Electrolysis of Aqueous Solutions Exam Questions

• In molten CuSO4, the only negative ions are sulfate ions (SO4^2-), which are discharged at the anode [1]
• In aqueous CuSO4, both sulfate ions and hydroxide ions (from water) are present at the anode [1]
• Hydroxide ions are preferentially discharged because they are easier to oxidise than sulfate ions, producing oxygen gas [1]

Examiner note: In aqueous solution, water provides H+ and OH- ions in addition to those from the dissolved compound. The discharge rules determine which ions are preferentially discharged at each electrode.

(a) Cathode: hydrogen gas [1] (H+ ions discharged in preference to water) Anode: oxygen gas [1] (OH- ions discharged in preference to SO4^2-)

(b) Cathode: copper metal [1] (Cu2+ ions discharged in preference to H+, because Cu is below H in reactivity series) Anode: oxygen gas [1] (OH- ions discharged in preference to SO4^2-)

Examiner note: At the cathode: if the metal is more reactive than hydrogen, hydrogen is produced. If less reactive, the metal is deposited. At the anode: halide ions (Cl-, Br-, I-) are discharged in preference to OH-; sulfate and nitrate are not.

(a) 2H+ + 2e- → H2 [1]

(b) 2Cl- → Cl2 + 2e- [1]

(c) Cathode: hydrogen gas. Anode: chlorine gas [1]. The solution left behind contains sodium hydroxide (NaOH) [1].

Examiner note: Sodium is too reactive to be discharged, so H+ (from water) is discharged instead. At the anode, Cl- is preferentially discharged over OH- because it is a halide. The remaining Na+ and OH- ions form NaOH solution.

• In concentrated solution, chloride ions are present in large numbers and are preferentially discharged, producing chlorine gas [1]
• In very dilute solution, there are far fewer chloride ions relative to hydroxide ions [1]
• Hydroxide ions are discharged instead, producing oxygen gas [1]

Examiner note: Concentration affects which ion is discharged at the anode. For halides, a concentrated solution produces the halogen; a very dilute solution produces oxygen. This exception applies only to halides at the anode.

(a) Cu2+ ions are discharged / deposited at the cathode [1]. The concentration of Cu2+ ions in solution decreases, so the blue colour fades [1].

(b) Cu2+ + 2e- → Cu [1]

(c) A brown/pink/orange coating of copper metal forms on the cathode [1]

Examiner note: The blue colour in copper sulfate solution is due to Cu2+ ions. As these are removed by electrolysis, the solution becomes paler and eventually colourless when all Cu2+ ions have been deposited. Eventually H+ ions are discharged instead.

(a) If the metal ion is less reactive than hydrogen (below hydrogen in the reactivity series), the metal is deposited. If the metal is more reactive, hydrogen gas is produced [1].

(b) If a halide ion (Cl-, Br-, I-) is present, the halogen is produced (in concentrated solution) [1]. If only sulfate, nitrate, or other non-halide ions are present, oxygen gas is produced from discharge of OH- ions [1].

Examiner note: These are the selective discharge rules. Memorise them — they are essential for predicting products. The reactivity series determines cathode products; the nature of the anion determines anode products.

Cathode: hydrogen gas [1] Half equation: 2H+ + 2e- → H2 [1] (Na is more reactive than H, so H+ is discharged)

Anode: oxygen gas [1] Half equation: 4OH- → 2H2O + O2 + 4e- [1] (SO4^2- is not discharged; OH- is discharged instead)

Examiner note: Dilute sodium sulfate effectively produces the same result as electrolysis of water — hydrogen at the cathode and oxygen at the anode. The Na+ and SO4^2- ions remain in solution and serve only to carry the current. A few drops of universal indicator can confirm: the solution near the cathode turns alkaline (OH- accumulates) and near the anode turns acidic (H+ accumulates).