5.5.4Physical Chemistry and Transition Elements
Electrode Potential Calculations — OCR A Level Chemistry A H432 Revision Notes & Topical Questions
5 min read 5 parts Lesson 32 of 40
Section 1 of 5Calculating E⦵cell
The cell potential is the difference between the two half-cell potentials.
Formula
- E⦵cell = E⦵(positive electrode) − E⦵(negative electrode), where the positive electrode has the more positive potential.
Positive electrode
- Reduction occurs here, and electrons arrive from the external circuit.
Negative electrode
- Oxidation occurs here, and electrons leave into the external circuit.
Cell notation
- Write the negative half-cell on the left and the positive on the right, with || for the salt bridge: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s).
Section 2 of 5Predicting Feasibility
A positive E⦵cell shows that a reaction is thermodynamically feasible under standard conditions.
1
Write the two half-equations as reductions with their E⦵ values.
2
Identify the more negative E⦵ value; this half-equation will run backwards as an oxidation.
3
Combine the half-equations after balancing electrons to produce the overall equation.
4
Calculate E⦵cell = E⦵(reduction) − E⦵(oxidation); if it is positive the reaction is feasible.
5
State the result as feasible, but remember that it does not mean the reaction is fast.
Section 3 of 5Limitations of Predictions
Limitations of Predictions
Non-standard conditions
- Changes in concentration, temperature or pressure alter electrode potentials, so a prediction may fail.
Rate of reaction
- A reaction with positive E⦵cell may be extremely slow because the activation energy is high.
Small E⦵cell
- Values near zero are the most affected by changes in concentration, so predictions are the least reliable.
Practical example
- MnO₄⁻ can oxidise chloride ions (E⦵cell = 1.51 − 1.36 = +0.15 V),
- which is why hydrochloric acid cannot be used to acidify manganate(VII) titrations.
Section 4 of 5Worked Example
Question Calculate E⦵cell for the Zn/Cu cell, then decide whether Fe³⁺(aq) can oxidise I⁻(aq), given E⦵(I₂/I⁻) = +0.54 V.
- Zn/CuE⦵cell = (+0.34) − (−0.76) = +1.10 V.
- Fe³⁺ + e⁻ ⇌ Fe²⁺ has E⦵ = +0.77 V; I₂ + 2e⁻ ⇌ 2I⁻ has E⦵ = +0.54 V.
- I⁻ is oxidised (the more negative E⦵), so E⦵cell = 0.77 − 0.54 = +0.23 V.
- The reaction 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂ is feasible.
Answer
- E⦵cell (Zn/Cu) = +1.10 V;
- Fe³⁺ can oxidise I⁻ (E⦵cell = +0.23 V).
Section 5 of 5Exam Checklist & Tips
5.5.4 – Key Points for the Exam
- E⦵cell The standard cell potential, E⦵(positive) − E⦵(negative).
- Cell notation A shorthand showing electrodes, species and the salt bridge in a cell.
- Feasible reaction One with a positive E⦵cell, so thermodynamically possible under standard conditions.
- Oxidising agent A species that is reduced in a reaction, taking electrons from another species.
Common Mistakes
- Subtracting the potentials the wrong way round and obtaining a negative E⦵cell for a working cell.
- Multiplying E⦵ values by the number of electrons when balancing equations.
- Treating a positive E⦵cell as proof that the reaction occurs quickly.
- Writing the salt bridge as a single line instead of ||.
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