This mini-lesson covers AQA 3.1.7 Oxidation, reduction and redox equations and 3.1.11 Electrode potentials and electrochemical cells: oxidation states, half-equations, the standard hydrogen electrode, E°cell and feasibility, and batteries and fuel cells.
Work through each screen, answer the questions as you go (some are wordy, most are calculations) and collect ⭐ stars. Everything here is A-level standard — the maths is done properly, not skipped. Press Start when you're ready.
Redox · oxidation states
Oxidation states and half-equations
The oxidation state is the charge an atom would have if all its bonds were ionic. Learn these rules in order:
An uncombined element is 0 (Cl₂, Na, O₂, S₈).
Group 1 = +1, Group 2 = +2, F = −1 always.
O = −2 (except peroxides, −1, and in OF₂, +2). H = +1 (except in metal hydrides, −1).
The oxidation states in a neutral compound sum to 0; in an ion they sum to the charge.
Oxidation Is Loss of electrons (oxidation state increases). Reduction Is Gain (oxidation state decreases). The oxidising agent is the species that is itself reduced.
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂Obalance atoms → then O with H₂O → then H with H⁺ → then charge with electrons. Check: left charge = −1 + 8 − 5 = +2 = right ✓
Combining half-equations: multiply each one so the electrons cancel exactly, then add. For manganate(VII) and iron(II): MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O (charge: +17 on both sides ✓).
Calculate
Your turn
1Calculate the oxidation state of manganese in MnO₄⁻.
Hint: Let it be x. x + 4(−2) = −1 (the overall charge).
Calculate
Your turn
2Calculate the oxidation state of chromium in Cr₂O₇²⁻.
Hint: 2x + 7(−2) = −2, so 2x = 12.
Sort it
Oxidised, reduced, or neither?
Tap a species or half-equation, then tap what is happening to it.
🟩 Oxidised (loses e⁻)
🟪 Reduced (gains e⁻)
🟦 Neither (spectator)
Redox · electrode potentials
Half-cells and the standard hydrogen electrode
A half-cell is a metal in a solution of its own ions. On its own it has an electrode potential we cannot measure absolutely — only relative to a reference. That reference is the standard hydrogen electrode (SHE), defined as exactly 0.00 V.
The SHE: hydrogen gas at 100 kPa, bubbled over a platinum electrode (inert, and catalyses the equilibrium), dipping into 1.00 mol dm⁻³ H⁺(aq), at 298 K.
Standard conditions for any half-cell: all solutions at 1.00 mol dm⁻³, any gas at 100 kPa, temperature 298 K.
A more negative E° means the species is a better reducing agent — it is more readily oxidised (loses electrons).
A more positive E° means it is a better oxidising agent — it is more readily reduced.
The salt bridge (e.g. KNO₃ in agar) completes the circuit and balances the charge without the solutions mixing.
A high-resistance voltmeter is used so that negligible current flows — otherwise the concentrations would change and the reading would drift.
Redox · E°cell
Calculating E°cell and judging feasibility
E°cell = E°(reduced half-cell) − E°(oxidised half-cell)equivalently: the more positive E° minus the less positive E° — so E°cell always comes out positive for a spontaneous cell
Method: the half-cell with the more positive E° runs as a reduction (it takes the electrons); the other runs backwards, as an oxidation. Reverse that half-equation, add, and cancel the electrons.
Worked example — the Daniell cell
Zn²⁺ + 2e⁻ ⇌ Zn, E° = −0.76 V · Cu²⁺ + 2e⁻ ⇌ Cu, E° = +0.34 V
Copper is more positive → copper is reduced; zinc is oxidised.
E°cell = (+0.34) − (−0.76) = +1.10 V
Overall: Zn + Cu²⁺ → Zn²⁺ + Cu (charge: +2 on each side ✓)
The two limits of feasibility. A positive E°cell means the reaction is thermodynamically feasible — but (1) it says nothing about the rate: a large activation energy can make a feasible reaction immeasurably slow; and (2) it only applies under standard conditions — change the concentrations and the potential shifts.
