l) }\end{array} \)

This is the required reaction equation.

20.What sorts of informations can you draw from the following reaction ?

\(\begin{array}{l}(CN)_{ 2 \; (g) } \; + \; 2 \; OH^{ – }_{ (aq) } \; \rightarrow \; CN_{ (aq) }^{ – } \; + \; CNO_{ (aq) }^{ – } \; + \; H_{ 2 }O_{ (l) }\end{array} \)

Answer:

The oxidation no. of C in

\(\begin{array}{l}(CN)_{ 2 }\end{array} \)

\(\begin{array}{l}CN^{ – }\end{array} \)

and

\(\begin{array}{l}CNO^{ – }\end{array} \)

are +3, +2 and +4 respectively.

Let the oxidation no. of C be y.

\(\begin{array}{l}(CN)_{ 2 }\end{array} \)

2(y – 3) = 0

Therefore, y = 3

\(\begin{array}{l}CN^{ – }\end{array} \)

y – 3 = -1

Therefore, y = 2

\(\begin{array}{l}CNO^{ – }\end{array} \)

y – 3 -2 = -1

Therefore, y = 4

The oxidation no. of C in the reaction is:

Oxidation no. of C in

\(\begin{array}{l}(CN)_{ 2 }\end{array} \)

is +3

Oxidation no. of C in

\(\begin{array}{l}CN^{ – }\end{array} \)

is +2

Oxidation no. of C in

\(\begin{array}{l}CNO^{ – }\end{array} \)

is +4

We can see that the same compound is oxidized and reduced simultaneously in the reaction.

The reactions in which the same compound is oxidized and reduced is known as disproportionation reaction. Then, we can say that the alkaline decomposition of cyanogens is a disproportionation reaction.

21. The

\(\begin{array}{l}Mn^{ 3+ }\end{array} \)

ion is unstable in solution and undergoes disproportionation to give Mn²⁺, MnO₂, and H⁺ ion. Write a balanced ionic equation for the reaction.

Answer:

The reaction is as given below:

\(\begin{array}{l}Mn^{ 3+ }_{ (aq) } \; \rightarrow \; Mn^{ 2+ }_{ (aq) } \; + \; MnO_{ 2 \; (s) } \; + \; H^{ + }_{ (aq) }\end{array} \)

The oxidation half-reaction:

\(\begin{array}{l}Mn^{ 3+ }_{ (aq) } \; \rightarrow \; MnO_{ 2 \; (s) }\end{array} \)

Add 1 electron to balance the oxidation no. :

\(\begin{array}{l}Mn^{ 3+ }_{ (aq) } \; \rightarrow \; MnO_{ 2 \; (s) } \;+ \; e^{ – }\end{array} \)

Add

\(\begin{array}{l}4 \; H^{ + }\end{array} \)

ions to balance the charge:

\(\begin{array}{l}Mn^{ 3+ }_{ (aq) } \; \rightarrow \; MnO_{ 2 \; (s) } \;+ \; e^{ – }\; + \; 4 \; H^{ + }_{ (aq) }\end{array} \)

Add 2

\(\begin{array}{l}H_{ 2 }O\end{array} \)

to balance O atoms and

\(\begin{array}{l}H^{ + }\end{array} \)

ions:

\(\begin{array}{l}Mn^{ 3+ }_{ (aq) } \; + \; 2 \; H_{ 2 }O_{ (l) } \; \rightarrow \; MnO_{ 2 \; (s) } \;+ \; e^{ – }\; + \; 4 \; H^{ + }_{ (aq) }\end{array} \)

———— (1)

The reduction half-reaction:

\(\begin{array}{l}Mn^{ 3+ }_{ (aq) } \; \rightarrow \; Mn^{ 2+ }_{ (aq) }\end{array} \)

Add 1 electron to balance oxidation no. :

\(\begin{array}{l}Mn^{ 3+ }_{ (aq) } \; + \; e^{ – } \; \rightarrow \; Mn^{ 2+ }_{ (aq) }\end{array} \)

——— (2)

Add equations (1) and (2) to get the balanced chemical equation:

\(\begin{array}{l}2 \; Mn^{ 3+ }_{ (aq) } \; + \; 2 \; H_{ 2 }O_{ (l) } \; \rightarrow \; MnO_{ 2 \;(s) } \; + \; 2 \; Mn^{ 2+ }_{ (aq) } \; + \; 4 \; H^{ + }_{ (aq) }\end{array} \)

22.Consider the elements:

Cs, Ne, I and F

(a) Identify the element that exhibits only negative oxidation.

(b) Identify the element that exhibits only positive oxidation.

(c) Identify the element that exhibits both negative and positive oxidation states.

(d) Identify the element that exhibits neither negative nor positive oxidation state?

Answer:

(a) F exhibits only negative oxidation no. That is -1.

(b) Cs exhibits only positive oxidation no. That is +1.

(d) Ne exhibits neither negative nor positive oxidation no. That is 0.

23. Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.

Answer:

The redox reaction is as given below:

\(\begin{array}{l}Cl_{ 2 \;(s)} \; + \; SO_{ 2\;(aq)} \; + \; H_{ 2 }O_{(l)} \; \rightarrow \; Cl^{ – }_{(aq)} \; + \; SO^{ 2- }_{ 4 \; (aq) }\end{array} \)

The oxidation half-reaction:

\(\begin{array}{l}SO_{ 2 \;(aq)} \;\rightarrow \; SO^{ 2- }_{ 4 \; (aq) }\end{array} \)

Add 2 electrons to balance the oxidation no. :

\(\begin{array}{l}SO_{ 2 \;(aq)} \;\rightarrow \; SO^{ 2- }_{ 4 \; (aq) } \; + \; 2 \; e^{ – }\end{array} \)

Add

\(\begin{array}{l}4 \; H^{ + }\end{array} \)

ions to balance the charge:

