JAM Electrochemistry — Cells and Conductance
IIT-JAM treats electrochemistry as two connected halves: cells — EMF, the Nernst equation and the link to thermodynamics — and conductance, how well a solution carries current and what that says about dissociation. Students usually revise the first and skim the second, then lose marks on a conductance numerical that was entirely mechanical. This article covers both, with every calculation done in full.
Cell notation and EMF
Anode on the left, cathode on the right, a single bar for a phase boundary and a double bar for the salt bridge. Oxidation is at the anode and reduction at the cathode in both galvanic and electrolytic cells; only the sign of the electrodes changes.
E°cell = E°cathode − E°anode (both as reduction potentials)
Never reverse the sign of a standard potential here — the subtraction has already done that. And E° is intensive: multiplying a half-reaction by 2 does not double it, though it does double n and therefore ΔG°.
The thermodynamic link
F = 96 485 C mol⁻¹, R = 8.314 J K⁻¹ mol⁻¹
Worked example 1 — a Daniell cell, three quantities from one number. Taking the standard reduction potentials as E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V, with n = 2 and T = 298 K.
EMF: E° = 0.34 − (−0.76) = +1.10 V. Positive, so the cell reaction is spontaneous as written.
Free energy: ΔG° = −nFE° = −2 × 96 485 × 1.10 = −212 267 J mol⁻¹ ≈ −212 kJ mol⁻¹
Equilibrium constant: ln K = −ΔG°/RT = 212 267 / (8.314 × 298.15) =
212 267 / 2478.8 = 85.63
log₁₀K = 85.63 / 2.303 = 37.19, so K ≈ 1.5 × 10³⁷
Cross-check by the shortcut route: log K = nE°/0.0592 = (2 × 1.10)/0.0592 = 2.20/0.0592 = 37.16 ✓. The two agree to within rounding, which is the point of doing both.
The Nernst equation
Q is the reaction quotient for the cell reaction as written: products over reactants, pure solids and liquids omitted, gases as partial pressures. The 0.0592 V figure is 2.303RT/F at 298 K only — at any other temperature go back to RT/nF.
Worked example 2 — non-standard concentrations. For the same Daniell cell with [Zn²⁺] = 0.10 M and [Cu²⁺] = 0.010 M.
Cell reaction: Zn(s) + Cu²⁺ → Zn²⁺ + Cu(s), so Q = [Zn²⁺]/[Cu²⁺] = 0.10/0.010 = 10
log Q = 1
E = 1.10 − (0.0592/2) × 1 = 1.10 − 0.0296 = 1.07 V
Sensible direction: raising the product concentration relative to the reactant pushes the cell closer to equilibrium, so the EMF falls slightly.
Worked example 3 — a concentration cell. Cu | Cu²⁺(0.0010 M) ‖ Cu²⁺(0.10 M) | Cu. Here E° = 0 because both electrodes are identical; the entire EMF comes from the concentration difference.
E = (0.0592/n)·log([Cu²⁺]cathode / [Cu²⁺]anode)
= (0.0592/2) × log(0.10 / 0.0010) = 0.0296 × log(100) = 0.0296 × 2
= 0.0592 V
The dilute half-cell is always the anode — the cell runs in the direction that equalises the two concentrations. This same idea is what makes a pH-sensitive glass electrode work.
Electrolysis and Faraday's laws
Worked example 4. A current of 2.00 A passes for 30.0 min through aqueous CuSO₄. What mass of copper is deposited? (M = 63.55 g mol⁻¹, n = 2)
Q = 2.00 × (30.0 × 60) = 2.00 × 1800 = 3600 C
moles of electrons = 3600 / 96 485 = 0.03731 mol
moles of Cu = 0.03731 / 2 = 0.018656 mol
m = 0.018656 × 63.55 = 1.19 g
Conductance: the definitions that decide the answer
Molar conductivity Λm = κ / c (c in mol m⁻³ for S m² mol⁻¹, or use κ in S cm⁻¹ and c in mol cm⁻³)
Behaviour on dilution is the standard conceptual question, and the two quantities move in opposite directions. κ falls, because there are fewer ions per cubic metre. Λm rises, because it is already normalised per mole and ion–ion interference decreases. For a strong electrolyte Λm extrapolates linearly against √c to Λ°m; for a weak one it rises steeply near infinite dilution and cannot be extrapolated — which is why Kohlrausch's law is needed.
