Nernst Equation-Expression, Derivation, Formula, Solved Problems, Determining Equilibrium Constant with Nernst Equation, Applications, Limitations

What Is Nernst Equation?

Even under non-standard conditions, the cell potentials of electrochemical cells can be determined with the help of the Nernst equation.

The Nernst equation is often used to calculate the cell potential of an electrochemical cell at any given temperature, pressure, and reactant concentration. The equation was introduced by a German chemist, Walther Hermann Nernst.

Expression of Nernst Equation

The Nernst equation is an equation relating the capacity of an atom/ion to take up one or more electrons (reduction potential) measured at any conditions to that measured at standard conditions (standard reduction potentials) of 298 K and one molar or one atmospheric pressure.

Nernst Equation for Single Electrode Potential

Ecell = E0 – [RT/nF] ln Q

Where,

  • Ecell = cell potential of the cell
  • E0 = cell potential under standard conditions
  • R = universal gas constant
  • T = temperature
  • n = number of electrons transferred in the redox reaction
  • F = Faraday constant
  • Q = reaction quotient

The calculation of single electrode reduction potential (Ered) from the standard single electrode reduction potential (E°red) for an atom/ion is given by the Nernst equation.

For a reduction reaction, the Nernst equation for a single electrode reduction potential for a reduction reaction

Mn+ + nenM is;

Ered = EMn+/M = EoMn+/M – [2.303RT/nF] log [1/[Mn+]]

Where,

  • R is the gas constant = 8.314 J/K Mole
  • T = absolute temperature,
  • n = number of mole of electron involved,
  • F = 96487 (≈96500) coulomb/mole = charge carried by one mole of electrons.
  • [Mn+] = active mass of the ions. For simplicity, it may be taken as equal to the molar concentration of the salt.

Nernst Equation at 25oC

For measurements carried out 298 K, the Nernst equation can be expressed as follows:

E = E0 – 0.0592/n log10 Q

Therefore, as per the Nernst equation, the overall potential of an electrochemical cell is dependent on the reaction quotient.

Derivation of Nernst Equation

Consider a metal in contact with its own salt-aqueous solution. Reactions of metal losing an electron to become an ion and the ion gaining an electron to return to the atomic state are equally feasible and are in an equilibrium state.

Mn+ + nenM

In the reduction reaction, ‘n’ moles of an electron is taken up by the ion against a reduction potential of Ered.

1. The work done in the movement of electron

Wred = nFEred

Where,

  • F is Faraday = 96487 coulomb = electrical charge carried by one mole of electrons

2. Change in the Gibbs free energy is an indication of spontaneity, and it is also equal to the maximum useful work (other than volume expansion) done in a process.

Combining work done and Gibbs free energy change:

Wred = nFEred = – ∆G or ∆G = – nFEred

3. Change in the free energy at standard conditions of 298 K and one molar /one atmospheric pressure conditions is ∆G°. From the above relation, it can be written that

∆G° = – nFE°red

Where,

  • red is the reduction potential measured at standard conditions.

4. During the reaction, concentration keeps changing, and the potential also will decrease with the rate of reaction.

To get the maximum work or maximum free energy change, the concentrations have to be maintained the same. This is possible only by carrying out the reaction under a reversible equilibrium condition.

For a reversible equilibrium reaction, Van’t Hoff isotherm says:

∆G = ∆G° + RT ln K

Where,

  • K is the equilibrium constant
  • K = Product/Reactant = [M]n/[M]n+
  • R is the Gas constant =8 .314J/K mole
  • T is the temperature on the Kelvin scale.

5. Substituting for free energy changes in Van’t Hoff equation,

– nFEred = – nFE°red + RT ln [M]/[Mn+] = – nFE°red + 2.303 RT log [M]n/[Mn+]

Dividing both sides by – nF,

\(\begin{array}{l}E_{red} = E^{\circ}_{red} -\frac{2.303 RT}{nF1}\log\frac{[M]^n}{[M^{n+}]}\end{array} \)

or,

\(\begin{array}{l}EM^{n+}/M =E^o M^{n+}/M -\frac{2.303 RT}{nF1}\log\frac{[M]^n}{[M^{n+}]}\end{array} \)

The activity of the metal is always considered as equal to unity.