\(\begin{array}{l}SO_{ 2 \;(aq)} \;\rightarrow \; SO^{ 2- }_{ 4 \; (aq) } \; + \;4 \; H^{ + }_{ (aq) } \;+ \; 2 \; e^{ – }\end{array} \)

Add 2

\(\begin{array}{l}H_{ 2 }O\end{array} \)

to balance O atoms and

\(\begin{array}{l}H^{ + }\end{array} \)

ions:

\(\begin{array}{l}SO_{ 2 \;(aq)} \; + \; 2 \; H_{ 2 }O \; \rightarrow \; SO^{ 2- }_{ 4 \; (aq) } \; + \;4 \; H^{ + }_{ (aq) } \;+ \; 2 \; e^{ – }\end{array} \)

——— (1)

The reduction half-reaction:

\(\begin{array}{l}Cl_{2\;(s)} \; \rightarrow Cl^{-}_{(aq)}\end{array} \)

Balance Cl atoms:

\(\begin{array}{l}Cl_{ 2 \; (s) } \; \rightarrow \; 2 \; Cl^{ – }_{ (aq) }\end{array} \)

Add 2 electrons to balance the oxidation no. :

\(\begin{array}{l}Cl_{ 2 \; (s) } \; + \; 2 \; e^{ – } \rightarrow \; 2 \; Cl^{ – }_{ (aq) }\end{array} \)

——— (2)

Add equations (1) and (2) to get the balanced chemical equation:

\(\begin{array}{l}Cl_{ 2 \; (s) }\; + \; SO_{ 2 \;(aq)} \; + \; 2 \; H_{ 2 }O_{ (l) } \; \rightarrow \; 2 \; Cl^{ – }_{ (aq) } \; + \; SO^{ 2- }_{ 4 \; (aq) } \; + \; 4 \; H^{ + }_{ (aq) }\end{array} \)

24. Refer to the periodic table given in your book and now answer the following questions:

(a) Select the possible non–metals that can show disproportionation reaction?

(b) Select three metals that show disproportionation reaction?

Answer:

One of the reacting elements always has an element that can exist in at least 3 oxidation numbers.

(i) The non–metals which can show disproportionation reactions are P, Cl and S.

(ii) The three metals which can show disproportionation reactions are Mn, Ga and Cu.

25. In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen?

Answer:

The balanced reaction is as given below:

\(\begin{array}{l}4 \; NH_{ 3 \; (g) } \; + \; 5 \; O_{ 2 \; (g) } \; \rightarrow \; 4 \; NO_{ (g) } \; + \; 6 \; H_{ 2 }O_{ (g) }\end{array} \)

\(\begin{array}{l}4 \; NH_{ 3 }\end{array} \)

= 4 × 17 g = 68 g

\(\begin{array}{l}5 \; O_{ 2 }\end{array} \)

= 5 × 32 g = 160 g

\(\begin{array}{l}4 \; NO\end{array} \)

= 4 × 30 g = 120 g

\(\begin{array}{l}6 \; H_{ 2 }O\end{array} \)

= 6 × 18 g = 108 g

Thus,

\(\begin{array}{l}NH_{ 3 }\end{array} \)

(68 g) reacts with

\(\begin{array}{l}O_{ 2 }\end{array} \)

( 20 g )

Therefore, 10 g of

\(\begin{array}{l}NH_{ 3 }\end{array} \)

reacts with

\(\begin{array}{l}\frac{ 160 \; \times \; 10 }{ 68 }\end{array} \)

g = 23.53 g of

\(\begin{array}{l}O_{ 2 }\end{array} \)

But only 20 g of

\(\begin{array}{l}O_{ 2 }\end{array} \)

is available.

Hence,

\(\begin{array}{l}O_{ 2 }\end{array} \)

is a limiting reagent.

Now, 160 g of

\(\begin{array}{l}O_{ 2 }\end{array} \)

gives

\(\begin{array}{l}\frac{ 120 \; \times \; 20}{ 160 }\end{array} \)

g of N = 15 g of NO.

Therefore, max of 15 g of nitric oxide can be obtained.

26. Using the standard electrode potentials given in Table 8.1, predict if the reaction between the following is feasible:

(a)

\(\begin{array}{l}Fe^{ 3+ }_{ (aq) }\end{array} \)

and

\(\begin{array}{l}I^{ – }_{ (aq) }\end{array} \)

(b)

\(\begin{array}{l}Ag^{ + }_{ (aq) }\end{array} \)

and

\(\begin{array}{l}Cu^{ }_{ (s) }\end{array} \)

(c)

\(\begin{array}{l}Fe^{ 3+ }_{ (aq) }\end{array} \)

and

\(\begin{array}{l}Cu^{ }_{ (s) }\end{array} \)

(d)

\(\begin{array}{l}Ag^{ }_{ (s) }\end{array} \)

and

\(\begin{array}{l}Fe^{ 3+ }_{ (aq) }\end{array} \)

(e)

\(\begin{array}{l}Br_{ 2 \; (aq) }\end{array} \)

and

\(\begin{array}{l}Fe^{ 2+ }_{ (aq) }\end{array} \)

Answer:

(a)

\(\begin{array}{l}Fe^{ 3+ }_{ (aq) }\end{array} \)

and

\(\begin{array}{l}I^{ – }_{ (aq) }\end{array} \)

\(\begin{array}{l}2 \; Fe_{ (aq) }^{ 3+ } \; + \; 2 \; I^{ – }_{ (aq) } \; \rightarrow \; 2 \; Fe_{ (aq) }^{ 2+ } \; + \; I_{ 2 \; (s) }\end{array} \)

Oxidation half reaction:

\(\begin{array}{l}2 \; I^{ – }_{ (aq) } \; \rightarrow \; I_{ 2 \; (s) } \; + \; 2 \; e^{ – }\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -0.54V\end{array} \)

Reduction half reaction:

\(\begin{array}{l}[Fe_{ (aq) }^{ 3+ } \; + \; e^{ – } \; \rightarrow \; Fe_{ (aq) }^{ 2+ }] \; \times \; 2;\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; +0.77V\end{array} \)

\(\begin{array}{l}2\; Fe_{ (aq) }^{ 3+ } \; + \;2 \; I^{ – } \; \rightarrow \; 2 \; Fe_{ (aq) }^{ 2+ } \; + \; I_{ 2 \; (s) };\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; +0.23V\end{array} \)

\(\begin{array}{l}E^{\circ}\end{array} \)

for the overall reaction is positive. Therefore, the reaction between

\(\begin{array}{l}Fe_{ (aq) }^{ 3+ }\end{array} \)

and

\(\begin{array}{l}I^{ – }_{ (aq) }\end{array} \)

is feasible.

