CUET UG Chemistry Physical Chemistry Electrochemistry EMF Nernst Equation Electrolytic Cells

Must-Know (15–20 detailed bullets)

• The electromotive force (EMF) of a cell is the potential difference between two electrodes when no current flows; unit is volt (V). Example: Zn-Cu cell has EMF ≈ 1.10 V at standard conditions.
• Standard electrode potential (E°) is measured at 298 K, 1 M concentration, and 1 atm pressure for gases. Example: E°(Cu²⁺/Cu) = +0.34 V, E°(Zn²⁺/Zn) = –0.76 V.
• The cell potential is calculated as E°cell = E°cathode – E°anode. For Daniel cell: E°cell = 0.34 – (–0.76) = 1.10 V.
• The Nernst equation relates electrode potential to concentration: E = E° – (RT/nF) lnQ, where Q is reaction quotient.
• At 298 K, the Nernst equation simplifies to E = E° – (0.059/n) logQ. Example: For Cu²⁺ + 2e⁻ → Cu, if [Cu²⁺] = 0.01 M, E = 0.34 – (0.059/2) log(1/0.01) = 0.28 V.
• In electrolytic cells, electrical energy drives non-spontaneous redox reactions. Example: Electrolysis of molten NaCl produces Na at cathode and Cl₂ at anode.
• Faraday’s first law: m = ZIt, where Z is electrochemical equivalent, I is current, t is time. Z = M/(nF), M = molar mass, n = number of electrons.
• Faraday’s second law: When same charge passes through different electrolytes, masses deposited are proportional to their chemical equivalent weights (E = M/n).
• 1 Faraday (F) = 96485 C mol⁻¹ ≈ 96500 C mol⁻¹. This is the charge of 1 mole of electrons.
• In electrolysis of aqueous NaCl, H₂ is liberated at cathode (not Na) because E°(H⁺/H₂) > E°(Na⁺/Na), so water gets reduced preferentially.
• Overpotential affects product of electrolysis; oxygen has high overpotential on Hg cathode, so in Castner-Kellner cell, Na is formed instead of H₂.
• Conductivity (κ) is the conductance of a 1 cm³ solution; unit is S cm⁻¹. Depends on number of ions and their mobility.
• Molar conductivity (Λₘ) = κ × 1000 / C, where C is concentration in mol L⁻¹; unit is S cm² mol⁻¹.
• Λₘ increases with dilution for both strong and weak electrolytes due to increased ion dissociation.
• For strong electrolytes, Λₘ vs √C is linear (Kohlrausch’s law): Λₘ = Λₘ⁰ – A√C. Example: KCl shows this linear behavior.
• Kohlrausch’s law: Limiting molar conductivity (Λₘ⁰) can be expressed as sum of ionic conductivities: Λₘ⁰ = ν₊λ₊⁰ + ν₋λ₋⁰. Example: Λₘ⁰(KCl) = λ⁰(K⁺) + λ⁰(Cl⁻).
• Electrolysis of acidified water produces H₂ and O₂ in 2:1 volume ratio at STP. Reaction: 2H₂O → 2H₂ + O₂.
• In a galvanic cell, electrons flow from anode (oxidation) to cathode (reduction) through external circuit.
• The standard hydrogen electrode (SHE) has E° = 0.00 V and serves as reference electrode; consists of Pt electrode in 1 M H⁺, H₂ at 1 atm.
• For concentration cells, E°cell = 0 but Ecell ≠ 0 due to concentration difference; E = (0.059/n) log(C₂/C₁) at 298 K.

Difficulty Level

Intermediate — Requires understanding of both conceptual electrochemistry and numerical application of Nernst and Faraday laws.

Common CUET Traps (3 bullets)

• Trap: Assuming Na is produced during electrolysis of aqueous NaCl.
  Avoid: Remember H⁺ gets reduced preferentially due to lower discharge potential; H₂ gas is formed.
• Trap: Using concentration in mol dm⁻³ directly in Nernst equation without converting to log form or missing 'n'.
  Avoid: Always write Q correctly and use E = E° – (0.059/n) logQ at 298 K.
• Trap: Confusing molar conductivity (Λₘ) with conductivity (κ).
  Avoid: Λₘ = (κ × 1000)/C — it's normalized for concentration.

Practice MCQs (5 questions)

Q1. What is the EMF of the cell: Zn | Zn²⁺ (1 M) || Ag⁺ (1 M) | Ag, given E°(Zn²⁺/Zn) = –0.76 V, E°(Ag⁺/Ag) = +0.80 V?
A. 0.04 V
B. 1.16 V
C. 1.56 V
D. 2.36 V
Answer: C
Explanation: E°cell = E°cathode – E°anode = 0.80 – (–0.76) = 1.56 V.
Why others fail: Option B is common error from 0.80 – 0.76 = 0.04, forgetting sign of Zn potential.

Q2. Which of the following represents the Nernst equation for the reaction: Fe²⁺ + Ag⁺ → Fe³⁺ + Ag?
A. E = E° – (0.059) log([Fe³⁺]/([Fe²⁺][Ag⁺]))
B. E = E° – (0.059) log([Fe²⁺][Ag⁺]/[Fe³⁺])
C. E = E° – (0.059/2) log([Fe³⁺]/([Fe²⁺][Ag⁺]))
D. E = E° – (0.059/2) log([Fe²⁺][Ag⁺]/[Fe³⁺])
Answer: A
Explanation: n = 1 (one electron transfer), Q = [Fe³⁺]/([Fe²⁺][Ag⁺]), so E = E° – (0.059/1) logQ.
Why others fail: Option C incorrectly assumes n = 2.

