Redox and electrochemistry — Question 27

27a)

$3\,\times \,(+1)\,+\,\mathrm{A}\mathrm{s}\,+\,3\,\times \,(–2)\,=\,0\,$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,+\,\mathrm{A}\mathrm{s}\,–\,6\,=\,0\,$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathrm{A}\mathrm{s}\,–\,3\,=\,0\,$

$\mathrm{A}\mathrm{s}\,=\,+3\,$

• provides +3 [1 mark]

Do not accept 3 or 3+.

Do not penalise for incorrect decimal places/significant figures.

27b)

Balanced oxidation half-equation: H₃AsO₃ + H₂O → H₃AsO₄ + 2H⁺ + 2e⁻

Balanced reduction half-equation: ⁻ + 2H⁺ + e⁻ → NO2 + H2O NO₃

Multiply by 2: ⁻ + 4H⁺ + 2e⁻ → 2NO2 + 2H2O 2NO₃

Balanced redox equation:

⁻ + 2H⁺ → H₃AsO₄ + 2NO₂ + H₂O

H₃AsO₃ + 2NO₃

• provides balanced oxidation half-equation [1 mark] • provides balanced reduction half-equation [1 mark] • uses multiplication factor to balance electrons [1 mark] • determines balanced redox equation [1 mark]

Allow FT error for multiplication factor and balanced equation. Award full marks for correctly balanced equation without full working.

27c)

Acceptable responses are: - nitrogen - N