Electrochemical cells worksheet answers for A Level chemistry (free PDF download)

Problem 1 Anode, cathode and salt bridge

1. The anode is the magnesium electrode, where oxidation happens (Mg → Mg²⁺ + 2e⁻).
2. Electrons flow from magnesium (anode) to copper (cathode) through the wire.
3. The salt bridge completes the circuit and lets ions move to keep both half-cells electrically neutral.

Problem 2 Writing cell notation

1. Magnesium has the more negative E°, so it is the anode; nickel is the cathode.
2. Mg(s) | Mg²⁺(aq) ‖ Ni²⁺(aq) | Ni(s).
3. E°cell = −0.26 − (−2.37) = +2.11 V.

Problem 3 The reference electrode

1. 298 K, 1.00 mol/dm³ H⁺ and 100 kPa hydrogen gas (with a platinum electrode).
2. 2H⁺(aq) + 2e⁻ ⇌ H₂(g).
3. E° = 0.00 V by definition.

Problem 4 Emf of a zinc–silver cell

1. E°cell = +0.80 − (−0.76) = +1.56 V.
2. Zn + 2Ag⁺ → Zn²⁺ + 2Ag.
3. Zinc dissolves, because it is oxidised at the anode.

Problem 5 Charge and moles of electrons

1. t = 90.0 × 60 = 5400 s; Q = 0.250 × 5400 = 1350 C.
2. n(e⁻) = 1350 ÷ 96 500 = 0.0140 mol.

Problem 6 Will Fe³⁺ or H⁺ oxidise the metal?

1. Reduction Fe³⁺ (+0.77), oxidation Cu (+0.34): E°cell = +0.43 V, feasible.
2. Reduction H⁺ (0.00), oxidation Ag (+0.80): E°cell = 0.00 − 0.80 = −0.80 V, not feasible.
3. 2Fe³⁺ + Cu → 2Fe²⁺ + Cu²⁺.

Problem 7 Halogens and halide ions

1. E°cell = 1.36 − 1.07 = +0.29 V, feasible.
2. E°cell = 0.54 − 1.36 = −0.82 V, not feasible.
3. Cl₂, then Br₂, then I₂, in order of decreasing E°.

Problem 8 Hydrogen fuel cell half-equations

1. H₂ → 2H⁺ + 2e⁻.
2. O₂ + 4H⁺ + 4e⁻ → 2H₂O.
3. Double the first equation: 2H₂ + O₂ → 2H₂O; E°cell = 1.23 − 0.00 = +1.23 V.

Problem 9 Hydrogen consumed by a fuel cell

1. Q = 5.0 × 3600 = 18 000 C.
2. n(e⁻) = 18 000 ÷ 96 500 = 0.1865 mol; each H₂ gives 2e⁻, so n(H₂) = 0.0933 mol.
3. m = 0.0933 × 2.016 = 0.188 g; V = 0.0933 × 24.0 = 2.24 dm³.

Problem 10 Plating copper

1. t = 1500 s; Q = 1.25 × 1500 = 1875 C.
2. n(e⁻) = 1875 ÷ 96 500 = 0.01943 mol; n(Cu) = 0.01943 ÷ 2 = 0.009715 mol.
3. m = 0.009715 × 63.5 = 0.617 g.

Problem 11 Electrolysis of brine

1. Cathode: hydrogen; anode: chlorine.
2. Cathode: 2H₂O + 2e⁻ → H₂ + 2OH⁻. Anode: 2Cl⁻ → Cl₂ + 2e⁻.
3. Water (−0.83 V) is reduced more easily than Na⁺ (−2.71 V), since the higher E° is reduced first.
4. Aqueous sodium hydroxide, from Na⁺ and OH⁻.

Problem 12 Diluting one half-cell

1. The emf decreases a little (it is about 0.06 V lower).
2. Lower [Cu²⁺] shifts the equilibrium to the left, so Cu²⁺ is less strongly attracted to electrons and the copper half-cell becomes less positive. The difference between the two half-cells is smaller, so the emf falls.

Problem 13 Identify metals from cell voltages

1. X: E° = 0.34 − 1.10 = −0.76 V. Y: 0.34 − 0.47 = −0.13 V. Z is the cathode against negative copper: E° = 0.34 + 0.46 = +0.80 V.
2. X is zinc, Y is lead, Z is silver.

Problem 14 A cell with a platinum electrode

1. Both Fe³⁺ and Fe²⁺ are in solution, so an inert conducting electrode is needed to carry electrons.
2. E°cell = 0.77 − (−0.76) = +1.53 V.
3. Electrons flow from Zn to Pt. Overall: Zn + 2Fe³⁺ → Zn²⁺ + 2Fe²⁺.

Problem 15 Ion movement in the salt bridge

1. Zn becomes Zn²⁺, so the zinc half-cell gains positive charge; the copper half-cell loses Cu²⁺ and becomes negative.
2. K⁺ moves towards the copper half-cell and NO₃⁻ moves towards the zinc half-cell.
3. A metal wire carries electrons only; the solutions need ions to flow to complete the circuit and cancel the charge build-up.

Problem 16 Silver nitrate electrolysis

1. t = 10 800 s; Q = 0.250 × 10 800 = 2700 C; n(e⁻) = 2700 ÷ 96 500 = 0.02798 mol.
2. Ag⁺ + e⁻ → Ag, so n(Ag) = 0.02798 mol; m = 0.02798 × 107.9 = 3.02 g.
3. 2H₂O → O₂ + 4H⁺ + 4e⁻; n(O₂) = 0.02798 ÷ 4 = 0.006995 mol; V = 0.006995 × 24.0 = 0.168 dm³ = 168 cm³.

Problem 17 Charge on a metal ion

1. t = 1500 s; Q = 0.800 × 1500 = 1200 C; n(e⁻) = 1200 ÷ 96 500 = 0.01244 mol.
2. n(M) = 0.699 ÷ 112.4 = 0.006219 mol.
3. Ratio = 0.01244 ÷ 0.006219 = 2.00, so each atom needs 2 electrons: the ion is M²⁺.

Problem 18 Efficiency of copper plating

1. Q = 2.00 × 1200 = 2400 C; n(e⁻) = 0.02487 mol; n(Cu) = 0.012435 mol; m = 0.790 g.
2. Efficiency = 0.760 ÷ 0.790 × 100 = 96.2 %.
3. Some current reduces H⁺ to hydrogen at the cathode; some copper may fall off the spoon or dissolve again; the current may not have been steady.

Problem 19 Alkaline hydrogen fuel cell

1. Double the anode equation: 2H₂ + 4OH⁻ → 4H₂O + 4e⁻. Add the cathode equation and cancel: 2H₂ + O₂ → 2H₂O.
2. E°cell = 0.40 − (−0.83) = +1.23 V, the same as in acid.
3. OH⁻ moves from the cathode to the anode, because OH⁻ is made at the cathode and used up at the anode.
4. A fuel cell converts chemical energy directly into electrical energy with fewer losses (typically 40–60 % efficient) than burning hydrogen to produce heat and then work.

Problem 20 Spot the errors

1. The formula is upside down: it must be E°(cathode) − E°(anode).
2. E°cell = 0.34 − (−0.76) = +1.10 V, so the cell works.
3. E° is a potential per unit charge, an intensive property. Doubling the amounts of Cu²⁺ and electrons doubles the energy but not the energy per coulomb, so E° stays +0.34 V and the emf stays 1.10 V.