Electrochemistry (Numericals) Solved Examples (Class 12 Chemistry)
Electrochemistry connects chemistry and electricity through oxidation-reduction reactions and electrochemical cells. These examples cover electrodes, stand
TL;DR: Electrochemistry connects chemistry and electricity through oxidation-reduction reactions and electrochemical cells. These examples cover electrodes,…
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Electrochemistry connects chemistry and electricity through oxidation-reduction reactions and electrochemical cells. These examples cover electrodes, stand
Electrochemistry (Numericals) — Solved Numerical Examples (Step by Step)
Example 1: Calculate the standard cell potential for the reaction: Zn + Cu²⁺ → Zn²⁺ + Cu. (E°Zn²⁺/Zn = -0.76 V, E°Cu²⁺/Cu = +0.34 V)
Solution: In the cell:
Zn is oxidized to Zn²⁺ (anode): Zn → Zn²⁺ + 2e⁻
Cu²⁺ is reduced to Cu (cathode): Cu²⁺ + 2e⁻ → Cu
Standard cell potential: E°cell = E°cathode - E°anode
E°cell = E°Cu²⁺/Cu - E°Zn²⁺/Zn
E°cell = 0.34 - (-0.76) = 0.34 + 0.76 = 1.10 V
Example 2: How much charge (in coulombs) is required to deposit 3.2 g of copper from a copper sulfate solution? (Atomic mass of Cu = 64, Faraday = 96500 C/mol)
Solution: Copper deposition: Cu²⁺ + 2e⁻ → Cu
Molar mass of Cu = 64 g/mol
Number of moles of Cu = 3.2 / 64 = 0.05 mol
Since each Cu atom requires 2 electrons:
Number of moles of electrons = 0.05 × 2 = 0.1 mol
Charge = moles of electrons × Faraday
Charge = 0.1 × 96500 = 9650 C
Example 3: An electric current of 2 A is passed through a copper sulfate solution for 30 minutes. Calculate the mass of copper deposited. (Atomic mass of Cu = 64, Faraday = 96500 C/mol)
Solution: Current I = 2 A, time t = 30 min = 1800 s
Total charge: Q = I × t = 2 × 1800 = 3600 C
Number of moles of electrons = Q / Faraday = 3600 / 96500 = 0.0373 mol
Cu²⁺ + 2e⁻ → Cu
Moles of Cu deposited = 0.0373 / 2 = 0.01865 mol
Mass of Cu = 0.01865 × 64 = 1.194 g ≈ 1.2 g
Example 4: Calculate the EMF of a galvanic cell using Nernst equation at 25°C: Zn | Zn²⁺(0.1 M) || Cu²⁺(0.01 M) | Cu. (E°cell = 1.10 V, R = 8.314 J/mol K, T = 298 K, F = 96500 C/mol)
Solution: Using Nernst equation: E_cell = E°cell - (RT/nF) × ln(Q)
For this cell, n = 2 (electrons transferred)
Q = [Zn²⁺] / [Cu²⁺] = 0.1 / 0.01 = 10
E_cell = 1.10 - (8.314 × 298 / (2 × 96500)) × ln(10)
E_cell = 1.10 - (2477.572 / 193000) × 2.303
E_cell = 1.10 - 0.01283 × 2.303
E_cell = 1.10 - 0.0295 = 1.07 V
Example 5: In the electrolysis of water, 100 mL of oxygen gas is produced at 1 atm and 25°C. Calculate the charge and current if this takes 1000 seconds. (F = 96500 C/mol, R = 0.082 L atm / mol K)
Solution: For water electrolysis: 2H2O → 2H2 + O2
Moles of O2 = PV / RT = (1 × 0.1) / (0.082 × 298) = 0.1 / 24.436 = 0.00409 mol
From reaction: 1 mol O2 requires 4 moles of electrons
Moles of electrons = 0.00409 × 4 = 0.01636 mol
Charge = moles of electrons × Faraday
Q = 0.01636 × 96500 = 1578 C
Current = Q / t = 1578 / 1000 = 1.578 A ≈ 1.58 A
Example 6: Calculate the number of Faradays of electricity required to produce 27 g of aluminum by electrolysis of molten Al2O3. (Atomic mass of Al = 27)
Solution: Reduction of Al³⁺: Al³⁺ + 3e⁻ → Al
Molar mass of Al = 27 g/mol
Moles of Al = 27 / 27 = 1 mol
Each mole of Al requires 3 moles of electrons
Moles of electrons = 1 × 3 = 3 mol
Number of Faradays = 3 (since 1 Faraday = 1 mole of electrons)
Tips
- Always identify the oxidation and reduction half-reactions to determine the number of electrons transferred.
- Cell potential is always E°cathode - E°anode, where cathode is reduction and anode is oxidation.
- Use Faraday's laws for electrolysis calculations: moles of substance = (charge × n) / (Faraday number), where n is electrons transferred.
- The Nernst equation shows that cell EMF decreases as the reaction proceeds (Q increases), which is why galvanic cells lose voltage over time.
Frequently Asked Questions
What is the difference between a galvanic cell and an electrolytic cell?
A galvanic cell (or voltaic cell) spontaneously generates electricity through a redox reaction. An electrolytic cell uses external electrical energy to drive a non-spontaneous redox reaction. In a galvanic cell, the anode is negative and cathode is positive; in an electrolytic cell, it is reversed.
Why does Faraday's constant equal 96500 C/mol?
Faraday's constant is the charge of one mole of electrons. Since one electron has charge 1.602 × 10⁻¹⁹ coulombs, and one mole contains 6.022 × 10²³ particles, F = 1.602 × 10⁻¹⁹ × 6.022 × 10²³ ≈ 96500 C/mol.
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