Search This Blog

Exhaustive Guide: Nernst Equation & Gibbs Energy | Class 12 Chemistry

Exhaustive Guide: Nernst Equation & Gibbs Energy | Class 12 Chemistry | ChemCA

Nernst Equation & Gibbs Energy

Module 2 | CBSE Class 12 Chemistry | Electrochemistry Chapter

1. The Need for the Nernst Equation

In the previous module, we calculated the Standard Cell Potential (E°cell). However, this value is only valid under standard conditions: when the concentration of all species is exactly 1 M, pressure is 1 bar, and temperature is 298 K.

In real-world applications (like batteries discharging), the concentrations of ions constantly change. Walther Nernst provided an equation to calculate the electrode potential and cell potential at any given concentration and temperature.

2. Nernst Equation for a Single Electrode

Consider a general reduction reaction at an electrode:

Mn+(aq) + ne- → M(s)

The Nernst equation for this half-cell is written as:

E(Mn+/M) = E°(Mn+/M) - (RT / nF) × ln [M(s)] / [Mn+(aq)]

Where:
E = Electrode potential at given concentration
E° = Standard electrode potential
R = Universal gas constant (8.314 J K-1 mol-1)
T = Temperature in Kelvin
n = Number of electrons participating in the reaction
F = Faraday's constant (96487 C mol-1 ≈ 96500 C mol-1)

Crucial Rule: By convention, the concentration of solid metals [M(s)] and pure liquids is taken as unity (1).

If we substitute the values of R, F, and standard temperature T = 298 K (25°C), and convert the natural logarithm (ln) to base-10 logarithm (log) by multiplying by 2.303, the equation simplifies to:

E = E° - (0.0591 / n) × log ( 1 / [Mn+] )

3. Nernst Equation for an Electrochemical Cell

For a general electrochemical reaction:

aA + bB → cC + dD

The Nernst equation for the complete cell potential (Ecell) at 298 K is given by:

Ecell = E°cell - (0.0591 / n) × log Q

Ecell = E°cell - (0.0591 / n) × log ( [C]c[D]d / [A]a[B]b )

Here, Q is the reaction quotient. Remember to omit pure solids and liquids from this expression.

Application to Daniell Cell:
Reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Here, n = 2 electrons.
Ecell = E°cell - (0.0591 / 2) × log ( [Zn2+] / [Cu2+] )

Observation: Ecell increases if the concentration of Cu2+ increases or the concentration of Zn2+ decreases.

4. Equilibrium Constant from Nernst Equation

As a galvanic cell operates, the reactants are consumed and products are formed. The concentration of the anode electrolyte increases, and the cathode electrolyte decreases. Consequently, the cell potential (Ecell) keeps decreasing.

Eventually, the concentrations stop changing, and the voltmeter reads zero. At this point, the reaction has reached equilibrium.

At equilibrium, Ecell = 0 and the reaction quotient Q = Kc (Equilibrium Constant).

Substituting these into the Nernst equation at 298 K:

0 = E°cell - (0.0591 / n) × log Kc

E°cell = (0.0591 / n) × log Kc

log Kc = (n × E°cell) / 0.0591

This is a highly useful equation because equilibrium constants for heavily product-favored redox reactions are often extremely large and difficult to measure chemically, but can be easily calculated by measuring E°cell.

5. Electrochemical Cell and Gibbs Energy of the Reaction

Electrical work done in one second is equal to electrical potential multiplied by total charge passed. To obtain maximum work from a galvanic cell, charge has to be passed reversibly.

The reversible work done by a galvanic cell is equal to the decrease in its Gibbs energy. If the EMF of the cell is E and nF is the amount of charge passed, then:

ΔG = -nFEcell

If the concentration of all the reacting species is unity, then Ecell = E°cell and we calculate the Standard Gibbs Energy:

ΔG° = -nFE°cell

Spontaneity Check: For a cell reaction to be spontaneous, ΔG must be negative. Since n and F are always positive, this means Ecell must be positive for a spontaneous reaction.

Since we also know ΔG° = -RT ln Kc, we can connect standard cell potential directly to the equilibrium constant:

ΔG° = -2.303 RT log Kc

6. NCERT Solved Examples (Step-by-Step)

NCERT Example 2.1: Represent the cell in which the following reaction takes place:
Mg(s) + 2Ag+(0.0001 M) → Mg2+(0.130 M) + 2Ag(s)
Calculate its Ecell if E°cell = 3.17 V.

Solution:
Step 1: Cell Representation
Mg undergoes oxidation (anode), Ag+ undergoes reduction (cathode).
Mg(s) | Mg2+(0.130 M) || Ag+(0.0001 M) | Ag(s)

Step 2: Apply Nernst Equation
n = 2 (Mg loses 2e-, 2Ag+ gain 2e-).
Ecell = E°cell - (0.0591 / 2) × log ( [Mg2+] / [Ag+]2 )
(Notice the square on Ag+ because its stoichiometric coefficient is 2).

