Electrochemistry for NEET

Electrochemistry joins three chapters you have already studied. The half-reactions come from redox reactions, ΔG and K come from thermodynamics and equilibrium, and ions in solution come from ionic equilibrium. The new material is how to measure a reaction's driving force as a voltage, and how ions carry current.

NEET questions here are mostly short calculations: E°cell from two electrode potentials, E from the Nernst equation, ΔG° from E°, molar conductivity from conductivity, or Λ° of a weak electrolyte from Kohlrausch's law. Add a few theory questions on the electrochemical series and batteries. Each calculation is one formula long, so the marks go to whoever gets the signs and units right.

Galvanic and electrolytic cells

A galvanic (voltaic) cell turns the free energy of a spontaneous redox reaction into electrical energy. The Daniell cell is the standard example:

Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)

• Anode on the left, cathode on the right; the double line is the salt bridge.
• Zinc is oxidised at the anode: Zn → Zn²⁺ + 2e⁻.
• Copper ions are reduced at the cathode: Cu²⁺ + 2e⁻ → Cu.
• The salt bridge (for example KCl or KNO₃ in agar) completes the circuit and keeps both solutions electrically neutral.

Electrode potentials and the electrochemical series

A single electrode potential cannot be measured alone, so everything is measured against the standard hydrogen electrode (SHE): platinum in 1 M H⁺ with H₂ gas at 1 bar. Its potential is taken as 0 V.

Standard electrode potentials are listed as reduction potentials. The more positive E°, the stronger the oxidising agent on the left; the more negative E°, the stronger the reducing agent on the right.

Cell potential: E°cell = E°cathode − E°anode, using the reduction potentials of both electrodes. A positive E°cell means the reaction as written is spontaneous.

The Nernst equation and ΔG

For a cell reaction with n electrons transferred and reaction quotient Q, at 298 K:

E = E° − (0.059/n) log Q

For a single electrode Mⁿ⁺ + ne⁻ → M, this becomes E = E° − (0.059/n) log(1/[Mⁿ⁺]). For the hydrogen electrode at 1 bar, E = −0.059 × pH.

Link to Gibbs energy and equilibrium:

• ΔG = −nFE and ΔG° = −nFE°, with F = 96,500 C mol⁻¹.
• At equilibrium, E = 0 and Q = K, so E° = (0.059/n) log K.
• ΔG° = −2.303RT log K, the same relation met in chemical thermodynamics.

Conductance in electrolytic solutions

Metals conduct by electrons (metallic conduction). Electrolytes conduct by moving ions (electrolytic conduction), and their conductance rises with temperature, while a metal's falls.

Variation with dilution:

• Conductivity (κ) decreases on dilution, because there are fewer ions per cm³.
• Molar conductivity (Λm) increases on dilution, because each mole of electrolyte has more room and, for weak electrolytes, ionises more.
• For strong electrolytes, Λm = Λ°m − A√C. The plot of Λm against √C is a straight line that can be extrapolated to zero concentration to get Λ°m.
• For weak electrolytes, Λm rises steeply at high dilution, so Λ°m cannot be found by extrapolation.

Kohlrausch's law: at infinite dilution, each ion contributes independently, so Λ°m = ν₊λ°₊ + ν₋λ°₋. For example, Λ°(CaCl₂) = λ°(Ca²⁺) + 2λ°(Cl⁻).

Applications:

• Find Λ° of a weak electrolyte from strong ones: Λ°(CH₃COOH) = Λ°(HCl) + Λ°(CH₃COONa) − Λ°(NaCl).
• Degree of dissociation: α = Λm/Λ°m.
• Dissociation constant of a weak acid: Ka = Cα²/(1 − α).

Batteries and fuel cells

Lead storage battery, discharge:

• Anode: Pb + SO₄²⁻ → PbSO₄ + 2e⁻
• Cathode: PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O
• Overall: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O. Sulphuric acid is used up, so its density falls as the battery discharges.

Fuel cell: anode 2H₂ + 4OH⁻ → 4H₂O + 4e⁻; cathode O₂ + 2H₂O + 4e⁻ → 4OH⁻; overall 2H₂ + O₂ → 2H₂O. Fuel cells convert fuel energy to electricity directly, with higher efficiency than burning the fuel in a heat engine.

Electrolysis and Faraday's laws

In electrolysis, the product at each electrode is the species that needs the least energy to discharge. Electrolysis of aqueous NaCl gives H₂ at the cathode, not Na, because water is reduced more easily than Na⁺.

