12 min readConcept sheet3-mark numerical on the EMF of a cell at given concentrations; 1-mark MCQ on how the EMF changes; often part of a 5-mark answer; also a NEET and JEE favourite

Nernst equation for electrodes and cells

The Nernst equation gives a cell's EMF at any concentration: E_cell = E°cell − (0.059/n) log Q at 298 K. Raising the reactant ion's concentration raises the EMF.

Nernst equation for an electrodeE(Mⁿ⁺/M) = E°(Mⁿ⁺/M) − (RT/nF) ln (1/[Mⁿ⁺]); at 298 K, E = E° − (0.059/n) log (1/[Mⁿ⁺])

Nernst equation for a cellE_cell = E°cell − (2.303RT/nF) log Q, where Q = [products]/[reactants]; solids and pure liquids are left out

2.303RT/F at 298 K0.059 V, using R = 8.314 J K⁻¹ mol⁻¹ and F = 96487 C mol⁻¹

Hydrogen electrodeE(H⁺/H₂) = −0.059 pH at 298 K, with H₂ at 1 bar

Why concentration matters

Standard potentials hold only when every ion is at 1 M. In a working cell, concentrations keep changing. Walther Nernst showed how the potential depends on concentration. For a metal electrode, Mⁿ⁺(aq) + ne⁻ → M(s): E = E° − (RT/nF) ln (1/[Mⁿ⁺]). The concentration of a pure solid is taken as 1, so the metal does not appear. Changing ln to log (multiply by 2.303) and putting T = 298 K, R = 8.314 J K⁻¹ mol⁻¹ and F = 96487 C mol⁻¹ gives E = E° − (0.059/n) log (1/[Mⁿ⁺]).

Nernst equation for the Daniell cell

1
Cathode: E(Cu²⁺/Cu) = E°(Cu²⁺/Cu) − (RT/2F) ln (1/[Cu²⁺])
2
Anode: E(Zn²⁺/Zn) = E°(Zn²⁺/Zn) − (RT/2F) ln (1/[Zn²⁺])
3
E_cell = E(Cu²⁺/Cu) − E(Zn²⁺/Zn)
4
E_cell = [E°(Cu²⁺/Cu) − E°(Zn²⁺/Zn)] − (RT/2F) [ln (1/[Cu²⁺]) − ln (1/[Zn²⁺])] = E°cell − (RT/2F) ln ([Zn²⁺]/[Cu²⁺])
5★
At 298 K: E_cell = 1.10 − (0.059/2) log ([Zn²⁺]/[Cu²⁺]) V
6
So, E_cell rises when [Cu²⁺] rises and falls when [Zn²⁺] rises.

★ key step

The general form

For a cell reaction aA + bB → cC + dD in which n electrons are transferred, E_cell = E°cell − (RT/nF) ln Q, where Q = [C]ᶜ[D]ᵈ ÷ ([A]ᵃ[B]ᵇ). At 298 K, E_cell = E°cell − (0.059/n) log Q. Always take n from the balanced equation, and raise each concentration to the power of its coefficient.

Worked example: EMF of a zinc-silver cell

Problem

Find the EMF at 298 K of the cell Zn(s) | Zn²⁺(0.1 M) || Ag⁺(0.01 M) | Ag(s). E°(Zn²⁺/Zn) = −0.76 V; E°(Ag⁺/Ag) = +0.80 V.

Working

Cell reaction: Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s); n = 2.
E°cell = 0.80 − (−0.76) = 1.56 V.
Q = [Zn²⁺] ÷ [Ag⁺]² = 0.1 ÷ (0.01)² = 0.1 ÷ 10⁻⁴ = 10³.
E_cell = 1.56 − (0.059/2) log 10³ = 1.56 − 0.0295 × 3.
E_cell = 1.56 − 0.0885 = 1.4715 V ≈ 1.47 V.

Answer

E_cell ≈ 1.47 V

Worked example: single electrodes

Problem

At 298 K, find (a) the potential of a zinc electrode in 0.01 M Zn²⁺ solution (E° = −0.76 V) and (b) the potential of a hydrogen electrode, with H₂ at 1 bar, in a solution of pH 3.

Working

(a) Zn²⁺ + 2e⁻ → Zn: E = E° − (0.059/2) log (1/[Zn²⁺]).
E = −0.76 − 0.0295 × log (1/0.01) = −0.76 − 0.0295 × 2 = −0.76 − 0.059 = −0.819 V.
(b) H⁺ + e⁻ → ½H₂: E = 0 − (0.059/1) log (1/[H⁺]) = −0.059 pH.
E = −0.059 × 3 = −0.177 V.

Answer

(a) −0.819 V (about −0.82 V); (b) −0.177 V

Activity: how the cell potential changes with concentration

You need zinc and copper strips, ZnSO₄ and CuSO₄ solutions of 1 M, 0.1 M and 0.01 M (made by diluting the 1 M solutions), beakers, salt bridges (agar U-tubes or filter-paper strips soaked in KNO₃ or KCl solution) and a high-resistance voltmeter or potentiometer.

1
Clean both strips with sandpaper. Dip the zinc strip in 1 M ZnSO₄ and the copper strip in 1 M CuSO₄.
2
Join the two solutions with a salt bridge. Connect the strips to the voltmeter, zinc to the negative terminal. Record E_cell.
3
Keep [Zn²⁺] at 1 M. Replace the CuSO₄ with 0.1 M and then 0.01 M solution, using a fresh salt bridge each time. Record E_cell each time.
4
Now keep [Cu²⁺] at 1 M and use 0.1 M and 0.01 M ZnSO₄. Record E_cell each time.
5
Observation: E_cell falls as [Cu²⁺] is lowered and rises as [Zn²⁺] is lowered. Expected values are 1.10 V, about 1.07 V and about 1.04 V for [Cu²⁺] = 1, 0.1 and 0.01 M with 1 M Zn²⁺.
6
Plot E_cell against log([Zn²⁺]/[Cu²⁺]). The points lie on a straight line with slope about −0.0295 V and intercept 1.10 V, as the Nernst equation predicts.

In real life

A pH meter used to test milk, soil or a swimming pool works on the Nernst equation. Its electrode potential changes by about 0.059 V for each unit of pH at 298 K.

Exam angle

3 marks: “Write the Nernst equation and calculate the EMF at 298 K of the cell Mg(s) | Mg²⁺(0.01 M) || Cu²⁺(0.1 M) | Cu(s). E°(Mg²⁺/Mg) = −2.36 V; E°(Cu²⁺/Cu) = +0.34 V.” E°cell = 0.34 − (−2.36) = 2.70 V; n = 2. E = E° − (0.059/2) log ([Mg²⁺]/[Cu²⁺]) = 2.70 − 0.0295 × log 0.1 = 2.70 + 0.0295 = 2.7295 V ≈ 2.73 V.

1 marks: “How does the EMF of a Daniell cell change if [Zn²⁺] is increased?” It decreases, because log([Zn²⁺]/[Cu²⁺]) increases in E = 1.10 − 0.0295 log([Zn²⁺]/[Cu²⁺]).

2 marks: “Calculate the potential of a hydrogen electrode (H₂ at 1 bar) in a solution of pH 4 at 298 K.” H⁺ + e⁻ → ½H₂: E = −0.059 × pH = −0.059 × 4 = −0.236 V.

Remember it

Nernst: E = E° − (0.059/n) log (products ÷ reactants). More reactant gives more push, and more EMF.