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II-14: Electrode potentials and the Nernst equation

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Module II-14 · Recommended before: Before electrochemistry

Electrode potentials and the Nernst equation

The Nernst equation corrects a standard cell potential for reaction conditions. Temperature, electron count and reaction quotient matter.

Learn the method

E = E° − (0.05916 V/n) log₁₀Q (25 °C)

  1. Identify E°, n, Q and temperature.
  2. At 25 °C use E = E° − (0.05916/n) log₁₀Q.
  3. Compute the logarithm.
  4. Apply the correct sign.
  5. Check E moves downward when Q is greater than one.

Common mistake: The 0.05916 coefficient is valid for base-10 logs at approximately 25 °C.

Worked example

Problem: At 25 °C, E° = 1.00 V, n = 1 and Q = 10. Find E.

Step-by-step solution: E = 1.00 − (0.05916/1)(1) = 0.941 V.

Check the units and whether the answer is reasonable before continuing.

More applications of this mathematics

Try each extension before expanding its solution.

Standard cell potential

E°cell = E°cathode − E°anode. If the cathode reduction potential is +0.80 V and anode reduction potential is −0.76 V, E°cell=1.56 V.

How Q changes voltage

At 25 °C, Q>1 makes log₁₀Q positive and reduces E relative to E° for n>0; Q<1 increases E.

Independent practice — five questions

Suggested mastery threshold: four out of five. Incorrect answers display a worked explanation after submission.

Choose one response for each question, then select “Check my answers.” No timer. Hints are optional. You can retry freely. Questions are formative, not proctored.

Question 1. At 25 °C, E°=1.40 V, n=1, log₁₀Q=-1. Find E using E=E°−(0.05916/n)log₁₀Q.
Hint for question 1

Do not forget the leading minus sign in the Nernst correction.

Question 2. At 25 °C, E°=0.90 V, n=2, log₁₀Q=-1. Find E using E=E°−(0.05916/n)log₁₀Q.
Hint for question 2

Do not forget the leading minus sign in the Nernst correction.

Question 3. At 25 °C, E°=1.10 V, n=1, log₁₀Q=1. Find E using E=E°−(0.05916/n)log₁₀Q.
Hint for question 3

Do not forget the leading minus sign in the Nernst correction.

Question 4. At 25 °C, E°=0.75 V, n=2, log₁₀Q=2. Find E using E=E°−(0.05916/n)log₁₀Q.
Hint for question 4

Do not forget the leading minus sign in the Nernst correction.

Question 5. At 25 °C, E°=1.30 V, n=1, log₁₀Q=-2. Find E using E=E°−(0.05916/n)log₁₀Q.
Hint for question 5

Do not forget the leading minus sign in the Nernst correction.

Supporting reference

OpenStax Chemistry 2e / NIST — reference for this subject. This lesson’s wording, examples and questions are original.

W3C WCAG 2.2 accessibility guidance.

Applied practice: Cell direction and electrochemical energy

These additional, ungraded problems cover skills beyond the scored five-question check above. Try each before expanding its solution. Your existing score and progress records are unchanged.

  1. Challenge 1. A galvanic cell uses Zn²⁺/Zn (E°red = −0.76 V) and Cu²⁺/Cu (E°red = +0.34 V). Identify the anode and cathode and calculate E°cell.

    Show worked solution for challenge 1

    Cu²⁺ is reduced at the cathode (+0.34 V); Zn is oxidized at the anode (−0.76 V is the reduction potential). E°cell = 0.34 − (−0.76) = 1.10 V.

  2. Challenge 2. For a cell with E°cell = 1.10 V and n = 2 mol e⁻ per mole of reaction, find ΔG° (F = 96,485 C/mol e⁻).

    Show worked solution for challenge 2

    ΔG° = −nFE° = −(2)(96,485 C/mol)(1.10 J/C) = −212,267 J/mol reaction ≈ −212 kJ/mol reaction.

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