Ered = E°red – or

\(\begin{array}{l}E M^{n+}/{M} =E^o M^{n+}/{M}- \frac{2.303 RT}{nF}\log\frac{1}{[Mn^{n+}]}\end{array} \)

This relation connecting reduction potential measurable at conditions other than standard conditions to the standard electrode potential is the Nernst equation.

For reactions conducted at 298 K but different concentrations, the Nernst equation is:

\(\begin{array}{l}E M^{n+}/{M}=E^o M^{n+}/{M}- \frac{2.303\times 8.314 \times293}{n96500}\log\frac{1}{[Mn^{n+}]}\end{array} \)

\(\begin{array}{l}=E^o M^{n+}/{M}-\frac{0.0591}{n}\log\frac{1}{[Mn^{n+}]}\end{array} \)

Determining Equilibrium Constant with Nernst Equation

When the reactants and the products of the electrochemical cell reach equilibrium, the value of ΔG becomes 0. At this point, the reaction quotient and the equilibrium constant (Kc) are the same. Since ΔG = -nFE, the cell potential at equilibrium is also 0.

Substituting the values of Q and E into the Nernst equation, the following equation is obtained.

0 = E0cell – (RT/nF) ln Kc

The relationship between the Nernst equation, the equilibrium constant, and Gibbs energy change are illustrated below.

Nernst equation provides the relation between the cell potential of an electrochemical cell, the standard cell potential, temperature, and the reaction quotient.

Nernst equation vs Equilibrium constant vs Gibbs energy change

Converting the natural logarithm into base-10 logarithm and substituting T = 298 K (standard temperature), the equation is transformed as follows.

E0cell = (0.0592V/n) log Kc

By rearranging this equation, the following equation can be obtained.

log Kc = (nE0cell)/0.0592V

Thus, the relationship between the standard cell potential and the equilibrium constant is obtained. When Kc is greater than 1, the value of E0cell will be greater than 0, implying that the equilibrium favours the forward reaction. Similarly, when Kc is less than 1, E0cell will hold a negative value which suggests that the reverse reaction will be favoured.

Nernst Equation Applications

The Nernst equation can be used to calculate the following:

  • Single electrode reduction or oxidation potential at any conditions
  • Standard electrode potentials
  • Comparing the relative ability as a reductive or oxidative agent
  • Finding the feasibility of combining such single electrodes to produce an electric potential
  • Emf of an electrochemical cell
  • Unknown ionic concentrations
  • The pH of solutions and solubility of sparingly soluble salts can be measured with the help of the Nernst equation.

Limitations of Nernst Equation

The activity of an ion in a very dilute solution is close to infinity and can, therefore, be expressed in terms of the ion concentration. However, the ion concentration is not equal to the ion activity for solutions having very high concentrations. In order to use the Nernst equation in such cases, experimental measurements must be conducted to obtain the true activity of the ion.

Another shortcoming of this equation is that it cannot be used to measure cell potential when a current flows through the electrode. This is because the current flow affects the ions’ activity on the surface of the electrode. Also, additional factors such as resistive loss and overpotential must be considered when a current flows through the electrode.

Solved Examples on NERNST Equation

1. The standard electrode potential of zinc ions is 0.76V. What will be the potential of a 2M solution at 300 K?

Solution:

The Nernst equation for the given conditions can be written as follows:

EMn+/M = Eo – [(2.303RT)/nF] × log 1/[Mn+]

Here,

  • E° = 0.76V
  • n = 2
  • F = 96500 C/mole
  • [Mn+] = 2 M
  • R =8.314 J/K mole
  • T =300 K

Substituting the given values in the Nernst equation, we get,

EZn2+/Zn = 0.76 – [(2.303×8.314×300)/(2×96500)] × log 1/2 = 0.76 – [0.0298 × (-0.301)]

= 0.76 + 0.009 = 0.769 V

Therefore, the potential of a 2 M solution at 300 K is 0.769 V.

