A Level · Cambridge (CIE) · 9701

Calculate electrode potentials with the Nernst equation

Apply the 298 K Nernst equation to aqueous redox couples using the correct electron number and concentration ratio.

Electrochemistry · Cambridge International AS & A Level Chemistry 9701 · 24.2

A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 24. Treat all tabulated electrode potentials as reduction potentials; state conditions, phases, signs and units. Written practice is self-checked

Before you begin

Use AS redox half-equations, oxidation states, mole ratios, current and charge.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

For Fe³⁺ + e⁻ ⇌ Fe²⁺, what is z in the Nernst equation?

What you will learn

Use E = E° + (0.059/z) log([Ox]/[Red]) at 298 K
Use the electron number z in the denominator
Apply the equation to metal-ion and same-element oxidation-state couples

Use the Nernst form supplied in the syllabus

At 298 K, for Ox + ze⁻ ⇌ Red, E = E° + (0.059/z) log([Ox]/[Red]) in volts, using base-10 logarithms and concentrations in the ratio.

For Cu²⁺(aq) + 2e⁻ ⇌ Cu(s), z=2 and the pure solid is omitted, so E = E° + (0.059/2)log[Cu²⁺]. For Fe³⁺ + e⁻ ⇌ Fe²⁺, z=1 and E = E° + 0.059log([Fe³⁺]/[Fe²⁺]).

Check ratio, logarithm and units

Use the ratio oxidised form over reduced form as written for the reduction half-equation. A ratio below one gives a negative logarithm; a ratio above one gives a positive contribution.

The factor 0.059 is for 298 K in this syllabus form. Use the equation and conditions given by the question; do not use it unchanged at another temperature.

Pause and explain

For Cu²⁺ + 2e⁻ ⇌ Cu(s), which species is omitted from the concentration ratio?

Put the idea to work

Worked example

For Cu²⁺/Cu, E°=+0.34 V and [Cu²⁺]=0.010 mol dm⁻³. Calculate E at 298 K.

Show the worked solution

Use z=2 and omit pure Cu(s).
E = 0.34 +(0.059/2)log(0.010).
log(0.010)=−2.

Answer E = 0.34 + 0.0295(−2) = +0.281 V.

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From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

For Fe³⁺/Fe²⁺, E°=+0.77 V, [Fe³⁺]=0.10 mol dm⁻³ and [Fe²⁺]=0.010 mol dm⁻³. Calculate E at 298 K.

Give me a hint

Use z=1 and the ratio [Fe³⁺]/[Fe²⁺].

Compare my reasoning

The ratio is 10.
E = 0.77 + 0.059 log(10).
log(10)=1.

E = +0.829 V.

Look for these in your work

Ratio correct
Electron number applied
Potential and unit correct

2 · Independent

For Cu²⁺/Cu, E°=+0.34 V and [Cu²⁺]=1.0×10⁻⁴ mol dm⁻³. Calculate E at 298 K.

Give me a hint

Use z=2 and log(10⁻⁴)=−4.

Compare my reasoning

E = 0.34 +(0.059/2)log(10⁻⁴).
The logarithm is −4.
Calculate the signed correction.

E = 0.34 + 0.0295(−4) = +0.222 V.

Look for these in your work

Correct ratio
Log value correct
Signed result correct

3 · Transfer

A student calculates Fe³⁺/Fe²⁺ potential using log([Fe²⁺]/[Fe³⁺]). Explain the reversal and state the correct ratio for the reduction equation.

Give me a hint

Write the half-equation as reduction and identify Ox and Red.

Compare my reasoning

Fe³⁺ is oxidised form in Fe³⁺ + e⁻ ⇌ Fe²⁺.
Fe²⁺ is reduced form.
Use [Fe³⁺]/[Fe²⁺].

The ratio was reversed. For the reduction half-equation Fe³⁺ + e⁻ ⇌ Fe²⁺, use [Fe³⁺]/[Fe²⁺].

Look for these in your work

Ox/Red identified
Ratio corrected
Reduction form linked

Common mistakes

Using z=1 for Cu²⁺/Cu
Putting reduced/oxidised in the logarithm
Including a pure solid in the concentration ratio
Using natural logarithm or the 298 K factor at another temperature

Recall without your notes

What can you explain now?

For Fe³⁺ + e⁻ ⇌ Fe²⁺, what is z in the Nernst equation?

Compare with the explanation

One electron appears in the half-equation.

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Make it stick

How do the reduction equation, electron number and oxidised-to-reduced ratio determine a Nernst calculation?

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Chemistry

Check the official syllabus 2028–2030

Course scope and assessment

Four hundred and forty-two original lessons map the published learning outcomes across AS Chemistry Topics 1–22 and A Level Topics 23–37, with guided, independent and transfer practice and explained answers. Every numbered subject-content topic and listed outcome tag now has a mapped lesson sequence. Topic 32 maps all 12 hydroxy-compound outcome tags in 11 lessons; Topic 33 maps all 15 carboxylic-acid and derivative outcome tags in 16 lessons; Topic 34 maps all 22 nitrogen-compound outcome tags in 20 lessons; Topic 35 maps all 12 polymerisation outcome tags in 16 lessons; Topic 36 maps all four organic-synthesis outcome tags in 14 lessons; Topic 37 maps all 22 chromatography and NMR outcome tags in 18 lessons. Cambridge's 2028–2030 Version 1 syllabus states that there are no significant changes which affect teaching from the preceding edition; the routes are aligned to that edition. This is 37 of the 37 numbered subject-content topics. Full cumulative exam assessment, Paper 3 laboratory competence, Paper 5 investigation practice and human subject-teacher review remain outstanding. Written practice is self-checked and does not certify exam readiness.