Electrolysis: Faraday's laws and products of electrolysis
In electrolysis, 1 F (96487 C) is the charge on one mole of electrons. Mass deposited is proportional to charge; the products depend on E° values and overpotential.
Electrolytic cells
In an electrolytic cell, an outside source of electricity drives a reaction that would not happen by itself. The cathode is joined to the negative terminal and the anode to the positive terminal. Refining of copper is an example. With CuSO₄ solution and copper electrodes, Cu²⁺(aq) + 2e⁻ → Cu(s) at the cathode and Cu(s) → Cu²⁺(aq) + 2e⁻ at the anode. So impure copper from the anode ends up as pure copper on the cathode. Michael Faraday found the laws that give the amounts.
Worked example: Faraday's first law
A current of 5.0 A is passed through CuSO₄ solution for 32 minutes 10 seconds. What mass of copper is deposited at the cathode? (Cu = 63.5 g mol⁻¹; F = 96500 C mol⁻¹)
- t = 32 × 60 + 10 = 1930 s.
- Q = I × t = 5.0 A × 1930 s = 9650 C.
- Moles of electrons = 9650 ÷ 96500 = 0.10 mol.
- Cu²⁺ + 2e⁻ → Cu: 2 mol of electrons give 1 mol of Cu, so moles of Cu = 0.10 ÷ 2 = 0.050 mol.
- Mass of Cu = 0.050 mol × 63.5 g mol⁻¹ = 3.175 g ≈ 3.18 g.
About 3.18 g of copper
Worked example: the second law, and the charge needed
(a) Cells of AgNO₃ and CuSO₄ solution are joined in series, and 1.08 g of silver is deposited. Find the mass of copper deposited (Ag = 108 g mol⁻¹, Cu = 63.5 g mol⁻¹). (b) How many coulombs are needed to obtain 27 g of aluminium from molten Al₂O₃ (Al = 27 g mol⁻¹)?
- (a) The same charge flows through both cells. Ag⁺ + e⁻ → Ag: 1.08 g = 0.010 mol of Ag, which needs 0.010 mol of electrons.
- Cu²⁺ + 2e⁻ → Cu: 0.010 mol of electrons gives 0.0050 mol of Cu = 0.0050 × 63.5 = 0.3175 g.
- Check with equivalent masses: 1.08 ÷ 0.3175 = 3.40, and 108 ÷ 31.75 = 3.40. They match.
- (b) Al³⁺ + 3e⁻ → Al: 27 g = 1 mol of Al, which needs 3 mol of electrons.
- Charge = 3 × 96500 = 289500 C = 2.895 × 10⁵ C.
(a) About 0.318 g of copper; (b) 2.895 × 10⁵ C, which is 3 F
What decides the products
The products at each electrode depend on three things: the state of the electrolyte (molten or in water), the kind of electrode (inert platinum or a reactive metal) and the E° values of the competing reactions. In water, H₂O itself can be reduced or oxidised, so it competes with the ions. Some electrode reactions are slow and need an extra voltage beyond E°, called overpotential. Oxygen evolution has a high overpotential, which is why Cl₂, not O₂, forms at the anode when brine is electrolysed.
Products of electrolysis
| Electrolyte (electrodes) | At the cathode | At the anode |
|---|---|---|
| Molten NaCl (inert) | Na⁺ + e⁻ → Na | Cl⁻ → ½Cl₂(g) + e⁻ |
| Aqueous NaCl (inert) | H₂O + e⁻ → ½H₂(g) + OH⁻ (E° = −0.83 V, easier than Na⁺/Na at −2.71 V) | Cl⁻ → ½Cl₂(g) + e⁻ (O₂ needs a high overpotential) |
| Dilute H₂SO₄ (Pt) | H⁺ + e⁻ → ½H₂(g) | 2H₂O → O₂(g) + 4H⁺ + 4e⁻ |
| Concentrated H₂SO₄ (Pt) | H⁺ + e⁻ → ½H₂(g) | 2SO₄²⁻ → S₂O₈²⁻ + 2e⁻ (peroxodisulphate) |
| Aqueous CuSO₄ (Pt) | Cu²⁺ + 2e⁻ → Cu(s) | 2H₂O → O₂(g) + 4H⁺ + 4e⁻ |
| Aqueous CuSO₄ (Cu) | Cu²⁺ + 2e⁻ → Cu(s) | Cu(s) → Cu²⁺ + 2e⁻ (the anode dissolves) |
Rules to predict the products
- Cathode (reduction): the species with the higher E° is reduced. Ions of very reactive metals such as Na⁺, K⁺, Ca²⁺ and Al³⁺ are not reduced from water; H₂ forms instead.
- Anode (oxidation) with inert electrodes: the species oxidised most easily is oxidised, allowing for overpotential. Cl⁻ gives Cl₂ from brine; solutions of sulphates and nitrates give O₂.
- With a reactive anode such as copper or silver, the anode metal itself dissolves.
- Overall for brine: NaCl(aq) + H₂O(l) → Na⁺(aq) + OH⁻(aq) + ½H₂(g) + ½Cl₂(g). This is how NaOH is made.
Electroplated steel taps and the gold shine on imitation jewellery come from electrolysis. Electrolysis of brine makes the NaOH and chlorine used to make soap and to disinfect drinking water.
3 marks: “How much charge, in coulombs, is needed to reduce (i) 0.5 mol of Cu²⁺ to Cu and (ii) 1 mol of Cr₂O₇²⁻ to Cr³⁺? (F = 96500 C mol⁻¹)” (i) Cu²⁺ + 2e⁻ → Cu: 0.5 mol needs 1 mol of electrons = 1 F = 96500 C. (ii) Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O: 6 F = 6 × 96500 = 5.79 × 10⁵ C.
2 marks: “Predict the products of electrolysis of aqueous NaCl with inert electrodes. Why is Cl₂, and not O₂, set free at the anode?” Cathode: H₂ (water is reduced, E° = −0.83 V, more easily than Na⁺, −2.71 V). Anode: Cl₂. By E° alone water should be oxidised, but O₂ evolution needs a high overpotential, so Cl⁻ is oxidised instead.
2 marks: “State Faraday's two laws of electrolysis.” (1) The mass of a substance deposited or set free at an electrode is proportional to the quantity of electricity passed. (2) For the same quantity of electricity, the masses of different substances set free are proportional to their equivalent masses.
One mole of electrons = one Faraday ≈ 96500 C. Count the electrons in the half-reaction first: Ag needs 1, Cu needs 2, Al needs 3.