2. From the following standard potentials, arrange the metals in the order of their increasing reducing power.

  • Zn2+(aq) + 2e → Zn(s): E° = -0.76 V
  • Ca2+(aq) + 2e → Ca(s): E° = -2.87 V
  • Mg2+(aq) + 2e → Mg(s): E° = -2.36 V
  • Ni2+(aq) + 2e → Ni(s): E° = -0.25 V
  • Ni(s) → Ni2+(aq) + 2e : E° = +0.25 V

Reducing power of a metal increases with its ability to give up electrons, i.e., lower standard potentials. Arranging the reduction potentials in decreasing order gives the increasing order of reducing the power of metals.

Increasing order of reduction potentials is Ni (-0.25V) < Zn (-0.76V) < Mg(-2.36V) < Ca(-2.87).

3. What is the Cell Potential of the electrochemical cell in which the cell reaction is: Pb2+ + Cd → Pb + Cd2+? Given that Eocell = 0.277 volts, temperature = 25oC, [Cd2+] = 0.02 M, and [Pb2+] = 0.2 M.

Solution

Since the temperature is equal to 25oC, the Nernst equation can be written as follows:

Ecell = E0cell – (0.0592/n) log10Q

Here, two moles of electrons are transferred in the reaction. Therefore, n = 2. The reaction quotient (Q) is given by [Cd2+]/[Pb2+] = (0.02M)/(0.2M) = 0.1.

The equation can now be rewritten as:

Ecell = 0.277 – (0.0592/2) × log10(0.1) = 0.277 – (0.0296)(-1) = 0.3066 Volts

Thus, the cell potential of this electrochemical cell at a temperature of 25oC is 0.3066 volts.

4. The Cu2+ ion concentration in a copper-silver electrochemical cell is 0.1 M. If Eo(Ag+/Ag) = 0.8 V, Eo(Cu2+/Cu) = 0.34V, and Cell potential (at 25oC) = 0.422 V. Find the silver ion concentration.

Solution

Here, the silver electrode acts as a cathode, whereas the copper electrode serves as the anode. This is because the standard electrode potential of the silver electrode is greater than that of the copper electrode. The standard electrode potential of the cell can now be calculated, as shown below:

Eocell = Eocathode – Eoanode = 0.8V – 0.34V = 0.46V

Since the charge on the copper ion is +2 and the charge on the silver ion is +1, the balanced cell reaction is:

2Ag+ + Cu → 2Ag + Cu2+

Since two electrons are transferred in the cell reaction, n = 2, the Nernst equation for this electrochemical cell can be written as follows:

Ecell = E0cell – (0.0592/2) × log(0.1/[Ag+]2)

0.422V = 0.46 – 0.0296 × (-1 – 2log[Ag+])

Therefore, -2log[Ag+] = 1.283 + 1 = 2.283

Or, log[Ag+] = -1.141

[Ag+] = antilog(-1.141) = 0.0722 M

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.

CBSE Class 12 Chemistry Syllabus Download PDF

Below is the CBSE Class 12 Syllabus along with the marking scheme and time duration of the Chemistry exam.

S.NoTitleNo. of PeriodsMarks
1Solutions107
2Electrochemistry129
3Chemical Kinetics107
4d -and f -Block Elements127
5Coordination Compounds127
6Haloalkanes and Haloarenes106
7Alcohols, Phenols and Ethers106
8Aldehydes, Ketones and Carboxylic Acids108
9Amines106
10Biomolecules127
Total70

CBSE Class 12 Chemistry Practical Syllabus along with Marking Scheme

The following is a breakdown of the marks for practical, project work, class records, and viva. The total number of marks for all parts is 15. The marks for both terms are provided in the table below.