(b)

\(\begin{array}{l}Ag^{ + }_{ (aq) }\end{array} \)

and

\(\begin{array}{l}Cu^{ }_{ (s) }\end{array} \)

\(\begin{array}{l}2 \; Ag^{ + }_{ (aq) } \; + \; Cu_{ (s) } \; \rightarrow \; 2 \; Ag_{ (s) } \; + \; Cu_{ (aq) }^{ 2+ }\end{array} \)

Oxidation half reaction:

\(\begin{array}{l}Cu_{ (s) } \; \rightarrow \; Cu^{ 2+ }_{ (aq) } \; + \; 2 \; e^{ – }\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -0.34V\end{array} \)

Reduction half reaction:

\(\begin{array}{l}[Ag^{ + }_{ (aq) } \; + \; e^{ – } \; \rightarrow \; Ag_{ (s) }] \; \times \; 2\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; +0.80V\end{array} \)

\(\begin{array}{l}2 \; Ag^{ + }_{ (aq) } \; + \; Cu_{ (s) } \; \rightarrow \; 2 \; Ag_{ (s) } \; + \; Cu^{ 2+ }\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; +0.46V\end{array} \)

\(\begin{array}{l}E^{\circ}\end{array} \)

for the overall reaction is positive. Therefore, the reaction between

\(\begin{array}{l}Ag_{ (aq) }^{ + }\end{array} \)

and

\(\begin{array}{l}Cu_{ (s) }\end{array} \)

is feasible.

(c)

\(\begin{array}{l}Fe^{ 3+ }_{ (aq) }\end{array} \)

and

\(\begin{array}{l}Cu^{ }_{ (s) }\end{array} \)

\(\begin{array}{l}2 \; Fe^{ 3+ }_{ (aq) } \; + \; Cu_{ (s) } \; \rightarrow \; 2 \; Fe^{ 2+ }_{ (s) } \; + \; Cu^{ 2+ }_{ (aq) }\end{array} \)

Oxidation half reaction:

\(\begin{array}{l}Cu_{ (s) } \; \rightarrow \; Cu^{ 2+ }_{ (aq) } \; + \; 2 \; e^{ – }\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -0.34V\end{array} \)

Reduction half reaction:

\(\begin{array}{l}[Fe_{ (aq) }^{ 3+ }\; + \; e^{ – } \; \rightarrow \; Fe_{ (s) }^{ 2+ }] \; \times \; 2\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; +0.77V\end{array} \)

\(\begin{array}{l}2 \; Fe_{ (aq) }^{ 3+ } \; + \; Cu_{ (s) } \; \rightarrow \; 2 \; Fe_{ (s) }^{ 2+ } \; + \; Cu_{ (aq) }^{ 2+ }\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; +0.43V\end{array} \)

\(\begin{array}{l}E^{\circ}\end{array} \)

for the overall reaction is positive. Therefore, the reaction between

\(\begin{array}{l}Fe^{ 3+ }_{ (aq) }\end{array} \)

and

\(\begin{array}{l}Cu^{ }_{ (s) }\end{array} \)

is feasible.

(d)

\(\begin{array}{l}Ag^{ }_{ (s) }\end{array} \)

and

\(\begin{array}{l}Fe^{ 3+ }_{ (aq) }\end{array} \)

\(\begin{array}{l}Ag_{ (s) } \; + \; 2 \; Fe_{ (aq) }^{ 3+ } \; \rightarrow \; Ag_{ (aq) }^{ + } \; + \; Fe_{ (aq) }^{ 2+ }\end{array} \)

Oxidation half reaction:

\(\begin{array}{l}Ag_{ (s) } \; \rightarrow \; Ag_{ (aq) }^{ + } \; + \; e^{ – }\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -0.80V\end{array} \)

Reduction half reaction:

\(\begin{array}{l}Fe_{ (aq) }^{ 3+ } \; + \; e^{ – } \; \rightarrow \; Fe_{ (aq) }^{ 2+ }\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; +0.77V\end{array} \)

\(\begin{array}{l}Ag_{ (s) } \; + \; Fe_{ (aq) }^{ 3+ } \; \rightarrow \; Ag_{ (aq) }^{ + } \; + \; Fe_{ (aq) }^{ 2+ }\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -0.03V\end{array} \)

\(\begin{array}{l}E^{\circ}\end{array} \)

for the overall reaction is positive. Therefore, the reaction between

\(\begin{array}{l}Ag^{ }_{ (s) }\end{array} \)

and

\(\begin{array}{l}Fe^{ 3+ }_{ (aq) }\end{array} \)

is feasible.