Ecell = 3.17 - 0.0295 × log [ 0.130 / (0.0001)2 ]
Ecell = 3.17 - 0.0295 × log [ 0.130 / 10-8 ]
Ecell = 3.17 - 0.0295 × log [ 1.3 × 107 ]
Ecell = 3.17 - 0.0295 × (7.11)
Ecell = 3.17 - 0.21 = 2.96 V

NCERT Example 2.2: Calculate the equilibrium constant of the reaction:
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
E°cell = 0.46 V

Solution:
Here, n = 2 electrons are exchanged.
Using the relation: log Kc = (n × E°cell) / 0.0591
log Kc = (2 × 0.46) / 0.0591 = 0.92 / 0.0591 = 15.6

Taking antilog on both sides:
Kc = Antilog (15.6) = 3.92 × 1015

NCERT Example 2.3: The standard electrode potential for Daniell cell is 1.1 V. Calculate the standard Gibbs energy for the reaction:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

Solution:
n = 2
F = 96487 C mol-1
E°cell = 1.1 V

ΔG° = -nFE°cell
ΔG° = - 2 × 96487 C mol-1 × 1.1 V
ΔG° = - 212271 J mol-1 = - 212.27 kJ mol-1
(Note: The negative sign indicates the reaction is highly spontaneous).

7. Previous Year Questions (PYQs) & Exhaustive Question Bank

Part A: Conceptual & Assertion-Reason (1 Mark)

[CBSE 2017, 2020]

Q1. What happens to the EMF of a cell when the system reaches equilibrium?

Answer: When the electrochemical cell reaches equilibrium, the rate of the forward reaction equals the rate of the backward reaction. The concentration of ions in both half-cells becomes constant, and the EMF of the cell (Ecell) drops exactly to zero.
[CBSE Sample Paper 2024]

Q2. Assertion (A): Ecell should have a positive value for the cell to function.
Reason (R): ΔG° = -nFE°cell. For a spontaneous reaction, ΔG° must be negative.

Answer: Both Assertion and Reason are correct, and Reason is the correct explanation for Assertion. Since n and F are always positive constants, the only way for the change in Gibbs free energy (ΔG) to be negative (which is the universal criteria for spontaneity) is if the cell potential (Ecell) is positive.

Part B: Application Based Problems (3 Marks)

[CBSE 2018, 2019]

Q3. Calculate the EMF of the following cell at 298 K:
Fe(s) | Fe2+(0.001 M) || H+(1 M) | H2(g) (1 bar) | Pt(s)
(Given: E°Fe2+/Fe = -0.44 V, E°H+/H2 = 0.00 V)

Answer:
Step 1: Calculate E°cell
E°cell = E°cathode - E°anode
E°cell = 0.00 - (-0.44) = +0.44 V

Step 2: Identify 'n' and reaction
Anode: Fe → Fe2+ + 2e-
Cathode: 2H+ + 2e- → H2
n = 2 electrons.

Step 3: Apply Nernst Equation
Ecell = E°cell - (0.0591 / 2) × log ( [Fe2+] × p(H2) / [H+]2 )
Ecell = 0.44 - 0.0295 × log ( 0.001 × 1 / 12 )
Ecell = 0.44 - 0.0295 × log (10-3)
Ecell = 0.44 - 0.0295 × (-3)
Ecell = 0.44 + 0.0885 = 0.5285 V
[CBSE 2016]

Q4. Calculate ΔG° and log Kc for the following reaction at 298 K:
2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd(s)
Given: E°Cr3+/Cr = -0.74 V; E°Cd2+/Cd = -0.40 V. (F = 96500 C mol-1)

Answer:
Step 1: Calculate E°cell and determine 'n'
E°cell = E°cathode (Cd) - E°anode (Cr)
E°cell = -0.40 - (-0.74) = +0.34 V
Total electrons exchanged (n) = 2 × 3 = 6.

Step 2: Calculate Standard Gibbs Energy (ΔG°)
ΔG° = -nFE°cell
ΔG° = -6 × 96500 × 0.34 = -196860 J mol-1 (or -196.86 kJ mol-1).

Step 3: Calculate log Kc
log Kc = (n × E°cell) / 0.0591
log Kc = (6 × 0.34) / 0.0591 = 2.04 / 0.0591 = 34.51

© 2026 ChemCA.in - The Premier Digital Resource for Chemistry.
Explore more video solutions and strategies at ChemCA Hindi.
Coming up in Module 3: Conductance of Electrolytic Solutions (Kohlrausch's Law).

Powered by

📚 Also Read

Lecture Notes

No comments:

Post a Comment

Subscribe to: Post Comments (Atom)

Featured Post

H₂O as a Ligand: Weak vs Strong Field Cases