Evaluation Scheme for ExaminationMarks
Volumetric Analysis08
Salt Analysis08
Content-Based Experiment06
Project Work and Viva04
Class record and Viva04
Total30

CBSE Class 12 Chemistry Syllabus (Chapter-wise)

Unit -1: Solutions

  • Raoult's law.
  • Colligative properties - relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, osmotic pressure, determination of molecular masses using colligative properties, abnormal molecular mass.
  • Solutions, Types of solutions, expression of concentration of solutions of solids in liquids, solubility of gases in liquids, solid solutions.
  • Van't Hoff factor.

Unit -2: Electrochemistry

  • Redox reactions, EMF of a cell, standard electrode potential
  • Nernst equation and its application to chemical cells
  • Relation between Gibbs energy change and EMF of a cell
  • Kohlrausch's Law
  • Electrolysis and law of electrolysis (elementary idea)
  • Dry cell-electrolytic cells and Galvanic cells
  • Conductance in electrolytic solutions, specific and molar conductivity, variations of conductivity with concentration.
  • Lead accumulator
  • Fuel cells

Unit -3: Chemical Kinetics

  • Rate of a reaction (Average and instantaneous)
  • Rate law and specific rate constant
  • Integrated rate equations and half-life (only for zerfirst-order order reactions)
  • Concept of collision theory (elementary idea, no mathematical treatment)
  • Factors affecting rate of reaction: concentration, temperature, catalyst;
  • Order and molecularity of a reaction
  • Activation energy
  • Arrhenius equation

Unit -4: d and f Block Elements  

  • Lanthanoids- Electronic configuration, oxidation states, chemical reactivity and lanthanoid contraction and its consequences.
  • Actinoids- Electronic configuration, oxidation states and comparison with lanthanoids.
  • General introduction, electronic configuration, occurrence and characteristics of transition metals, general trends in properties of the first-row transition metals – metallic character, ionization enthalpy, oxidation states, ionic radii, color, catalytic property, magnetic properties, interstitial compounds, alloy formation, preparation and properties of K2Cr2O7 and KMnO4.

Unit -5: Coordination Compounds  

  • Coordination compounds - Introduction, ligands, coordination number, color, magnetic properties and shapes
  • The importance of coordination compounds (in qualitative analysis, extraction of metals and biological system).
  • IUPAC nomenclature of mononuclear coordination compounds.
  • Bonding
  • Werner's theory, VBT, and CFT; structure and stereoisomerism

Unit -6: Haloalkanes and Haloarenes  

  • Haloarenes: Nature of C–X bond, substitution reactions (Directive influence of halogen in monosubstituted compounds only). Uses and environmental effects of - dichloromethane, trichloro methane, tetrachloromethane, iodoform, freons, DDT.
  • Haloalkanes: Nomenclature, nature of C–X bond, physical and chemical properties, optical rotation mechanism of substitution reactions.

Unit -7: Alcohols, Phenols and Ethers   

  • Phenols: Nomenclature, methods of preparation, physical and chemical properties, acidic nature of phenol, electrophilic substitution reactions, uses of phenols.
  • Ethers: Nomenclature, methods of preparation, physical and chemical properties, uses.
  • Alcohols: Nomenclature, methods of preparation, physical and chemical properties (of primary alcohols only), identification of primary, secondary and tertiary alcohols, mechanism of dehydration, and uses with special reference to methanol and ethanol.

Unit -8: Aldehydes, Ketones and Carboxylic Acids   

  • Carboxylic Acids: Nomenclature, acidic nature, methods of preparation, physical and chemical properties; uses.
  • Aldehydes and Ketones: Nomenclature, nature of carbonyl group, methods of preparation, physical and chemical properties, mechanism of nucleophilic addition, the reactivity of alpha hydrogen in aldehydes, uses.

Unit -9: Amines    

  • Diazonium salts: Preparation, chemical reactions and importance in synthetic organic chemistry.
  • Amines: Nomenclature, classification, structure, methods of preparation, physical and chemical properties, uses, and identification of primary, secondary and tertiary amines.