(e)

\(\begin{array}{l}Br_{ 2 \; (aq) }\end{array} \)

and

\(\begin{array}{l}Fe^{ 2+ }_{ (aq) }\end{array} \)

\(\begin{array}{l}Br_{ 2 \; (s) } \; + \; 2 \; Fe_{ (aq) }^{ 2+ } \; \rightarrow \; 2 \; Br_{ (aq) }^{ – } \; + \; 2 \; Fe_{ (aq) }^{ 3+ }\end{array} \)

Oxidation half reaction:

\(\begin{array}{l}[Fe_{ (aq) }^{ 2+ } \; \rightarrow \; Fe_{ (aq) }^{ 3+ } \; + \; e^{ – } ] \; \times \; 2\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -0.77V\end{array} \)

Reduction half reaction:

\(\begin{array}{l}Br_{2\; (aq) } \; + \; 2 \; e^{ – } \; \rightarrow \; 2 \; Br_{ (aq) }^{ – }\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; +1.09V\end{array} \)

\(\begin{array}{l}Br_{2\; (s) } \; + \; 2 \; Fe_{ (aq) }^{ 2+ } \; \rightarrow \; 2 \; Br_{(aq)}^{ – } \; + \; 2 \; Fe_{ (aq) }^{3+}\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -0.32V\end{array} \)

\(\begin{array}{l}E^{\circ}\end{array} \)

for the overall reaction is positive. Therefore, the reaction between

\(\begin{array}{l}Br_{2\; (aq) }\end{array} \)

and

\(\begin{array}{l}Fe^{2+}_{ (aq) }\end{array} \)

is feasible.

27. Predict the products of electrolysis in each of the following:

(i) An aqueous solution of

\(\begin{array}{l}AgNO_{3}\end{array} \)

with silver electrodes

(ii) An aqueous solution

\(\begin{array}{l}AgNO_{3}\end{array} \)

with platinum electrodes

(iii) A dilute solution of

\(\begin{array}{l}H_{2}SO_{4}\end{array} \)

with platinum electrodes

(iv) An aqueous solution of

\(\begin{array}{l}CuCl_{2}\end{array} \)

with platinum electrodes.

Answer:

(i)

\(\begin{array}{l}AgNO_{3}\end{array} \)

ionizes in aqueous solution to form

\(\begin{array}{l}Ag^{+}\end{array} \)

and

\(\begin{array}{l}NO^{-}_{3}\end{array} \)

ions.

On electrolysis, either

\(\begin{array}{l}Ag^{+}\end{array} \)

ion or

\(\begin{array}{l}H_{2}O\end{array} \)

molecule can be decreased at cathode. But the reduction potential of

\(\begin{array}{l}Ag^{+}\end{array} \)

ions is higher than that of

\(\begin{array}{l}H_{2}O\end{array} \)

\(\begin{array}{l}Ag^{+}_{(aq)} \; + \; e^{-} \; \rightarrow Ag_{\left ( s \right )}\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; +0.80V\end{array} \)

\(\begin{array}{l}2\; H_{2}O_{\left ( l \right )} \; +\; 2 \;e^{-} \; \rightarrow \;H_{2\;\left ( g \right )} \;+ \; 2 \;OH^{-}_{\left ( aq \right )}\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -0.83V\end{array} \)

Therefore,

\(\begin{array}{l}Ag^{+}\end{array} \)

ions are decreased at the cathode. Same way, Ag metal or

\(\begin{array}{l}H_{2}O\end{array} \)

molecules can be oxidized at the anode. But the oxidation potential of Ag is greater than that of

\(\begin{array}{l}H_{2}O\end{array} \)

molecules.

\(\begin{array}{l}Ag_{\left ( s \right )} \; \rightarrow \;Ag^{+}_{\left ( aq \right )} \;+ \;e^{-}\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -0.80V\end{array} \)

\(\begin{array}{l}2\; H_{2}O_{\left ( l \right )}\; \rightarrow \;O_{2\;\left ( g \right )}\; +\; 4 \;H^{+}_{\left ( aq \right )}\; +\; 4\; e^{-}\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -1.23V\end{array} \)

Hence, Ag metal gets oxidized at the anode.

(ii) Pt cannot be oxidized very easily. Therefore, at the anode, oxidation of water occurs to liberate

\(\begin{array}{l}O_{2}\end{array} \)

. At the cathode,

\(\begin{array}{l}Ag^{+}\end{array} \)

ions are decreased and get deposited.

(iii)

\(\begin{array}{l}H_{2}SO_{4}\end{array} \)

ionizes in aqueous solutions to give

\(\begin{array}{l}H^{+}\end{array} \)

and

\(\begin{array}{l}SO_{4}^{2-}\end{array} \)

ions.

\(\begin{array}{l}H_{2}SO_{4\;\left ( aq \right )} \;\rightarrow \; 2\; H^{+}_{\left ( aq \right )} \; +\; SO^{2-}_{4 \left ( aq \right )}\end{array} \)

On electrolysis, either of

\(\begin{array}{l}H_{2}O\end{array} \)

molecules or

\(\begin{array}{l}H^{+}\end{array} \)

ions can get decreased at cathode. But the decreased potential of

\(\begin{array}{l}H^{+}\end{array} \)

ions is higher than that of

\(\begin{array}{l}H_{2}O\end{array} \)

molecules.

\(\begin{array}{l}2 \;H^{+}_{\left ( aq \right )} \;+ \;2 \;e^{-} \;\rightarrow \;H_{2\;\left ( g \right )}\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; 0.0V\end{array} \)

\(\begin{array}{l}2 \;H_{2}O_{\left ( aq \right )} \;+ \;2 \;e^{-} \;\rightarrow \;H_{2\;\left ( g \right )} \;+ \;2 \;OH^{-}_{\left ( aq \right )}\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -0.83V\end{array} \)

Therefore, at cathode,

\(\begin{array}{l}H^{+}\end{array} \)

ions are decreased to free

\(\begin{array}{l}H_{2}\end{array} \)

gas.

