Eduqas PI1 puts numbers on redox. You will assign oxidation numbers, build half-equations, measure standard electrode potentials against the hydrogen electrode, calculate E°cell and use it to judge feasibility — then apply it to fuel cells and redox titrations.
Work through each screen, answer every question as you go — the multiple-choice checks, the calculations and the sorting games — and collect ⭐ stars. Press Start when you are ready.
PI1 · oxidation numbers
Oxidation numbers
Rules, applied in order of priority:
Uncombined element = 0. Simple ion = its charge.
Group 1 = +1; Group 2 = +2; F = −1 always.
H = +1 (but −1 in metal hydrides such as NaH).
O = −2 (but −1 in peroxides, and +2 in F₂O).
The sum equals 0 for a neutral compound, or the overall charge for an ion.
Worked example — sulfur in SO₄²⁻
x + 4(−2) = −2 → x − 8 = −2 → x = +6
Oxidation = loss of electrons = oxidation number increases. Reduction = gain = oxidation number decreases. An oxidising agent is itself reduced.
Calculate
Your turn — manganate(VII)
1What is the oxidation number of manganese in the manganate(VII) ion, MnO₄⁻ ?
(sign + number)
Hint: x + 4(−2) = −1, so x − 8 = −1.
Calculate
Your turn — dichromate(VI)
2What is the oxidation number of chromium in the dichromate(VI) ion, Cr₂O₇²⁻ ?
(sign + number)
Hint: 2x + 7(−2) = −2, so 2x − 14 = −2 and 2x = 12.
PI1 · half-equations
Building half-equations in acid
Balance a half-equation in this order — O, H, charge:
To combine two half-equations, scale them so the electrons cancel, then add.
Quick check
Check the half-equation
?Which is the correctly balanced half-equation for the reduction of MnO₄⁻ to Mn²⁺ in acid?
PI1 · electrode potentials
Standard electrode potentials
Every half-cell has an electrode potential, but you can only ever measure a difference. So all values are quoted against the standard hydrogen electrode (SHE), defined as exactly 0.00 V.
Standard conditions:298 K, 100 kPa, all solutions at 1.00 mol dm⁻³, and an inert platinum electrode where there is no metal (e.g. for Fe³⁺/Fe²⁺).
Zn(s) | Zn²⁺(aq) ‖ Cu²⁺(aq) | Cu(s)cell notation: oxidation on the left, reduction on the right; ‖ is the salt bridge
A more negative E° means the system sits further to the left — it releases electrons more readily, so it is the better reducing agent. A more positive E° means it is the better oxidising agent.
Zn²⁺ + 2e⁻ ⇌ Zn, E° = −0.76 V · Cu²⁺ + 2e⁻ ⇌ Cu, E° = +0.34 V
Copper has the more positive E°, so copper is reduced and zinc is oxidised.
E°cell = (+0.34) − (−0.76) = +1.10 V
Never multiply an E° value by the number of electrons. E° is an intensive property — doubling the half-equation does not double the voltage.
Calculate
Your turn — E°cell
3Calculate E°cell for a cell made from Mg²⁺/Mg (E° = −2.37 V) and Ag⁺/Ag (E° = +0.80 V). Give the value for the spontaneous (feasible) direction.
V
Hint: silver has the more positive E°, so it is reduced. E°cell = (+0.80) − (−2.37).
Calculate
Your turn — a redox pair
4MnO₄⁻/Mn²⁺ has E° = +1.51 V and Fe³⁺/Fe²⁺ has E° = +0.77 V. Calculate E°cell when manganate(VII) oxidises iron(II).
V
Hint: MnO₄⁻ is reduced (more positive E°), so E°cell = (+1.51) − (+0.77).
Quick check
Is it feasible?
?What does a positive E°cell tell you about a reaction?
Sort it
Rank the redox systems
Tap a half-cell, then the box that describes it. Look only at the sign and size of E°.
