KCSE 2025 Chemistry P2 Q6 — Electrolysis, Faraday's Law and Electrochemical Cells
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The Question
“(a) Copper(II) sulphate solution can be electrolysed to give different products depending on the electrodes used. State the type of electrodes needed to obtain (i) oxygen and copper and (ii) copper only. (b) In an electrolysis experiment, a current of 15 A was passed for 2 hours (7200 s) through a solution of a metal salt, depositing 19.4 g of metal A at the cathode according to A³⁺(aq) + 3e⁻ → A(s). Using 1 F = 96 500 C, calculate the relative atomic mass of A. (c) A cell is represented with a double-line symbol between the half-cells. (i) State what the double line represents. (ii) Write the overall equation for a cell made from nickel and silver electrodes. (iii) Given E° of the silver half-cell is +0.80 V and the cell emf is 1.05 V, calculate E° for the nickel half-cell. (iv) Explain how the cell emf predicts whether a reaction is feasible. (d) Explain why aluminium resists corrosion while iron does not.”
(a)(i) Electrodes for oxygen and copper
To get oxygen and copper you use inert electrodes — carbon, graphite or platinum. Being inert, the electrodes themselves do not react. At the anode the water is oxidised to release oxygen gas, while copper is deposited at the cathode.
(a)(ii) Electrodes for copper only
To get copper only, with no oxygen released, you use active copper electrodes. Now the copper anode itself dissolves into solution instead of oxygen being given off, and copper is deposited at the cathode — copper dissolves at one electrode and plates onto the other.
(b) Relative atomic mass of A from Faraday's law
First find the total charge passed, Q = It. Each mole of A needs three moles of electrons (from A³⁺ + 3e⁻ → A), so one mole of A is discharged by 3 Faradays = 3 × 96 500 C. The 19.4 g deposited corresponds to Q coulombs, so scale up to the charge for one mole (3F) to find the mass of one mole — that mass is the relative atomic mass.
(c)(i) What the double line represents
In the cell diagram the double line drawn between the two half-cells stands for the salt bridge. It is the link that completes the circuit by allowing ions to flow between the two half-cells while keeping their solutions separate.
(c)(ii) Overall cell equation for nickel and silver
Nickel is the more reactive metal, so it gives up electrons and dissolves as nickel(II) ions. Each nickel atom loses two electrons but each silver ion needs only one, so two silver ions are reduced for every nickel atom oxidised. The two electrons then balance on each side.
(c)(iii) Standard electrode potential of nickel
The cell emf is the more positive electrode potential minus the more negative one. Silver is the positive electrode at +0.80 V and nickel is the negative electrode. So the emf equals the silver value minus the nickel value; substitute the emf of 1.05 V and solve for the nickel potential.
(c)(iv) How emf predicts feasibility
The rule is simple: if the overall cell emf works out positive, the reaction is feasible and will happen on its own. If the emf comes out negative, the reaction is not feasible. Here the emf is +1.05 V (positive), so the nickel–silver reaction is feasible.
(d) Corrosion of aluminium versus iron
Both metals form an oxide, but the two behave in opposite ways. Aluminium forms a thin, non-porous oxide layer that clings tightly to the surface and seals the metal underneath, so corrosion stops — the layer is protective. Iron's oxide, rust, is porous and flaky; it does not stick but peels away, exposing fresh iron that rusts in turn, so the corrosion keeps eating deeper into the metal.
Final Result
Use inert electrodes (carbon/graphite/platinum) to get oxygen and copper, and active copper electrodes to get copper only. For metal A: Q = 15 × 7200 = 108 000 C; 3F = 289 500 C; RAM = (19.4/108 000) × 289 500 = 52.0. The double line is the salt bridge. The cell reaction is Ni(s) + 2Ag⁺(aq) → Ni²⁺(aq) + 2Ag(s). E° of nickel = 0.80 − 1.05 = −0.25 V. A positive cell emf means the reaction is feasible. Aluminium resists corrosion because its oxide is thin, non-porous and adherent (protective), whereas iron's rust is porous and flaky and peels off, so corrosion continues.
Why this method works
The whole question turns on the movement of electrons. In electrolysis the electrode material decides whether the anode reaction is oxidation of water (inert electrodes) or dissolution of the metal itself (active electrodes), which is why the same electrolyte gives different products. Faraday's law then links charge to moles of electrons: knowing how many electrons each ion needs lets you convert coulombs straight into moles of metal, and hence its relative atomic mass. In a spontaneous cell the same electron flow produces a voltage; a positive emf means electrons flow in the direction written, so the reaction is feasible, and the sign of an unknown electrode potential falls out of E_cell = E_positive − E_negative. Corrosion is the reverse story — whether an oxide protects or destroys depends entirely on whether it forms a coherent barrier (aluminium) or a porous flaky one (iron).
Check the RAM by ratio: 108 000 C is 108 000/96 500 ≈ 1.119 F, which discharges 1.119/3 ≈ 0.373 mol of A; 19.4 g ÷ 0.373 mol ≈ 52 g/mol, confirming RAM ≈ 52.