CBSE Class 12 Electrochemistry — Every Formula You Need
Electrochemistry is one of the most numerical-heavy chapters in the Class 12 syllabus, and almost every numerical in it comes from one of six relations. Learn these six, know what each symbol means, and the chapter becomes arithmetic rather than memory. Below is every formula with its units, followed by a fully worked example for each.
1. Standard cell potential
Both values are standard reduction potentials. Do not flip the sign of the anode value first and then subtract — the subtraction already does that job. Oxidation happens at the anode, reduction at the cathode, and a positive E°cell means the cell reaction is spontaneous as written.
Daniell cell: E°(Zn²⁺/Zn) = −0.76 V, E°(Cu²⁺/Cu) = +0.34 V.
Copper has the higher reduction potential, so copper is the cathode.
E°cell = 0.34 − (−0.76) = +1.10 V
2. Gibbs energy from cell potential
n = number of electrons transferred in the balanced cell reaction, F = 96500 C mol⁻¹ (Faraday constant). With E° in volts and F in coulombs, ΔG° comes out in joules — divide by 1000 for kJ.
For the Daniell cell above, Zn + Cu²⁺ → Zn²⁺ + Cu, so n = 2.
ΔG° = − 2 × 96500 × 1.10 = − 212300 J = − 212.3 kJ mol⁻¹
The negative sign confirms a spontaneous reaction.
3. The Nernst equation
Standard potentials assume every solution is 1 M. Real cells are not, and the Nernst equation corrects for that.
At 298 K this simplifies to: Ecell = E°cell − (0.0591 / n) log Q
Q is the reaction quotient, written from the balanced cell reaction the same way you write an equilibrium constant — products over reactants, and pure solids and liquids are left out. The 0.0591 shortcut is valid only at 298 K.
Q. Find Ecell for Zn | Zn²⁺ (0.1 M) || Cu²⁺ (0.01 M) | Cu at 298 K.
Cell reaction: Zn + Cu²⁺ → Zn²⁺ + Cu, n = 2, E° = 1.10 V
Q = [Zn²⁺] / [Cu²⁺] = 0.1 ÷ 0.01 = 10, so log Q = 1
E = 1.10 − (0.0591 ÷ 2) × 1 = 1.10 − 0.0296 = 1.07 V
Note the direction: the product ion is more concentrated than the reactant ion, so the cell potential drops slightly below E°. That is exactly what should happen.
4. Cell potential and the equilibrium constant
At equilibrium the cell is dead — Ecell = 0 and Q = Kc. Putting that into the Nernst equation gives:
equivalently, ΔG° = − RT ln Kc
Q. Calculate Kc for the Daniell cell at 298 K.
log Kc = (2 × 1.10) ÷ 0.0591 = 2.20 ÷ 0.0591 = 37.2
Kc = 1037.2 ≈ 1.7 × 10³⁷
A huge K means the reaction goes essentially to completion — which is why a zinc rod in copper sulphate solution keeps reacting until one reactant runs out.
5. Faraday's laws of electrolysis
First law: the mass deposited is proportional to the charge passed.
moles of electrons = Q / F = Q / 96500
m = (Q × M) / (n × F) where n = electrons per ion
Q. A current of 1.5 A is passed through CuSO₄ solution for 10 minutes. What mass of copper is deposited? (Cu = 63.55)
t = 10 × 60 = 600 s
Q = 1.5 × 600 = 900 C
moles of electrons = 900 ÷ 96500 = 9.326 × 10⁻³ mol
The cathode reaction is Cu²⁺ + 2e⁻ → Cu, so 2 mol of electrons give 1 mol of Cu:
n(Cu) = 9.326 × 10⁻³ ÷ 2 = 4.663 × 10⁻³ mol
m = 4.663 × 10⁻³ × 63.55 = 0.296 g
Second law: when the same charge is passed through different electrolytes, the masses deposited are in the ratio of their equivalent masses.
Q. The same 900 C is passed through AgNO₃ solution. What mass of silver is deposited? (Ag = 107.87)
E(Cu) = 63.55 ÷ 2 = 31.78 E(Ag) = 107.87 ÷ 1 = 107.87
m(Ag) = 0.296 × (107.87 ÷ 31.78) = 0.296 × 3.394 = 1.005 g
6. Conductance, conductivity and molar conductivity
Cell constant G* = l / A (unit: cm⁻¹)
Conductivity κ = G × G* = (1/R) × (l/A) (unit: S cm⁻¹)
Molar conductivity Λm = (κ × 1000) / c (unit: S cm² mol⁻¹, c in mol L⁻¹)
The factor of 1000 exists only to convert litres to cubic centimetres. Forgetting it is the single most common arithmetic error in this chapter.
Q. A 0.1 M KCl solution in a cell of cell constant 1.29 cm⁻¹ has a resistance of 100 Ω. Find κ and Λm.
κ = (1 ÷ 100) × 1.29 = 0.0129 S cm⁻¹
Λm = (0.0129 × 1000) ÷ 0.1 = 12.9 ÷ 0.1 = 129 S cm² mol⁻¹
7. Kohlrausch's law and degree of dissociation
At infinite dilution, each ion contributes independently:
α = Λm / Λ°m and Ka = cα² / (1 − α)
This is how the limiting molar conductivity of a weak electrolyte is found — you cannot measure it directly, so you build it from strong electrolytes.
Q. Given Λ°m(CH₃COONa) = 91.0, Λ°m(HCl) = 426.2 and Λ°m(NaCl) = 126.5 S cm² mol⁻¹, find Λ°m for acetic acid. If Λm at 0.01 M is 16.5 S cm² mol⁻¹, find α and Ka.
Λ°m(CH₃COOH) = 91.0 + 426.2 − 126.5 = 390.7 S cm² mol⁻¹
α = 16.5 ÷ 390.7 = 0.0422 (about 4.2% dissociated)
Ka = (0.01 × 0.0422²) ÷ (1 − 0.0422)
= (0.01 × 0.001781) ÷ 0.9578 = 1.781 × 10⁻⁵ ÷ 0.9578 = 1.86 × 10⁻⁵
Common mistakes that cost marks
- Flipping the anode sign twice. E°cell = E°cathode − E°anode uses reduction potentials as printed in the table.
- Wrong n in the Nernst equation. n is the number of electrons in the balanced overall reaction, not in one half-cell as written.
- Including solids in Q. In Zn + Cu²⁺ → Zn²⁺ + Cu, only the two ions appear in Q. Zn(s) and Cu(s) do not.
- Leaving time in minutes. Q = It needs seconds. 10 minutes is 600 s.
- Dropping the 1000 in Λm. κ is per centimetre, c is per litre; the 1000 reconciles them.
- Using 0.0591 away from 298 K. At any other temperature go back to the full RT/nF form.
All formulas in one table
| Quantity | Formula | Unit |
|---|---|---|
| Standard cell potential | E° = E°cathode − E°anode | V |
| Gibbs energy | ΔG° = − nFE° | J mol⁻¹ |
| Nernst (298 K) | E = E° − (0.0591/n) log Q | V |
| Equilibrium constant | log Kc = nE°/0.0591 | — |
| Charge passed | Q = It | C |
| Mass deposited | m = QM / nF | g |
| Conductivity | κ = (1/R)(l/A) | S cm⁻¹ |
| Molar conductivity | Λm = 1000κ / c | S cm² mol⁻¹ |
| Degree of dissociation | α = Λm / Λ°m | — |
Check your Nernst answers instantly. The free Nernst Equation calculator takes E°, n and the ion concentrations and returns Ecell at 298 K, so you can confirm both the number and the direction of the shift before an exam.
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