On the other hand, at the anode, either of

\(\begin{array}{l}H_{2}O\end{array} \)

molecules or

\(\begin{array}{l}SO_{4}^{2-}\end{array} \)

ions can be oxidized. But the oxidation of

\(\begin{array}{l}SO_{4}^{2-}\end{array} \)

involves breaking of more bonds than that of

\(\begin{array}{l}H_{2}O\end{array} \)

molecules. Therefore,

\(\begin{array}{l}SO_{4}^{2-}\end{array} \)

ions have lower oxidation potential than

\(\begin{array}{l}H_{2}O\end{array} \)

. Hence,

\(\begin{array}{l}H_{2}O\end{array} \)

is oxidized at anode to free

\(\begin{array}{l}O_{2}\end{array} \)

molecules.

(iv) In aqueous solutions,

\(\begin{array}{l}CuCl_{2}\end{array} \)

ionizes to give

\(\begin{array}{l}Cu^{2+}\end{array} \)

and

\(\begin{array}{l}Cl^{-}\end{array} \)

ions as:

\(\begin{array}{l}CuCl_{2\; \left ( aq \right )} \;\rightarrow \;Cu^{2+}_{\left ( aq \right )} \;+ \;2 \;Cl_{\left ( aq \right )}^{-}\end{array} \)

\(\begin{array}{l}CuCl_{2\;\left ( aq \right )} \;\rightarrow \;Cu^{2+}_{\left ( aq \right )} \;+ \;2 \;Cl_{\left ( aq \right )}^{-}\end{array} \)

On electrolysis, either of

\(\begin{array}{l}Cu^{2+}\end{array} \)

ions or

\(\begin{array}{l}H_{2}O\end{array} \)

molecules can get decreased at cathode. But the decreased potential of

\(\begin{array}{l}Cu^{2+}\end{array} \)

is more than that of

\(\begin{array}{l}H_{2}O\end{array} \)

molecules.

\(\begin{array}{l}Cu^{2+}_{\left ( aq \right )} \;+ \;2 \;e^{-} \;\rightarrow \;Cu_{\left ( aq \right )}\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; +0.34V\end{array} \)

\(\begin{array}{l}H_{2}O_{\left ( l \right )} \;+ \;2 \;e^{-} \;\rightarrow \;H_{2\;\left ( g \right )} \;+ \; 2 \;OH^{-}\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -0.83V\end{array} \)

Therefore,

\(\begin{array}{l}Cu^{2+}\end{array} \)

ions are decreased at the cathode and get deposited. In the same way, at the anode, either of

\(\begin{array}{l}Cl^{-}\end{array} \)

\(\begin{array}{l}H_{2}O\end{array} \)

is oxidized. The oxidation potential of

\(\begin{array}{l}H_{2}O\end{array} \)

is higher than that of

\(\begin{array}{l}Cl^{-}\end{array} \)

\(\begin{array}{l}2 \;Cl^{-}_{\left ( aq \right )} \;\rightarrow \;Cl_{2\;\left ( g \right )} \;+ \;2 \;e^{-}\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; +0.34V\end{array} \)

\(\begin{array}{l}2\; H_{2}O_{\left ( l \right )} \;\rightarrow \;O_{2\;\left ( g \right )} \;+ \;4 \;H^{+}_{\left ( aq \right )} \;+ \;4 \;e^{-}\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -1.23V\end{array} \)

But oxidation of

\(\begin{array}{l}H_{2}O\end{array} \)

molecules occurs at a lower electrode potential compared to that of

\(\begin{array}{l}Cl^{-}\end{array} \)

ions because of over-voltage (extra voltage required to liberate gas). As a result,

\(\begin{array}{l}Cl^{-}\end{array} \)

ions are oxidized at the anode to liberate

\(\begin{array}{l}Cl_{2}\end{array} \)

gas.

28. Arrange the given metals in the order in which they displace each other from the solution of their salts.

Al, Fe, Cu, Zn, Mg

Answer:

A metal with stronger reducing power displaces another metal with weaker reducing power from its solution of salt.

The order of the increasing reducing power of the given metals is as given below:

Cu < Fe < Zn < Al < Mg

Therefore, Mg can displace Al from its salt solution, but Al cannot displace Mg. Thus, the order in which the given metals displace each other from the solution of their salts is as given below: Mg >Al>Zn> Fe >Cu

29. Given the standard electrode potentials,

\(\begin{array}{l}K^{+}\end{array} \)

/K = –2.93V

\(\begin{array}{l}Ag^{+}\end{array} \)

/Ag = 0.80V

\(\begin{array}{l}Hg^{2+}\end{array} \)

/Hg = 0.79V

\(\begin{array}{l}Mg^{2+}\end{array} \)

/Mg = –2.37V

\(\begin{array}{l}Cr^{3+}\end{array} \)

/Cr = –0.74V

Arrange these metals in their increasing order of reducing power.

Answer:

The reducing agent is stronger as the electrode potential decreases. Hence, the increasing order of the reducing power of the given metals is as given below:

Ag < Hg < Cr < Mg < K

30. Depict the galvanic cell in which the reaction is:

\(\begin{array}{l}Zn_{\left ( s \right )} \;+ \;2 \;Ag_{\left ( aq \right )}^{+} \;\rightarrow \;Zn^{2+}_{\left ( aq \right )} \;+ \;2 \;Ag_{\left ( s \right )}\end{array} \)

Further show:

(i) which of the electrode is negatively charged?

(ii) the carriers of the current in the cell.

(iii) individual reaction at each electrode.

Answer:

The galvanic cell corresponding to the given redox reaction can be shown as:

Zn|

\(\begin{array}{l}Zn^{2+}_{\left ( aq \right )}\end{array} \)

\(\begin{array}{l}Ag^{+}_{\left ( aq \right )}\end{array} \)

|Ag

(i) Zn electrode is negatively charged because at this electrode, Zn oxidizes to

\(\begin{array}{l}Zn^{2+}\end{array} \)

and the leaving electrons accumulate on this electrode.

(ii) The carriers of current are ions in the cell.

