Join two half-cells and the more positive one pulls electrons from the other. Subtract the two E values to get the cell potential, use its sign to predict whether a reaction is feasible, and convert it into a free energy change with ΔG = −nFE.
When two half-cells are joined, the one with the more positive E⊖ is the positive electrode: reduction happens there. The other is oxidised, and electrons flow through the wire from the negative to the positive electrode. The standard cell potential is E⊖cell = E⊖(positive) − E⊖(negative), e.g. Zn/Cu: +0.34 − (−0.76) = +1.10 V. A reaction is feasible when E⊖cell for it, as written, is positive. E values say nothing about rate (a feasible reaction may have a high activation energy) and apply only under standard conditions. The cell potential links to free energy: ΔG⊖ = −nFE⊖cell, so a positive Ecell gives a negative ΔG.
Pair two half-cells in the simulator and watch which electrode gives electrons to which. Compare what you see with the E⊖ ladder.
| Cell | Positive electrode (reduction) | Negative electrode (oxidation) | E⊖cell / V |
|---|---|---|---|
| Zn | Zn2+ || Cu2+ | Cu | Cu2+ + 2e− → Cu | Zn → Zn2+ + 2e− | +0.34 − (−0.76) = +1.10 |
| Cu | Cu2+ || Ag+ | Ag | Ag+ + e− → Ag | Cu → Cu2+ + 2e− | +0.80 − 0.34 = +0.46 |
| Pt | I−, I2 || Fe3+, Fe2+ | Pt | Fe3+ + e− → Fe2+ | 2I− → I2 + 2e− | +0.77 − 0.54 = +0.23 |
Four quick checks. Each correct answer earns XP and records this skill on your profile.
What is E⊖cell for a cell made from Zn2+/Zn (−0.76 V) and Cu2+/Cu (+0.34 V)?
In the Cu2+/Cu and Ag+/Ag cell, which electrode is positive?
When is a redox reaction predicted to be feasible from E⊖ values?
Which relationship links the cell potential to the free energy change?
For a proposed reaction, put the species being reduced as the positive electrode. If E⊖cell comes out positive, the reaction is feasible.
| Proposed reaction | E⊖cell / V | Feasible? | ΔG⊖ = −nFE⊖cell |
|---|---|---|---|
| Zn + Cu2+ → Zn2+ + Cu | +1.10 | yes | −2 × 96 500 × 1.10 = −212 kJ mol−1 |
| 2Fe3+ + 2I− → 2Fe2+ + I2 | +0.23 | yes | −2 × 96 500 × 0.23 = −44.4 kJ mol−1 |
| MnO4− oxidising Cl− to Cl2 | 1.52 − 1.36 = +0.16 | yes: so KMnO4 is not acidified with HCl | n = 10: −154 kJ mol−1 |
| Cr2O72− oxidising Cl− to Cl2 | 1.33 − 1.36 = −0.03 | no (under standard conditions) | positive |
E⊖cell = E⊖(positive) − E⊖(negative): never add them, and never multiply E⊖ by the coefficients used to balance electrons. Electrons flow in the external wire, from negative to positive; ions, not electrons, carry the current through the salt bridge. In ΔG = −nFE, n is moles of electrons, and the answer in J must be converted to kJ. A feasible reaction may still be too slow to see.
Unlocks once the four checks above are done. Worth more XP, written to A Level Paper 4 standard.
Using E⊖(Fe3+/Fe2+) = +0.77 V and E⊖(I2/I−) = +0.54 V, what is ΔG⊖ for 2Fe3+ + 2I− → 2Fe2+ + I2?
Acidified potassium manganate(VII) is not acidified with hydrochloric acid in titrations. Using E⊖ values, why?
E⊖cell for a redox reaction X is +0.50 V, yet no change is seen when the reagents are mixed at room temperature. What is the most likely reason?
Which cell diagram has E⊖cell = +0.46 V?
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