Worked example 5 — cell constant, then molar conductivity. A conductivity cell filled with a standard KCl solution of conductivity κ = 1.29 S m⁻¹ (value supplied with the question) has resistance 100 Ω. The same cell filled with 0.020 mol dm⁻³ of another electrolyte reads 520 Ω.
Cell constant: κ = G × (l/A) = (1/R) × (l/A), so l/A = κ × R = 1.29 × 100 = 129 m⁻¹
Conductivity of the test solution: κ = 129 / 520 = 0.2481 S m⁻¹
Concentration in SI: 0.020 mol dm⁻³ = 0.020 × 1000 = 20 mol m⁻³
Molar conductivity: Λm = 0.2481 / 20 = 0.01240 S m² mol⁻¹
Converting to the units most tables use: 1 S m² mol⁻¹ = 10⁴ S cm² mol⁻¹, so Λm = 124.0 S cm² mol⁻¹.
Kohlrausch's law and the Ka of a weak acid
At infinite dilution each ion contributes independently, so limiting molar conductivities add and subtract like algebra — which is how Λ°m of a weak acid is obtained even though it cannot be measured directly.
α = Λm / Λ°m Ka = cα² / (1 − α)
Worked example 6. Using the values given in a question as Λ°m(CH₃COONa) = 91.0, Λ°m(HCl) = 426.2 and Λ°m(NaCl) = 126.5 S cm² mol⁻¹, and Λm = 48.15 S cm² mol⁻¹ for 0.0010 mol dm⁻³ acetic acid.
Limiting value: 91.0 + 426.2 − 126.5 = 390.7 S cm² mol⁻¹
(The sodium and chloride contributions cancel, leaving H⁺ + CH₃COO⁻ — check that the
combination you write actually does cancel.)
Degree of dissociation: α = 48.15 / 390.7 = 0.1232, i.e. 12.3%
Dissociation constant:
α² = 0.1232² = 0.015178
cα² = 0.0010 × 0.015178 = 1.5178 × 10⁻⁵
1 − α = 0.8768
Ka = 1.5178 × 10⁻⁵ / 0.8768 = 1.73 × 10⁻⁵
Check: the approximate form Ka ≈ cα² gives 1.52 × 10⁻⁵, about 12% low — which tells you that dropping the (1 − α) term is only safe when α is genuinely small.
- Reversing the sign of E°(anode) and also subtracting it. E°cell = E°cathode − E°anode, using reduction potentials as tabulated.
- Multiplying E° when you balance the equation. E° is intensive; only n and ΔG° scale.
- Writing Q upside down. Products over reactants, for the cell reaction as written.
- Using 0.0592 away from 298 K. Go back to RT/nF.
- Concentration units in Λm. Either κ in S m⁻¹ with c in mol m⁻³, or κ in S cm⁻¹ with c in mol cm⁻³. Mixing them gives an answer out by 10⁶.
- Saying conductivity increases on dilution. κ falls; it is Λm that rises.
- Extrapolating Λm against √c for a weak electrolyte — that linear extrapolation is valid for strong electrolytes only.
Exam relevance
| Question style | What it is testing |
|---|---|
| Find Ecell at given concentrations | Nernst equation and a correctly written Q |
| Find K or ΔG° from E° | ΔG° = −nFE° = −RT ln K |
| Concentration cell EMF | E° = 0; the whole EMF is the log of a ratio |
| Cell constant and Λm | Unit handling more than anything else |
| α and Ka of a weak acid | Kohlrausch's law then Ostwald's dilution law |
| Mass deposited in electrolysis | Faraday's laws and the value of n |
| Conceptual: κ vs Λm on dilution | The definitions, not memory |
Check the current official notification for the syllabus and pattern for your session.
Confirm every EMF before you commit it. The Nernst equation calculator takes E°, n and the reaction quotient and returns the cell potential at your chosen temperature, so a concentration-cell or non-standard-condition answer can be verified in seconds instead of re-doing the logarithm by hand.
Open the Nernst Equation Calculator →Preparing for IIT-JAM, GATE, CSIR-NET or CUET-PG? ABC Chemistry runs dedicated chemical-sciences batches — coaching centre and online classes across India. Details at abcchemistry.in.