Short on time? This page is the recap. For explanations, a potential ladder diagram and worked examples, use the full study guide.
Recap
- A standard reduction potential, E°, measures how easily a species is reduced, compared with the hydrogen half-cell (0.00 V by definition).
- Higher (more positive) E° = more easily reduced = stronger oxidising agent.
- In a cell, the half-reaction with the higher E° is the reduction (cathode); the lower one runs backwards as an oxidation (anode).
- E°cell > 0: favoured, galvanic, gives a positive voltage. E°cell < 0: unfavoured, needs an externally applied potential (electrolytic).
- Free energy and cell potential are linked: ΔG° = −nFE°, so the two always have opposite signs.
Key relationships
| Relationship | Symbols and units | Use it to |
|---|---|---|
| E°cell = E°(cathode) − E°(anode) | all in V, from a reduction-potential table | find the standard cell potential |
| ΔG° = −nFE° | n = mol e⁻ per mol reaction; F = 96,485 C mol⁻¹; E° in V; ΔG° in J mol⁻¹ | turn a cell potential into free energy (or back: E° = −ΔG° / nF) |
| Reverse the reaction | E° and ΔG° both change sign | judge the opposite direction |
| Multiply the equation by k | E° unchanged; n and ΔG° multiplied by k | handle different ways of writing the equation |
Assumptions behind the numbers
- Standard conditions: 1 M solutions, 1 atm gases, pure solids and liquids; tables are usually for 298 K. Other concentrations change E (Topic 9.10).
- The tabulated values are all relative to the hydrogen half-cell, so only differences between them have meaning in a cell.
- A favourable E° says nothing about rate.
Mistakes to avoid
- Multiplying E° by coefficients. Balance electrons for the equation and for n, but leave E° alone.
- Double sign change. Either subtract the anode’s reduction potential or add its oxidation potential, never both.
- Wrong n. Use the electrons that cancel in the balanced overall equation.
- J vs kJ. −nFE° gives joules; divide by 1000 for kJ.
- Sign slips. Positive E° must give negative ΔG°. If your signs agree, check again.
- Favoured = fast. No: E° and ΔG° say nothing about rate.
Quick self-check
- Calculate E°cell for Mg(s) + 2Ag⁺(aq) → Mg²⁺(aq) + 2Ag(s). (Mg²⁺/Mg −2.37 V; Ag⁺/Ag +0.80 V) (+3.17 V)
- A reaction with n = 1 has E° = +0.50 V. What is ΔG°? (−48.2 kJ mol⁻¹)
- Is Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g) favoured at standard conditions? (Zn²⁺/Zn −0.76 V) (Yes: E° = +0.76 V, ΔG° = −147 kJ mol⁻¹)
Next: practice questions.