(iii) Reaction at Zn electrode is shown as:

\(\begin{array}{l}Zn_{\left ( s \right )} \;\rightarrow \;Zn^{2+}_{\left ( aq \right )} \;+ \;2 \;e^{-}\end{array} \)

Reaction at Ag electrode is shown as:

\(\begin{array}{l}Ag^{+}_{\left ( aq \right )} \;+ \;e^{-}\;\rightarrow \;Ag_{\left ( s \right )}\end{array} \)

Some of the important topics covered in NCERT Solutions for Class 11 Chemistry Chapter 8 are given below:

Classical Idea of Redox Reactions – Oxidation and Reduction Reactions
Redox Reactions in Terms of Electron Transfer Reactions
Oxidation Number
Redox Reactions and Electrode Processes

Frequently Asked Questions on NCERT Solutions for Class 11 Chemistry Chapter 8

What are the main topics covered in NCERT Solutions for Class 11 Chemistry Chapter 8?

The main topics covered in NCERT Solutions for Class 11 Chemistry Chapter 8 are given below:
Classical Idea of Redox Reactions – Oxidation and Reduction Reactions
Redox Reactions in Terms of Electron Transfer Reactions
Competitive Electron Transfer Reactions
Oxidation Number
Types of Redox Reactions
Balancing of Redox Reactions
Redox Reactions as the Basis for Titrations
Limitations of Concept of Oxidation Number
Redox Reactions and Electrode Processes.

How many questions are there in NCERT Solutions for Class 11 Chemistry Chapter 8?

NCERT Solutions for Class 11 Chemistry Chapter 8 contains a total of 30 questions. Practising these exercises helps you in scoring high in the board exams and also helps to ease the subject. These solutions are explained by subject matter experts to help you clear your doubts.

What is the meaning of oxidation number according to NCERT Solutions for Class 11 Chemistry Chapter 8?

According to NCERT Solutions for Class 11 Chemistry Chapter 8, oxidation number can be described as the number that is allocated to elements in a chemical combination. The oxidation number is basically the count of electrons that atoms in a molecule can share, lose or gain while forming chemical bonds with other atoms of a different element.

Q4.

List the Subtopics of Class 11 Chemistry Chapter 8 Redox Reactions

Classical Idea of Redox Reactions – Oxidation And Reduction Reactions
Redox Reactions in Terms of Electron Transfer Reactions
- Competitive Electron Transfer Reactions
Oxidation Number
- Types of Redox Reactions
- Balancing of Redox Reactions
- Redox Reactions as the Basis for Titrations
- Limitations of the Concept of Oxidation Number
Redox Reactions and Electrode Processes.

Neeraj Anand, Param Anand

Er. Neeraj K.Anand is a freelance mentor and writer who specializes in Engineering & Science subjects. Neeraj Anand received a B.Tech degree in Electronics and Communication Engineering from N.I.T Warangal & M.Tech Post Graduation from IETE, New Delhi. He has over 30 years of teaching experience and serves as the Head of Department of ANAND CLASSES. He concentrated all his energy and experiences in academics and subsequently grew up as one of the best mentors in the country for students aspiring for success in competitive examinations. In parallel, he started a Technical Publication "ANAND TECHNICAL PUBLISHERS" in 2002 and Educational Newspaper "NATIONAL EDUCATION NEWS" in 2014 at Jalandhar. Now he is a Director of leading publication "ANAND TECHNICAL PUBLISHERS", "ANAND CLASSES" and "NATIONAL EDUCATION NEWS". He has published more than hundred books in the field of Physics, Mathematics, Computers and Information Technology. Besides this he has written many books to help students prepare for IIT-JEE and AIPMT entrance exams. He is an executive member of the IEEE (Institute of Electrical & Electronics Engineers. USA) and honorary member of many Indian scientific societies such as Institution of Electronics & Telecommunication Engineers, Aeronautical Society of India, Bioinformatics Institute of India, Institution of Engineers. He has got award from American Biographical Institute Board of International Research in the year 2005.

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CBSE Class 11 Chemistry Syllabus

CBSE Class 11 Chemistry Syllabus is a vast which needs a clear understanding of the concepts and topics. Knowing CBSE Class 11 Chemistry syllabus helps students to understand the course structure of Chemistry.

Unit-wise CBSE Class 11 Syllabus for Chemistry

Below is a list of detailed information on each unit for Class 11 Students.

UNIT I – Some Basic Concepts of Chemistry

General Introduction: Importance and scope of Chemistry.

Nature of matter, laws of chemical combination, Dalton’s atomic theory: concept of elements,
atoms and molecules.

Atomic and molecular masses, mole concept and molar mass, percentage composition, empirical and molecular formula, chemical reactions, stoichiometry and calculations based on stoichiometry.

UNIT II – Structure of Atom

Discovery of Electron, Proton and Neutron, atomic number, isotopes and isobars. Thomson’s model and its limitations. Rutherford’s model and its limitations, Bohr’s model and its limitations, concept of shells and subshells, dual nature of matter and light, de Broglie’s relationship, Heisenberg uncertainty principle, concept of orbitals, quantum numbers, shapes of s, p and d orbitals, rules for filling electrons in orbitals – Aufbau principle, Pauli’s exclusion principle and Hund’s rule, electronic configuration of atoms, stability of half-filled and completely filled orbitals.

UNIT III – Classification of Elements and Periodicity in Properties

Significance of classification, brief history of the development of periodic table, modern periodic law and the present form of periodic table, periodic trends in properties of elements -atomic radii, ionic radii, inert gas radii, Ionization enthalpy, electron gain enthalpy, electronegativity, valency. Nomenclature of elements with atomic number greater than 100.

