We came across electrolysis of molten metal salts and of solutions at GCSE, and the good news is, nothing has changed 😀. You do need to know the half equations for the oxidation and reduction of water through, as well as the rules predicting the products of electrolysis of a metal salt in solution.
Electrolysis uses electricity (energy) to drive a non-spontaneous reaction which is useful for recharging batteries, splitting water to make hydrogen and in producing chlorine from brine.
Let’s start by considering the electrolysis of water.
Water is stable with respect to hydrogen and oxygen under standard conditions. Splitting water to make hydrogen is not a feasible / spontaneous reaction as can be seen by the positive change in Gibbs energy and the negative Ecell value.
2H2O(l) ⇾ 2H2(g) + O2(g) ΔrG⦵ = + 474 kJ mol-1 ; Ecell = -1.23V
However, if we supply more than +1.23V to the reaction, we can make it happen.
Now, moving on to the electrolysis of brine, sodium chloride solution. The salt for the solution comes from the mining of rock salt rather than sea water in most cases. This process is the basis of the chlor-alkali industry.
In the industrial set-up, the anode is titanium and the cathode is nickel.
The overall reaction is: 2H2O(l) + 2Cl–(aq) ⇾ Cl2(g) + H2(g) + 2OH–(aq) E⦵cell = -2.19V
We need to supply a potential difference of more than +2.19V to drive this reaction in a non-spontaneous direction.
We can electrolyse the solutions of many metal salts. The products at each electrode can be predicted based on the following rules:
At the anode, either the anion or water will be oxidised
- If the salt is a halide, the halide ion is oxidised in preference to water
2Br–(aq) ⇾ Br2(l) + 2e–
- If the salt is a sulphate or a nitrate, water is oxidised
2H2O(l) ⇾ 4H+(aq) + O2(g) + 4e–
- If the salt is a hydroxide, the hydroxide ion is oxidised
4OH–(aq) ⇾ O2(g) + 2H2O(l) + 2e–
At the cathode, either the cation or water will be reduced
- If the metal is from group 1, group 2 or is aluminium, water is reduced and the metal ion remains in solution
2H2O(l) + 2e– ⇾ H2(g) + 2OH–(aq)
- For all other metals, the metal ion is reduced and deposited on the cathode
Cu2+(aq) + 2e– ⇾ Cu(s)
- If we are electrolysing an acid, the H+ ions are reduced
2H+(aq) + 2e– ⇾ H2(g)
Practice questions
1. Use the rules above to predict the products at the anode and cathode for the electrolysis of the following solutions with graphite electrodes. Write half equations for the reactions happening.
(a) KBr
(b) Al(NO3)3
(c) HNO3
(d) Mg(OH)2
(e) CuI2
2. Which of the following is correct?
| Electrolyte | Product at anode | Product at cathode | |
| A | MgI2 (l) | iodine | magensium |
| B | AgNO3 (aq) | oxygen | hydrogen |
| C | NaCl (aq) | chlorine | sodium |
| D | PbS (l) | hydrogen sulphide | lead |
3. The apparatus shown below was used to investigate the electrolysis of the sodium halides in solution.
(a) What would be observed at the anode and cathode during the electrolysis of aqueous sodium iodide?
(b) Write half equations for the reactions happening at each electrode during the electrolysis of sodium bromide.
4. Chlorine is manufactured by the electrolysis of sodium chloride solution.
2NaCl(aq) + 2H2O(l) ⇾ Cl2(g) + H2(g) + NaOH(aq)
Give the half equation for the reaction at the negative electrode and explain why this is a reduction reaction.
Answers
1.
(a) 2H2O(l) + 2e– ⇾ H2(g) + 2OH–(aq) and 2Br–(aq) ⇾ Br2(l) + 2e–
(b) 2H2O(l) + 2e– ⇾ H2(g) + 2OH–(aq) and 2H2O(l) ⇾ 4H+(aq) + O2(g) + 4e–
(c) 2H+(aq) + 2e– ⇾ H2(g) and 2H2O(l) ⇾ 4H+(aq) + O2(g) + 4e–
(d) 2H2O(l) + 2e– ⇾ H2(g) + 2OH–(aq) and 4OH–(aq) ⇾ O2(g) + 2H2O(l) + 2e–
(e) Cu2+(aq) + 2e– ⇾ Cu(s) and 2I–(aq) ⇾ I2(aq) + 2e–
2. A
3. (a) at the anode you would see a yellow / orange / brown colour in solution
(b) at the anode: 2Br–(aq) ⇾ Br2(l) + 2e–
at the cathode: 2H2O(l) + 2e– ⇾ H2(g) + 2OH–(aq)
4.  2H2O(l) + 2e– ⇾ H2(g) + 2OH–(aq) ; this is reduction because water gains two electrons
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