Calculating E°cell
\[\boxed{E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}}\]
💡 Interpreting the Result
- E°cell > 0 → Reaction is spontaneous (thermodynamically feasible)
- E°cell < 0 → Reaction is non-spontaneous
Link to Gibbs Free Energy
\[\Delta G^\circ = -nFE^\circ_{\text{cell}}\]
n = moles of electrons, F = Faraday constant (96485 C mol⁻¹)
Worked Example
Q: Will Zn reduce Cu²⁺ spontaneously?
Cathode (reduction): Cu²⁺ + 2e⁻ → Cu, E° = +0.34 V
Anode (oxidation): Zn → Zn²⁺ + 2e⁻, E° = -0.76 V
\(E^\circ_{\text{cell}} = (+0.34) - (-0.76) = \textbf{+1.10 V}\)
Positive E°cell, so the reaction IS spontaneous. ✓
⚠️ Common Mistake
Do not reverse the sign of E° when writing the oxidation half-equation. The formula E°cell = E°cathode - E°anode accounts for this automatically.
📋 Exam Tip
E°cell predicts thermodynamic feasibility only, not rate. A positive E°cell does not mean the reaction is fast (it may have a high activation energy).