Faraday's first law says the mass deposited is proportional to the charge passed. One faraday (96,500 C) is the charge on one mole of electrons, so:

Moles deposited = Q/(nF), and mass = (M × I × t)/(n × F)

Worked NEET-style questions

Example 1: E°cell and ΔG°

For the Daniell cell, E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V. Find E°cell and ΔG°.

• E°cell = 0.34 − (−0.76) = 1.10 V.
• n = 2, so ΔG° = −2 × 96,500 × 1.10 = −2.123 × 10⁵ J ≈ −212.3 kJ mol⁻¹.

Example 2: the Nernst equation

Find E for Zn | Zn²⁺(0.1 M) || Cu²⁺(0.01 M) | Cu at 298 K.

• Cell reaction: Zn + Cu²⁺ → Zn²⁺ + Cu, so Q = [Zn²⁺]/[Cu²⁺] = 0.1/0.01 = 10.
• E = 1.10 − (0.059/2) log 10 = 1.10 − 0.0295 ≈ 1.07 V.

Example 3: equilibrium constant from E°

Find K for the Daniell cell reaction at 298 K.

• log K = nE°/0.059 = (2 × 1.10)/0.059 ≈ 37.3.
• K ≈ 2 × 10³⁷. The reaction goes essentially to completion.

Example 4: Kohlrausch's law and Ka

A 0.001 M solution of acetic acid has Λm = 39.05 S cm² mol⁻¹. Λ°(HCl) = 425.9, Λ°(CH₃COONa) = 91.0 and Λ°(NaCl) = 126.4 S cm² mol⁻¹. Find α and Ka.

• Λ°(CH₃COOH) = 425.9 + 91.0 − 126.4 = 390.5 S cm² mol⁻¹.
• α = 39.05/390.5 = 0.1.
• Ka = Cα²/(1 − α) = (0.001 × 0.01)/0.9 ≈ 1.1 × 10⁻⁵.

Practice MCQs

1. E°(Ag⁺/Ag) = +0.80 V and E°(Cu²⁺/Cu) = +0.34 V. E°cell for the spontaneous cell is: (a) 1.14 V (b) 0.46 V (c) −0.46 V (d) 1.26 V
2. A cell with cell constant 0.5 cm⁻¹ has a resistance of 50 Ω. The conductivity is: (a) 0.01 S cm⁻¹ (b) 25 S cm⁻¹ (c) 100 S cm⁻¹ (d) 0.1 S cm⁻¹
3. κ = 0.01 S cm⁻¹ for a 0.1 M solution. Λm is: (a) 0.1 (b) 10 (c) 100 (d) 1000 S cm² mol⁻¹
4. On dilution, for a strong electrolyte: (a) κ and Λm both increase (b) κ decreases and Λm increases (c) κ increases and Λm decreases (d) both decrease
5. In a lead storage battery during discharge, the anode is: (a) PbO₂ (b) PbSO₄ (c) Pb (d) H₂SO₄
6. For a reaction with n = 1 and E° = 0.5 V, ΔG° is about: (a) −48.25 kJ (b) +48.25 kJ (c) −96.5 kJ (d) −24.1 kJ
7. The potential of a hydrogen electrode (1 bar H₂) in a solution of pH 2 at 298 K is: (a) 0 V (b) +0.118 V (c) −0.118 V (d) −0.059 V
8. Λ°m cannot be found by extrapolating Λm to zero concentration for: (a) KCl (b) NaCl (c) HCl (d) CH₃COOH

Answers

1. (b) Silver is the cathode: 0.80 − 0.34 = 0.46 V.
2. (a) κ = cell constant/R = 0.5/50 = 0.01 S cm⁻¹.
3. (c) Λm = 0.01 × 1000/0.1 = 100 S cm² mol⁻¹.
4. (b) Fewer ions per cm³, but each mole conducts better.
5. (c) Lead is oxidised to PbSO₄.
6. (a) ΔG° = −1 × 96,500 × 0.5 = −48,250 J.
7. (c) E = −0.059 × 2 = −0.118 V.
8. (d) Acetic acid is a weak electrolyte.

What to do next

• Make one card with E°cell, Nernst, ΔG° = −nFE°, E° = (0.059/n) log K and Λm = κ × 1000/C, with units.
• Solve five Nernst questions where you write the cell reaction and Q before substituting.
• Draw the Λm against √C graph for a strong and a weak electrolyte, and practise three Kohlrausch questions.
• Revise the reaction quotient in chemical equilibrium, then go on to chemical kinetics.