UNIT IV – Chemical Bonding and Molecular Structure

Valence electrons, ionic bond, covalent bond, bond parameters, Lewis structure, polar character of covalent bond, covalent character of ionic bond, valence bond theory, resonance, geometry of covalent molecules, VSEPR theory, concept of hybridization, involving s, p and d orbitals and shapes of some simple molecules, molecular orbital theory of homonuclear diatomic molecules(qualitative idea only), Hydrogen bond.

UNIT V – Chemical Thermodynamics

Concepts of System and types of systems, surroundings, work, heat, energy, extensive and intensive properties, state functions. First law of thermodynamics – internal energy and enthalpy, measurement of U and H, Hess’s law of constant heat summation, enthalpy of bond dissociation, combustion, formation, atomization, sublimation, phase transition, ionization, solution and dilution. Second law of Thermodynamics (brief introduction)
Introduction of entropy as a state function, Gibb’s energy change for spontaneous and nonspontaneous processes.
Third law of thermodynamics (brief introduction).

UNIT VI – Equilibrium

Equilibrium in physical and chemical processes, dynamic nature of equilibrium, law of mass action, equilibrium constant, factors affecting equilibrium – Le Chatelier’s principle, ionic equilibrium- ionization of acids and bases, strong and weak electrolytes, degree of ionization,
ionization of poly basic acids, acid strength, concept of pH, hydrolysis of salts (elementary idea), buffer solution, Henderson Equation, solubility product, common ion effect (with illustrative examples).

UNIT VII – Redox Reactions

Concept of oxidation and reduction, redox reactions, oxidation number, balancing redox reactions, in terms of loss and gain of electrons and change in oxidation number, applications of redox reactions.

UNIT VIII – Organic Chemistry: Some basic Principles and Techniques

General introduction, classification and IUPAC nomenclature of organic compounds. Electronic displacements in a covalent bond: inductive effect, electromeric effect, resonance and hyper conjugation. Homolytic and heterolytic fission of a covalent bond: free radicals, carbocations, carbanions, electrophiles and nucleophiles, types of organic reactions.

UNIT IX – Hydrocarbons

Classification of Hydrocarbons
Aliphatic Hydrocarbons:
Alkanes – Nomenclature, isomerism, conformation (ethane only), physical properties, chemical reactions.
Alkenes – Nomenclature, structure of double bond (ethene), geometrical isomerism, physical properties, methods of preparation, chemical reactions: addition of hydrogen, halogen, water, hydrogen halides (Markovnikov’s addition and peroxide effect), ozonolysis, oxidation, mechanism of electrophilic addition.
Alkynes – Nomenclature, structure of triple bond (ethyne), physical properties, methods of preparation, chemical reactions: acidic character of alkynes, addition reaction of – hydrogen, halogens, hydrogen halides and water.

Aromatic Hydrocarbons:

Introduction, IUPAC nomenclature, benzene: resonance, aromaticity, chemical properties: mechanism of electrophilic substitution. Nitration, sulphonation, halogenation, Friedel Craft’s alkylation and acylation, directive influence of functional group in monosubstituted benzene. Carcinogenicity and toxicity.

To know the CBSE Syllabus for all the classes from 1 to 12, visit the Syllabus page of CBSE. Meanwhile, to get the Practical Syllabus of Class 11 Chemistry, read on to find out more about the syllabus and related information in this page.

CBSE Class 11 Chemistry Practical Syllabus with Marking Scheme

In Chemistry subject, practical also plays a vital role in improving their academic scores in the subject. The overall weightage of Chemistry practical mentioned in the CBSE Class 11 Chemistry syllabus is 30 marks. So, students must try their best to score well in practicals along with theory. It will help in increasing their overall academic score.

CBSE Class 11 Chemistry Practical Syllabus

The experiments will be conducted under the supervision of subject teacher. CBSE Chemistry Practicals is for 30 marks. This contribute to the overall practical marks for the subject.

The table below consists of evaluation scheme of practical exams.

Evaluation Scheme	Marks
Volumetric Analysis	08
Salt Analysis	08
Content Based Experiment	06
Project Work	04
Class record and viva	04
Total	30

CBSE Syllabus for Class 11 Chemistry Practical

Micro-chemical methods are available for several of the practical experiments. Wherever possible such techniques should be used.

A. Basic Laboratory Techniques
1. Cutting glass tube and glass rod
2. Bending a glass tube
3. Drawing out a glass jet
4. Boring a cork

B. Characterization and Purification of Chemical Substances
1. Determination of melting point of an organic compound.
2. Determination of boiling point of an organic compound.
3. Crystallization of impure sample of any one of the following: Alum, Copper Sulphate, Benzoic Acid.

C. Experiments based on pH

1. Any one of the following experiments:

Determination of pH of some solutions obtained from fruit juices, solution of known and varied concentrations of acids, bases and salts using pH paper or universal indicator.
Comparing the pH of solutions of strong and weak acids of same concentration.
Study the pH change in the titration of a strong base using universal indicator.

2. Study the pH change by common-ion in case of weak acids and weak bases.

D. Chemical Equilibrium
One of the following experiments:

1. Study the shift in equilibrium between ferric ions and thiocyanate ions by increasing/decreasing the concentration of either of the ions.
2. Study the shift in equilibrium between [Co(H₂O)₆] ²⁺ and chloride ions by changing the concentration of either of the ions.

E. Quantitative Estimation
i. Using a mechanical balance/electronic balance.
ii. Preparation of standard solution of Oxalic acid.
iii. Determination of strength of a given solution of Sodium hydroxide by titrating it against standard solution of Oxalic acid.
iv. Preparation of standard solution of Sodium carbonate.
v. Determination of strength of a given solution of hydrochloric acid by titrating it against standard Sodium Carbonatesolution.

