AP | A-Level | IB | NCERT 11 + 12 – FREE NOTES, RESOURCES AND VIDEOS!
1 Solutions 2 Electrochemistry 3 Chemical Kinetics 4 The d-and f-Block Elements 5 Coordination Compounds 6 Haloalkanes and Haloarenes 7 Alcohols, Phenols and Ethers 8 Aldehydes, Ketones and Carboxylic Acids 9 Amines 10 Biomolecules

1 Solutions

1.1 Types of Solutions 1.2 Expressing Concentration of Solutions 1.3 Solubility 1.4 Vapour Pressure of Liquid Solutions 1.5 Ideal and Non-ideal Solutions 1.6 Colligative Properties and Determination of Molar Mass 1.7 Abnormal Molar Masses 2.1 Electrochemical Cells 2.2 Galvanic Cells 2.3 Nernst Equation 2.4 Conductance of Electrolytic Solutions 2.5 Electrolytic Cells and Electrolysis 2.6 Batteries 2.7 Fuel Cells 2.8 Corrosion

Nernst Equation

NCERT Reference: Chapter 2 – Electrochemistry – Page 39–42 (Part I)

Quick Notes

  • Nernst Equation: NCERT Class 12 Chemistry diagram showing the Nernst equation form and terms for calculating non-standard electrode or cell potential.
  • E° is standard cell potential and n is the number of electrons exchanged.
  • Q is the reaction quotient (ratio of product to reactant concentrations).
  • At equilibrium, E = 0 and E° = (0.0591/n) log K.
  • Gibbs energy link: ΔG = −nFE and ΔG° = −nFE°
  • Sign of E° helps determine reaction spontaneity.

Full Notes

Introduction to the Nernst Equation

In real-world conditions, concentrations of ions and gases often vary from their standard values. The Nernst equation provides a tool to calculate the actual electrode potential or cell potential under such non-standard conditions.

Nernst Equation (Qualitative Form)

NCERT Class 12 Chemistry diagram showing the Nernst equation form and terms for calculating non-standard electrode or cell potential.

The key idea is:

Recap – What Q Tells Us About the Reaction

What is the Reaction Quotient (Q)?
The reaction quotient (Q) is a snapshot of a reaction’s progress. It is calculated by using concentration values at a specific point in time, which might not be equilibrium values.

NCERT Class 12 Chemistry general reaction form aA + bB ⇌ cC + dD for defining Q and K. NCERT Class 12 Chemistry expression for the reaction quotient Q in terms of concentrations of products and reactants.

This explains why a cell generates electricity: it's operating to get to equilibrium, and the further it is from equilibrium, the greater the “push” (E) driving electrons through the circuit.

Equilibrium Constant from Nernst Equation

At equilibrium, the cell stops working, i.e., E = 0. This condition helps derive a link between cell potential and equilibrium constant (K) of the redox reaction.

NCERT Class 12 Chemistry derivation linking Nernst equation to equilibrium constant K at 298 K.

(note that this form comes from Converting natural log (ln) to log base 10:
ln Kc = 2.303 × log Kc
Substitute this into the equation:
E = E° − (2.303 × RT / nF) × log Kc

This expression allows us to calculate the equilibrium constant of a redox reaction from its standard electrode potential.

Interpretation:

Electrochemical Cell and Gibbs Energy of the Reaction

The Gibbs free energy change (ΔG) helps us determine whether a reaction can occur without external energy input. This section shows how cell potential relates directly to Gibbs energy.

The energy available to drive an electric current (E°cell) comes directly from the energy change in the chemical reaction (ΔG°).

These two quantities are linked by the equation:

NCERT Class 12 Chemistry relation ΔG = −nFE and ΔG° = −nFE° that links Gibbs free energy change to cell potential.

This allows us to connect the electrical world of electrochemical cells with the thermodynamic world of Gibbs energy and predict spontaneity from either.

How to Use the Equation

Interpretation:

Thus, positive cell potential indicates a thermodynamically favorable process.

We can also use Gibbs free energy change to determine an equilibrium constant using the equation:

NCERT Class 12 Chemistry equation linking ΔG° and equilibrium constant K to predict reaction position.
Worked Example

Zn + Cu2+ → Zn2+ + Cu
From data booklet:
Zn2+/Zn = −0.76 V
Cu2+/Cu = +0.34 V
F = 96,500 C mol⁻¹

  1. Calculate E°cell
    cell = 0.34 − (−0.76) = +1.10 V
  2. Electrons transferred
    n = 2 electrons (Zn loses 2 e⁻; Cu2+ gains 2 e⁻).
  3. Calculate ΔG°
    ΔG° = −nFE°cell = −(2)(96,500)(1.10) = −212,300 J mol⁻¹ = −212.3 kJ mol⁻¹

Summary