Chemistry 5070/41 — October/November 2017

Walkthrough

The question states two facts about the replacement electrolyte:

1. The gas at the cathode is unchanged — still hydrogen. This means H⁺ ions are still being discharged, so the replacement must be an acid (providing H⁺).
2. A different gas is produced at the anode. In dilute HCl, Cl⁻ was discharged to give chlorine. For a different gas to form, the acid must not contain chloride ions. Instead, the anion must be one that is not discharged in dilute solution, so OH⁻ from water is discharged instead, giving oxygen.

Dilute sulfuric acid, H₂SO₄(aq), fits perfectly. It provides H⁺ (so hydrogen still forms at the cathode) and SO₄²⁻ ions. Sulfate ions are not discharged in dilute solution — OH⁻ from water is discharged at the anode instead, producing oxygen gas.

$$4\text{OH}^-(\text{aq}) \rightarrow 2\text{H}_2\text{O}(\text{l}) + \text{O}_2(\text{g}) + 4\text{e}^-$$

Key Takeaways

• The cathode product tells you what cation is present: hydrogen means H⁺ is still there, so the electrolyte must be an acid.
• The anode product changes when the anion is not discharged. Sulfate, nitrate, and other oxyanions are not discharged in dilute solutions; OH⁻ from water is discharged instead, giving oxygen.
• The electrolyte must still be dilute — concentrated solutions can change which ions discharge.

Common Mistakes

• Suggesting dilute nitric acid — while technically it would work (NO₃⁻ is not discharged), sulfuric acid is the standard expected answer at O Level. Nitric acid is an oxidising acid and can complicate the chemistry.
• Forgetting "dilute" — concentrated sulfuric acid behaves very differently in electrolysis.
• Suggesting sodium chloride solution — this is not an acid, so the cathode product would still be hydrogen (from water), but the question says "another dilute acid".
• Suggesting dilute hydrofluoric acid — HF is a weak acid and not a standard 5070 answer.

Things to Be Careful About

• The mark scheme specifically accepts "dilute sulfuric acid". Include "dilute" — concentrated gives different results.
• Do not write the formula H₂SO₄ alone — the name "dilute sulfuric acid" is what is asked for.
• This is a reasoning question: the student must connect the observations to the composition of the electrolyte.

Walkthrough

With dilute sulfuric acid as the electrolyte, the ions present are H⁺, SO₄²⁻, and H⁺, OH⁻ from water. At the anode, SO₄²⁻ and OH⁻ are attracted. Sulfate ions are not discharged in dilute solution (they are too stable). Instead, hydroxide ions from water are discharged:

$$4\text{OH}^-(\text{aq}) \rightarrow 2\text{H}_2\text{O}(\text{l}) + \text{O}_2(\text{g}) + 4\text{e}^-$$

Oxygen gas is produced. The test for oxygen is to insert a glowing splint (a splint that has been lit and then blown out so it is just glowing) into the test tube. The oxygen supports combustion and the splint relights (rekindles).

Key Takeaways

• In the electrolysis of dilute acids with non-halide anions (sulfate, nitrate), oxygen is produced at the anode from discharge of OH⁻ ions.
• Oxygen is identified by a glowing splint relighting.
• This is the opposite test to hydrogen: burning splint pops for hydrogen, glowing splint relights for oxygen.

Common Mistakes

• Writing "burning splint" for the oxygen test — it must be a glowing splint. A burning splint will just burn more brightly, which is not the distinctive test.
• Saying "splint goes out" — that is the test for carbon dioxide.
• Naming the gas as "air" or "atmospheric gas" — must be specifically oxygen.
• Writing the half-equation wrong: forgetting the 4 electrons or the water molecules.

Things to Be Careful About

• Both "relights" and "rekindles" are accepted by the mark scheme.
• The test must be described as "glowing splint" — not "lighted splint" or "burning splint".
• This question tests whether the student can connect the change in electrolyte to a change in anode product, and then correctly identify and test the new gas.