💪 Powerful oxidising agent (E° > +1.0 V)
⚖️ Moderate oxidising agent (0 to +1.0 V)
🔋 Reducing agent (E° negative)
PI1 · limitations
The limitations of E° predictions
E°cell tells you whether a reaction can go, never whether it will. Three big caveats:
Kinetics: a large activation energy can make a feasible reaction immeasurably slow. Thermodynamics is silent about rate.
Non-standard conditions: real concentrations are rarely 1.00 mol dm⁻³, and changing them shifts the electrode potential (Le Chatelier applied to the half-cell equilibrium).
Temperature: E° values are only valid at 298 K.
The classic example: the oxidation of glucose by oxygen has a hugely positive E°cell, but sugar sits happily in a bowl for years — the activation energy is far too high without an enzyme.
Quick check
Feasible but nothing happens
?A reaction has E°cell = +1.20 V, yet no reaction is observed when the reagents are mixed at room temperature. What is the most likely explanation?
PI1 · fuel cells
Cells, batteries and fuel cells
In a hydrogen–oxygen fuel cell the reactants are fed in continuously, so it never goes flat.
In alkaline conditions:
Negative electrode: H₂ + 2OH⁻ → 2H₂O + 2e⁻ Positive electrode: ½O₂ + H₂O + 2e⁻ → 2OH⁻negative electrode charge: (−2) on the left = (−2) on the right ✓ · positive electrode: O 1 + 1 = 2 · H 2 = 2 · charge (−2) = (−2) ✓
Overall: 2H₂ + O₂ → 2H₂OH: 4 = 4 · O: 2 = 2 — the only product is water
Advantages: the only product is water; higher efficiency than a combustion engine; no CO₂ at the point of use.
Drawbacks: hydrogen is hard to store and transport; it is usually made from methane or by electrolysis, so the CO₂ may simply be moved elsewhere; expensive catalysts are needed.
Quick check
The fuel cell equation
?What is the overall equation for a hydrogen–oxygen fuel cell?
PI1 · redox titrations
Redox titrations
Manganate(VII) titrations are self-indicating: MnO₄⁻ is deep purple and Mn²⁺ is almost colourless, so the end point is the first permanent pale pink.
Acidify with dilute sulfuric acid — never with HCl (Cl⁻ would be oxidised to Cl₂) and never with nitric acid (it is itself an oxidising agent).
Iodine–thiosulfate titrations use starch, added near the end point, which turns blue-black then colourless:
I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻the ratio is 1 I₂ : 2 thiosulfate
Calculate
Your turn — manganate(VII) titration
525.0 cm³ of an Fe²⁺ solution needed 21.6 cm³ of 0.0200 mol dm⁻³ KMnO₄ for the first permanent pink. Using the 1 : 5 ratio, calculate the concentration of the Fe²⁺, to 3 significant figures.
6The iodine liberated in a flask needed 24.0 cm³ of 0.100 mol dm⁻³ sodium thiosulfate. Using I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻, calculate the amount of I₂ present, in millimoles (mmol).
mmol
Hint: n(S₂O₃²⁻) = 0.0240 × 0.100 = 2.40 × 10⁻³ mol = 2.40 mmol. The ratio is 2 : 1, so halve it.
Match it
Electrochemistry vocabulary
Tap a definition on the left, then the term on the right.
Definition
Term
Recap
The big ideas to know
Oxidation numbers: sum to 0 (compound) or to the charge (ion); O = −2, H = +1 (with exceptions)
Half-equations: balance O with H₂O, H with H⁺, then charge with electrons
E° values: measured against the SHE (0.00 V) at 298 K, 100 kPa, 1.00 mol dm⁻³
E°cell: E°(reduced) − E°(oxidised). Never multiply E° by the number of electrons
Feasibility: positive E°cell means feasible (ΔG negative) — but says NOTHING about rate
Fuel cells: 2H₂ + O₂ → 2H₂O; only water at the point of use
Titrations: MnO₄⁻/Fe²⁺ is 1 : 5, self-indicating, acidify with dilute H₂SO₄ · I₂/thiosulfate is 1 : 2
That is the whole of PI1. Press Finish to see your score.
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