Unit -10: Biomolecules     

  • Proteins -Elementary idea of - amino acids, peptide bond, polypeptides, proteins, structure of proteins - primary, secondary, tertiary structure and quaternary structures (qualitative idea only), denaturation of proteins; enzymes. Hormones - Elementary idea excluding structure.
  • Vitamins - Classification and functions.
  • Carbohydrates - Classification (aldoses and ketoses), monosaccharides (glucose and fructose), D-L configuration oligosaccharides (sucrose, lactose, maltose), polysaccharides (starch, cellulose, glycogen); Importance of carbohydrates.
  • Nucleic Acids: DNA and RNA.

The syllabus is divided into three parts: Part A, Part B, and Part C. Part A consist of Basic Concepts of Chemistry, which covers topics such as atomic structure, chemical bonding, states of matter, and thermochemistry. Part B consists of Topics in Physical Chemistry, which includes topics such as chemical kinetics, equilibrium, and electrochemistry. Part C consists of Topics in Organic Chemistry, which covers topics such as alkanes, alkenes, alkynes, and aromatic compounds.

Basic Concepts of Chemistry:

  • Atomic structure: This section covers the fundamental concepts of atomic structure, including the electronic configuration of atoms, the Bohr model of the atom, and the wave nature of matter.
  • Chemical bonding: This section covers the different types of chemical bonds, including ionic, covalent, and metallic bonds, as well as the concept of hybridization.
  • States of the matter: This section covers the three states of matter - solid, liquid, and gas - and the factors that influence their properties.
  • Thermochemistry: This section covers the principles of thermochemistry, including the laws of thermodynamics and the concept of enthalpy.

Chapters in Physical Chemistry:

  • Chemical kinetics: This section covers the study of the rate of chemical reactions and the factors that influence it, including the concentration of reactants, temperature, and the presence of catalysts.
  • Equilibrium: This section covers the principles of chemical equilibrium, including the concept of Le Chatelier's principle and the equilibrium constant.
  • Electrochemistry: This section covers the principles of electrochemistry, including the concept of half-cell reactions, galvanic cells, and electrolysis.

Chapters in Organic Chemistry:

  • Alkanes: This section covers the properties and reactions of alkanes, including their structure, isomerism, and combustion.
  • Alkenes: This section covers the properties and reactions of alkenes, including their structure, isomerism, and addition reactions.
  • Alkynes: This section covers the properties and reactions of alkynes, including their structure, isomerism, and addition reactions.
  • Aromatic compounds: This section covers the properties and reactions of aromatic compounds, including their structure, isomerism, and electrophilic substitution reactions.

In addition to the topics covered in the syllabus, the CBSE Class 12 Chemistry exam also tests students on their analytical and problem-solving skills, as well as their ability to apply the concepts learned in the classroom to real-world situations.

Students can also check out the Tips for the Class 12 Chemistry Exam. They can easily access the Class 12 study material in one place by visiting the CBSE Class 12 page at ANAND CLASSES (A School Of Competitions). Moreover, to get interactive lessons and study videos, download the ANAND CLASSES (A School Of Competitions) App.

Frequently Asked Questions on CBSE Class 12 Chemistry Syllabus

Q1

How many chapters are there in the CBSE Class 12 Chemistry as per the syllabus?

There are 10 chapters in the CBSE Class 12 Chemistry as per Syllabus. Students can learn all these chapters efficiently using the study materials provided at ANAND CLASSES (A School Of Competitions).

Q2

What is the marking scheme for CBSE Class 12 Chemistry practical exam according to the syllabus?

The marking scheme for CBSE Class 12 Chemistry practical exam, according to the syllabus, is 8 marks for volumetric analysis, 8 marks for salt analysis, 6 marks for the content-based experiment, 4 marks for the project and viva and 4 marks for class record and viva.

Q3

Which is the scoring chapter in Chemistry as per CBSE Class 12 syllabus?

The chapter Electrochemistry in Chemistry is the scoring chapter as per CBSE Class 12 syllabus.