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Cell EMF and the Nernst equation: Class 12 practice

E°cell = E°cathode − E°anode. Read the method, follow the three worked examples, then try the six questions.

What you should know first

  • Logarithms
  • Redox

The idea behind cell emf and the nernst equation

E°cell = E°cathode − E°anode. At 298 K, E = E° − (0.0591/n) log Q. ΔG° = −nFE° with F = 96500 C/mol.

Step-by-step method

  1. Identify cathode (higher E°) and anode.
  2. E°cell = E°cathode − E°anode.
  3. Apply Nernst with Q from concentrations; ΔG° = −nFE°.

Worked example 1 (Basic)

Find E°cell for a cell made from Fe²⁺/Fe (E° = -0.44 V) and Cu²⁺/Cu (E° = 0.34 V).

  1. Cathode: Cu²⁺/Cu, anode: Fe²⁺/Fe
  2. E° = 0.34 − (-0.44) = 0.78 V
  3. Answer: 0.78 V

Worked example 2 (Standard)

For the Ni²⁺/Ni | Cu²⁺/Cu cell (E° = 0.59 V, n = 2), find ΔG°.

  1. ΔG° = −nFE° = −2 × 96500 × 0.59
  2. = -113.87 kJ
  3. Answer: -113.87 kJ

Worked example 3 (Challenge)

For the Ni²⁺/Ni | Cu²⁺/Cu cell with Q = 0.1, find E at 298 K.

  1. E = 0.59 − (0.0591/2) log 0.1
  2. = 0.62 V
  3. Answer: 0.62 V

Common mistakes and misconceptions

  • Subtracting the wrong way
  • Wrong n

Practice questions

  1. (Basic) Find E°cell for a cell made from Ni²⁺/Ni (E° = -0.25 V) and Cu²⁺/Cu (E° = 0.34 V).
  2. (Basic) Find E°cell for a cell made from Zn²⁺/Zn (E° = -0.76 V) and Cu²⁺/Cu (E° = 0.34 V).
  3. (Standard) For the Zn²⁺/Zn | Fe²⁺/Fe cell (E° = 0.32 V, n = 2), find ΔG°.
  4. (Standard) For the Mg²⁺/Mg | Cu²⁺/Cu cell (E° = 2.71 V, n = 2), find ΔG°.
  5. (Challenge) For the Mg²⁺/Mg | Cu²⁺/Cu cell with Q = 10, find E at 298 K.
  6. (Challenge) For the Fe²⁺/Fe | Cu²⁺/Cu cell with Q = 0.1, find E at 298 K.

Answer key with full working

  • Q1: Cathode: Cu²⁺/Cu, anode: Ni²⁺/Ni → E° = 0.34 − (-0.25) = 0.59 V → Answer: 0.59 V
  • Q2: Cathode: Cu²⁺/Cu, anode: Zn²⁺/Zn → E° = 0.34 − (-0.76) = 1.1 V → Answer: 1.1 V
  • Q3: ΔG° = −nFE° = −2 × 96500 × 0.32 → = -61.76 kJ → Answer: -61.76 kJ
  • Q4: ΔG° = −nFE° = −2 × 96500 × 2.71 → = -523.03 kJ → Answer: -523.03 kJ
  • Q5: E = 2.71 − (0.0591/2) log 10 → = 2.68 V → Answer: 2.68 V
  • Q6: E = 0.78 − (0.0591/2) log 0.1 → = 0.81 V → Answer: 0.81 V

Syllabus note

This skill is listed under "Electrochemistry" for Class 12 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

Related practice and notes

Frequently asked questions

Which chapter is cell emf and the nernst equation in?

"Electrochemistry", Class 12 Chemistry (NCERT/CBSE). Other boards may place it differently.

How were the answers checked?

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

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