Calculate
Your turn
3Zn²⁺ + 2e⁻ ⇌ Zn has E° = −0.76 V; Cu²⁺ + 2e⁻ ⇌ Cu has E° = +0.34 V. Calculate E°cell.
V
Hint: E°cell = (+0.34) − (−0.76).
Calculate
Your turn
4Fe³⁺ + e⁻ ⇌ Fe²⁺ has E° = +0.77 V; I₂ + 2e⁻ ⇌ 2I⁻ has E° = +0.54 V. Calculate E°cell for the reaction that occurs.
V
Hint: The more positive half-cell (Fe³⁺/Fe²⁺) is reduced: E°cell = 0.77 − 0.54.
Quick check
Quick check
?In the cell above, iodide ions and iron(III) ions are mixed. Which species is oxidised?
Quick check
Quick check
?A reaction has E°cell = +1.20 V but nothing happens when the reagents are mixed at room temperature. Why?
Redox · cells & fuel cells
Batteries and fuel cells
Non-rechargeable cells (e.g. alkaline zinc–manganese) run until a reactant is used up and the e.m.f. falls to zero. Rechargeable cells (lithium-ion, lead–acid) run the cell reaction in reverse when an external voltage is applied.
A fuel cell is different: the reactants are supplied continuously from outside, so it never goes flat and does not need recharging.
Overall: 2H₂ + O₂ → 2H₂Othe same reaction as burning hydrogen — but the energy comes out as electricity, not heat
Evaluate honestly: a hydrogen fuel cell produces only water at the point of use and is more efficient than a combustion engine. But hydrogen is hard to store and transport (it is flammable, and liquefying or compressing it costs energy), and it is usually manufactured from fossil fuels — so the overall carbon saving depends entirely on how the hydrogen was made.
Match it
Match the electrochemistry term to its meaning
Tap a term on the left, then its meaning on the right.
Term
Meaning
Quick check
Quick check
?Why is the voltmeter used to measure E°cell a high-resistance voltmeter?
Quick check
Quick check
?What is the main environmental caveat about hydrogen fuel cells?
Redox · exam traps
Half-equations, E° values and what NOT to multiply
E° values are NEVER multiplied. When you scale a half-equation to cancel the electrons, the E° value stays exactly the same — it is an intensive property (a potential per electron), not a total energy.
Balancing a half-equation, in order: balance the main atoms → balance O with H₂O → balance H with H⁺ → balance the charge with electrons. Then check that the charges match.
The oxidising agent is the one that is reduced. Say what happens to the other species, not to itself.
Feasible ≠ fast. A positive E°cell is a thermodynamic statement only.
Conventional cell diagrams are written with the most negative half-cell on the left, and E°cell = E°(right) − E°(left). Set it up that way and E°cell always comes out positive for the spontaneous reaction.
Quick check
Quick check
?A half-equation with E° = +0.77 V is multiplied by 5 to balance the electrons. What is the new E° value?
Calculate
Your turn
5Calculate the oxidation state of sulfur in the thiosulfate ion, S₂O₃²⁻.
Hint: 2x + 3(−2) = −2, so 2x = 4.
Recap
The big ideas to know
Oxidation state: elements = 0 · O usually −2 · H usually +1 · the sum equals the overall charge
OIL RIG: Oxidation Is Loss, Reduction Is Gain — of electrons
Half-equations: balance atoms, then O with H₂O, then H with H⁺, then charge with e⁻
SHE: the zero point of the scale: 1.00 mol dm⁻³ H⁺, H₂ at 100 kPa, 298 K, platinum electrode
E°_cell: = E°(more positive half-cell) − E°(less positive). Positive E°_cell → the reaction is feasible
Feasibility limits: E°_cell says nothing about the RATE — a high activation energy can make a feasible reaction immeasurably slow
Fuel cells: 2H₂ + O₂ → 2H₂O — continuous supply of fuel, only water as waste, no discharge
That is the whole of AQA 3.1.7 and 3.1.11. Press Finish.
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