F. Qualitative Analysis
1) Determination of one anion and one cation in a given salt
Cations‐ Pb²⁺, Cu²⁺, As³⁺, Al³⁺, Fe³⁺, Mn²⁺, Ni²⁺, Zn²⁺, Co²⁺, Ca²⁺, Sr²⁺, Ba²⁺, Mg²⁺, NH₄ ⁺
Anions – (CO₃)^2‐ , S^2‐, NO₂^‐ , SO₃^2‐, SO^2‐ , NO ^‐ , Cl^‐ , Br^‐, I‐, PO₄^3‐ , C₂O²‐ ,CH₃COO^‐
(Note: Insoluble salts excluded)

2) Detection of ‐ Nitrogen, Sulphur, Chlorine in organic compounds.

G) PROJECTS
Scientific investigations involving laboratory testing and collecting information from other sources.

A few suggested projects are as follows:

Checking the bacterial contamination in drinking water by testing sulphide ion
Study of the methods of purification of water.
Testing the hardness, presence of Iron, Fluoride, Chloride, etc., depending upon the regional
variation in drinking water and study of causes of presence of these ions above permissible
limit (if any).
Investigation of the foaming capacity of different washing soaps and the effect of addition of
Sodium carbonate on it.
Study the acidity of different samples of tea leaves.
Determination of the rate of evaporation of different liquids Study the effect of acids and
bases on the tensile strength of fibres.
Study of acidity of fruit and vegetable juices.

Note: Any other investigatory project, which involves about 10 periods of work, can be chosen with the approval of the teacher.

Practical Examination for Visually Impaired Students of Class 11

Below is a list of practicals for the visually impaired students.

A. List of apparatus for identification for assessment in practicals (All experiments)
Beaker, tripod stand, wire gauze, glass rod, funnel, filter paper, Bunsen burner, test tube, test tube stand,
dropper, test tube holder, ignition tube, china dish, tongs, standard flask, pipette, burette, conical flask, clamp
stand, dropper, wash bottle
• Odour detection in qualitative analysis
• Procedure/Setup of the apparatus

B. List of Experiments A. Characterization and Purification of Chemical Substances
1. Crystallization of an impure sample of any one of the following: copper sulphate, benzoic acid
B. Experiments based on pH
1. Determination of pH of some solutions obtained from fruit juices, solutions of known and varied
concentrations of acids, bases and salts using pH paper
2. Comparing the pH of solutions of strong and weak acids of same concentration.

C. Chemical Equilibrium
1. Study the shift in equilibrium between ferric ions and thiocyanate ions by increasing/decreasing
the concentration of eitherions.
2. Study the shift in equilibrium between [Co(H2O)6]2+ and chloride ions by changing the
concentration of either of the ions.

D. Quantitative estimation
1. Preparation of standard solution of oxalic acid.
2. Determination of molarity of a given solution of sodium hydroxide by titrating it against standard
solution of oxalic acid.

E. Qualitative Analysis
1. Determination of one anion and one cation in a given salt
2. Cations – NH+4
Anions – (CO3)2-, S2-, (SO3)2-, Cl-, CH3COO-
(Note: insoluble salts excluded)
3. Detection of Nitrogen in the given organic compound.
4. Detection of Halogen in the given organic compound.

Note: The above practicals may be carried out in an experiential manner rather than recording observations.

We hope students must have found this information on CBSE Syllabus useful for their studying Chemistry. Learn Maths & Science in interactive and fun loving ways with ANAND CLASSES (A School Of Competitions) App/Tablet.

Frequently Asked Questions on CBSE Class 11 Chemistry Syllabus

How many units are in the CBSE Class 11 Chemistry Syllabus?

There are 9 units in the CBSE Class 11 Chemistry Syllabus. Students can access various study materials for the chapters mentioned in this article for free at ANAND CLASSES (A School Of Competitions).

What is the total marks for practicals examination as per the CBSE Class 11 Chemistry Syllabus?

The total marks for the practicals as per the CBSE Class 11 Chemistry Syllabus is 30. It includes volumetric analysis, content-based experiment, salt analysis, class record, project work and viva.

Which chapter carries more weightage as per the CBSE Syllabus for Class 11 Chemistry?

The organic chemistry chapter carries more weightage as per the CBSE Syllabus for Class 11 Chemistry.

NCERT Solutions-Redox Reactions for Class 11 Chemistry Chapter 8

NCERT Solutions for Class 11 Chemistry Chapter Redox Reactions

Some of the important topics covered in NCERT Solutions for Class 11 Chemistry Chapter 8 are given below:

Frequently Asked Questions on NCERT Solutions for Class 11 Chemistry Chapter 8

What are the main topics covered in NCERT Solutions for Class 11 Chemistry Chapter 8?

How many questions are there in NCERT Solutions for Class 11 Chemistry Chapter 8?

What is the meaning of oxidation number according to NCERT Solutions for Class 11 Chemistry Chapter 8?

List the Subtopics of Class 11 Chemistry Chapter 8 Redox Reactions

CBSE Class 11 Chemistry Syllabus

Unit-wise CBSE Class 11 Syllabus for Chemistry

UNIT I – Some Basic Concepts of Chemistry

UNIT II – Structure of Atom

UNIT III – Classification of Elements and Periodicity in Properties

UNIT IV – Chemical Bonding and Molecular Structure

UNIT V – Chemical Thermodynamics

UNIT VI – Equilibrium

UNIT VII – Redox Reactions

UNIT VIII – Organic Chemistry: Some basic Principles and Techniques

UNIT IX – Hydrocarbons

CBSE Class 11 Chemistry Practical Syllabus with Marking Scheme

CBSE Class 11 Chemistry Practical Syllabus

CBSE Syllabus for Class 11 Chemistry Practical

Practical Examination for Visually Impaired Students of Class 11

Frequently Asked Questions on CBSE Class 11 Chemistry Syllabus

How many units are in the CBSE Class 11 Chemistry Syllabus?

What is the total marks for practicals examination as per the CBSE Class 11 Chemistry Syllabus?

Which chapter carries more weightage as per the CBSE Syllabus for Class 11 Chemistry?

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