The electrolysis of sodium chloride solution
4 marks · NormalThe table shows some possible half-equations for reduction at the cathode during the electrolysis of aqueous sodium chloride.
| Equation | Product formed at cathode |
|---|---|
| 2Na^+(aq) + 2e^- → 2Na(s) | |
| 2H^+(aq) + 2e^- → H2(g) | |
| 2H2O(l) + 2e^- → H2(g) + 2OH^-(aq) |
Identify which equation(s) correctly represent the cathode reaction for the electrolysis of sodium chloride solution and explain your choice.
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(a) 2Na^+(aq) + 2e^- → 2Na(s): sodium metal; 2H^+(aq) + 2e^- → H2(g): hydrogen gas; 2H2O(l) + 2e^- → H2(g) + 2OH^-(aq): hydrogen gas + hydroxide ions
(b) Equations 2 and 3 are correct for the cathode in sodium chloride solution (hydrogen produced); sodium is not formed because it is more reactive than hydrogen and remains as ions.
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Marking points
B1 (a) · 2 marks · Correctly fills in all products: (i) sodium metal, (ii) hydrogen gas, (iii) hydrogen gas + hydroxide ions.
B2 (b) · 1 mark · Selects equations 2 and/or 3 as correct for the cathode in this context.
B3 (b) · 1 mark · Gives reason: hydrogen is formed because it is less reactive than sodium; sodium is not formed as it is more reactive.
Worked solution
| Equation | Product formed at cathode |
|---|---|
| 2Na^+(aq) + 2e^- → 2Na(s) | sodium metal |
| 2H^+(aq) + 2e^- → H2(g) | hydrogen gas |
| 2H2O(l) + 2e^- → H2(g) + 2OH^-(aq) | hydrogen gas and hydroxide ions |
Equations 2 and 3 are correct for the cathode reaction in sodium chloride solution because hydrogen is produced, not sodium. Hydrogen ions (or water molecules) are reduced because hydrogen is less reactive than sodium.