Edexcel A-Level Chemistry AS Paper 1, June 2025: Question 6

✅ Correct Answer

• Mark 1: Identify that a toxic, poisonous, or corrosive gas is produced: specifically H₂S (hydrogen sulfide) or I₂ (iodine vapour).
• Mark 2: State that the fume cupboard prevents this toxic gas being released into the laboratory / prevents it from being inhaled/breathed in.

🧠 Exam Technique

• Always name the specific species from the equation responsible for the hazard. Simply saying "toxic fumes are made" is too vague.
• Connect the hazard directly to the laboratory precaution: "so it is not inhaled by occupants".

❌ Common Errors

• Stating only that H₂S smells like rotten eggs without mentioning that it is toxic or hazardous.
• Claiming that sulfuric acid itself evaporates or causes the fume cupboard requirement, ignoring the products indicated by the equation.

✅ Correct Answer (Any Two)

• Formation of a black / grey solid (I₂)
• Formation of a purple vapour / gas (I₂)
• Effervescence / bubbling / fizzing (production of H₂S gas)
• Condensation / water droplets on sides of tube (formation of H₂O)

💡 Key Knowledge

Deduce observations straight from physical states of the products:

• I₂ : Solid iodine is shiny grey-black crystals. Due to the exothermic reaction, iodine sublimes into a distinct purple vapour.
• H₂S : Colourless gas causing visible fizzing/effervescence.
• H₂O : Visible liquid mist or condensation on tube walls.

❌ Common Errors

• Saying "a purple solution" (solid or gas is formed, not a solution).
• Saying "a yellow solid is seen" (sulfur S is formed in an alternative reduction route, but the provided equation explicitly produces H₂S, not S).
• Giving "bad egg smell" — smelling the gas is not an acceptable observation when the question explicitly states it is performed in a fume cupboard!

✅ Correct Option

A: from +6 to −2

1 mark is awarded for choosing A.

📐 Step-by-Step Calculation

• Reactant (H₂SO₄):
  H = +1, O = −2
  2(+1) + S + 4(−2) = 0  ⇒  +2 + S − 8 = 0  ⇒  S = +6
• Product (H₂S):
  H = +1
  2(+1) + S = 0  ⇒  +2 + S = 0  ⇒  S = −2
• Change: from +6 to −2 (a reduction involving a gain of 8 electrons).

❌ Distractor Breakdown

• B (+6 to +2): Incorrectly assumes each H in H₂S has an oxidation state of −1.
• C (+2 to −2): Incorrectly assumes O in H₂SO₄ has an oxidation state of −1 (like in peroxides).
• D (no change): Mistakenly assumes this is purely an acid-base neutralization rather than redox.

✅ Correct Answer

• Oxidation half-equation:
  2I⁻ → I₂ + 2e⁻  (or 2I⁻ − 2e⁻ → I₂) [1 Mark]
• Reduction half-equation:
  H₂SO₄ + 8H⁺ + 8e⁻ → H₂S + 4H₂O
  OR
  SO₄²⁻ + 10H⁺ + 8e⁻ → H₂S + 4H₂O [1 Mark]

📐 Balancing the Reduction Half-Equation

1. Sulfur balance: H₂SO₄ → H₂S (both have 1 S).
2. Oxygen balance: Add 4H₂O to the right to balance 4 oxygens:
   H₂SO₄ → H₂S + 4H₂O
3. Hydrogen balance: Right side has 2 + 8 = 10 H atoms. Left side already has 2 H in H₂SO₄, so add 8H⁺:
   H₂SO₄ + 8H⁺ → H₂S + 4H₂O
4. Charge balance: Left side has +8 charge; right is 0. Add 8e⁻ to the left:
   H₂SO₄ + 8H⁺ + 8e⁻ → H₂S + 4H₂O

❌ Common Errors

• Writing electrons on the wrong side or forgetting charge balance.
• Writing I⁻ → I + e⁻ instead of diatomic iodine I₂ .
• Confusing the reduction to H₂S (8e⁻ transfer) with reduction to SO₂ (2e⁻ transfer) or S (6e⁻ transfer).

🧠 Exam Tip

State symbols are explicitly not required and will be ignored even if incorrect. Multiples are allowed.

✅ Correct Answer

Hydrogen chloride / HCl(g)

Also allow: hydrochloric acid or HCl(aq) (as moisture in the air hydrates the gas to form misty droplets).

💡 Key Knowledge

When KCl reacts with concentrated sulfuric acid, only an acid-base reaction occurs because Cl⁻ is not a strong enough reducing agent:

KCl + H₂SO₄ → KHSO₄ + HCl(g)

The HCl gas reacts with atmospheric moisture to form tiny droplets of hydrochloric acid, observed as misty white fumes.

✅ Correct Answer

• Mark 1: The chloride ion (Cl⁻) does not reduce concentrated sulfuric acid / no redox reaction occurs / chloride remains at oxidation state −1.
• Mark 2: Therefore, iodide is a stronger reducing agent than chloride (or chloride is a weaker reducing agent than iodide).

🧠 Exam Technique: Dependent Marks

Mark 2 is dependent on scoring Mark 1. You cannot simply state "iodide is a stronger reducing agent" without explaining the evidence: that chloride cannot reduce sulfuric acid at all, whereas iodide reduces sulfur from +6 to −2.

💡 Why does reducing power increase down Group 7?

• Ionic radius: Iodide (I⁻) is much larger than chloride (Cl⁻).
• Shielding: The outermost electrons in I⁻ experience more shielding from inner electron shells.
• Attraction: The attraction between the nucleus and the outermost electrons is weaker in I⁻.
• Electron loss: Therefore, I⁻ loses an electron much more easily than Cl⁻, making it a far more powerful reducing agent.

❌ Common Errors

• Referring to chlorine and iodine instead of chloride ions (Cl⁻) and iodide ions (I⁻). Remember: the ions act as reducing agents, not the elemental halogens!
• Stating chloride is oxidized to chlorine (Cl₂ is never formed in this reaction).