A2 · Practice questions · The Nernst equation

Put the concentrations in.

Six original Cambridge-style questions on the Nernst equation and non-standard cells.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

Keep these straight

Oxidised on top, divide by z, base-10 log.

E = E^⊖ + (0.059 ÷ z) log ([ox] ÷ [red]) at 298 K.
Solids (metals) are left out of the fraction.
More oxidised form: E more positive.
Tenfold change: 0.059 ÷ z volts.

Calculation

[2 marks]

Calculate the electrode potential of a Ag⁺/Ag half-cell in which [Ag⁺] = 0.050 mol dm⁻³. (E^⊖ = +0.80 V)

z = 1: E = 0.80 + 0.059 log 0.050. ✓
E = 0.80 + 0.059 × (−1.301) = +0.723 V. ✓

Calculation

[3 marks]

Calculate E_cell for the cell Zn | Zn²⁺ (0.50 mol dm⁻³) || Ag⁺ (0.20 mol dm⁻³) | Ag.

E(Ag) = 0.80 + 0.059 log 0.20 = 0.80 − 0.0412 = +0.759 V. ✓
E(Zn) = −0.76 + 0.0295 log 0.50 = −0.76 − 0.0089 = −0.769 V. ✓
E_cell = 0.759 − (−0.769) = +1.53 V. ✓

Calculation

[2 marks]

Calculate E for the Fe³⁺/Fe²⁺ half-cell when [Fe³⁺] = 0.10 mol dm⁻³ and [Fe²⁺] = 1.0 mol dm⁻³. Would Fe³⁺ still oxidise I⁻ under standard conditions for the iodine half-cell?

E = 0.77 + 0.059 log (0.10 ÷ 1.0) = +0.711 V. ✓
E_cell = 0.711 − 0.54 = +0.171 V, still positive, so yes. ✓

Explain

[3 marks]

Explain, in terms of the equilibrium Cl₂ + 2e⁻ ⇌ 2Cl⁻, why chlorine rather than oxygen is the main anode product when concentrated sodium chloride solution is electrolysed.

A high [Cl⁻] shifts the equilibrium to the left (reduced form increased). ✓
So E(Cl₂/Cl⁻) becomes less positive than +1.36 V, approaching or below E(O₂/H₂O). ✓
Cl⁻ is therefore oxidised in preference to water, giving chlorine. ✓

Calculation

[2 marks]

Calculate E for the Cl₂/Cl⁻ half-cell when [Cl⁻] = 2.0 mol dm⁻³ and chlorine is at 100 kPa. (E^⊖ = +1.36 V)

E = 1.36 + (0.059 ÷ 2) log (1 ÷ 2.0²) = 1.36 − 0.059 log 2.0. ✓
E = 1.36 − 0.0178 = +1.34 V. ✓

Analysis

[3 marks]

A cell is set up from two Cu²⁺/Cu half-cells, one with [Cu²⁺] = 1.00 mol dm⁻³ and one with 0.010 mol dm⁻³. Calculate the e.m.f., identify the positive electrode and state what happens to the e.m.f. as the cell runs.

E(dilute) = 0.34 + 0.0295 log 0.010 = +0.281 V; E(concentrated) = +0.340 V. ✓
E_cell = 0.340 − 0.281 = 0.059 V; the 1.00 mol dm⁻³ electrode is positive (Cu²⁺ is reduced there). ✓
As it runs, the concentrated side becomes more dilute and the dilute side more concentrated, so the e.m.f. falls towards zero. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.