Q3. During electrolysis of dilute H₂SO₄, what gases are evolved at anode and cathode respectively?
A. H₂ and O₂
B. O₂ and H₂
C. SO₂ and H₂
D. H₂ and SO₂
Answer: B
Explanation: Water is electrolyzed: H₂ at cathode, O₂ at anode.
Why others fail: Option A reverses electrode positions.

Q4. The molar conductivity of 0.01 M KCl solution is 150 S cm² mol⁻¹. What is its conductivity (κ)?
A. 0.0015 S cm⁻¹
B. 0.015 S cm⁻¹
C. 1.5 S cm⁻¹
D. 15 S cm⁻¹
Answer: A
Explanation: κ = (Λₘ × C)/1000 = (150 × 0.01)/1000 = 0.0015 S cm⁻¹.
Why others fail: Option B omits division by 1000.

Q5. A current of 1.608 A is passed through molten AlCl₃ for 1 hour. What mass of Al is produced? (F = 96500 C mol⁻¹, M(Al) = 27 g mol⁻¹)
A. 0.54 g
B. 1.62 g
C. 4.86 g
D. 0.18 g
Answer: B
Explanation: Q = It = 1.608 × 3600 = 5788.8 C; n = Q/F = 5788.8/96500 ≈ 0.06 mol e⁻; Al³⁺ + 3e⁻ → Al, so moles Al = 0.06/3 = 0.02; mass = 0.02 × 27 = 0.54 g — wait, recalculate: 5788.8 / 96500 = 0.06 exactly? 1.608 × 3600 = 5788.8; 5788.8 / 96500 = 0.05998 ≈ 0.06; 0.06 / 3 = 0.02 mol; 0.02 × 27 = 0.54 g — but options suggest error. Wait: verify calculation.
Actually: 1.608 A × 3600 s = 5788.8 C
Moles of e⁻ = 5788.8 / 96500 = 0.05998 ≈ 0.06
For Al: 3F → 1 mol Al → 27 g
So, mass = (27 / 3×96500) × 5788.8 = (9 / 96500) × 5788.8 ≈ (9 × 5788.8)/96500 = 52099.2 / 96500 ≈ 0.5398 ≈ 0.54 g
So correct answer is A, but option B is 1.62 — likely typo in question or options.
Wait — recheck: Is current 1.608 A? 1.608 × 3600 = 5788.8 C
F = 96500 C/mol e⁻ → mol e⁻ = 5788.8 / 96500 = 0.05998
n(Al) = 0.05998 / 3 = 0.01999 mol
mass = 0.01999 × 27 = 0.5397 g ≈ 0.54 g
Answer: A
Explanation: Mass = (I × t × M) / (n × F) = (1.608 × 3600 × 27) / (3 × 96500) = 0.54 g.
Why others fail: Option B (1.62 g) comes from using n = 1 instead of n = 3.

Last‑Minute Revision (15–20 one-liners)

• ⚠️ EMF = E°cathode – E°anode — always subtract anode from cathode.
• ⚠️ Nernst equation at 298 K: E = E° – (0.059/n) logQ — 'n' is moles of electrons.
• ⚠️ In electrolytic cells, anode is positive, cathode is negative — opposite of galvanic.
• ⚠️ Faraday constant F = 96500 C mol⁻¹ — charge of 1 mole electrons.
• ⚠️ For dilute solutions, Λₘ increases with dilution — more dissociation.
• ⚠️ Kohlrausch’s law: Λₘ⁰ = λ₊⁰ + λ₋⁰ — sum of ionic conductivities.
• ⚠️ Electrolysis of molten NaCl: 2NaCl → 2Na + Cl₂.
• ⚠️ Aqueous NaCl electrolysis: cathode → H₂, anode → Cl₂ (if concentrated).
• ⚠️ Overpotential favors Na in Castner-Kellner cell — Hg cathode.
• ⚠️ Standard conditions: 298 K, 1 M, 1 atm.
• ⚠️ SHE: Pt, 1 M H⁺, 1 atm H₂, E° = 0.00 V.
• ⚠️ Molar conductivity Λₘ = (κ × 1000)/C — C in mol L⁻¹.
• ⚠️ For strong electrolytes, Λₘ vs √C is linear.
• ⚠️ In concentration cell, E°cell = 0, but Ecell ≠ 0.
• ⚠️ Faraday’s first law: m = ZIt; Z = M/(nF).
• ⚠️ Electrolysis of acidified water: 2H₂O → 2H₂ + O₂ — 2:1 volume ratio.
• ⚠️ Anode = oxidation, cathode = reduction — same in both cells.
• ⚠️ For Ag⁺/Ag, E = E° – 0.059 log(1/[Ag⁺]) — Nernst for reduction.
• ⚠️ Discharge order: Cations with higher E° reduced first — H⁺ over Na⁺.
• ⚠️ Mnemonic: "Red Cat" — Reduction at Cathode; "An Ox" — Anode for Oxidation.