2023 H2 Chem P1 Q14 — Redox Solution

The answer is C (3).
Total amount of \(\text{Br}_2 = \frac{30.0}{1000} \times 0.500 = 0.0150\) mol
Amount of KBr \(= \frac{2.98}{39.1 + 79.9} = \frac{2.98}{119.0} = 0.0250\) mol
Reduction: \(\text{Br}_2 + 2\text{e}^- \rightarrow 2\text{Br}^-\), so 0.0250 mol of electrons are gained and \(\frac{0.0250}{2} = 0.0125\) mol of \(\text{Br}_2\) is reduced.
The rest of the \(\text{Br}_2\) is oxidised: \(0.0150 - 0.0125 = 0.00250\) mol of \(\text{Br}_2\), which gives \(2 \times 0.00250 = 0.00500\) mol of \(\text{BrO}_x^-\).
Electrons lost = electrons gained = 0.0250 mol, so each Br atom that is oxidised loses \(\frac{0.0250}{0.00500} = 5\) electrons. The oxidation state of Br rises from 0 in \(\text{Br}_2\) to +5.
In \(\text{BrO}_x^-\): \(+5 + x(-2) = -1\), so \(x = 3\), i.e. \(\text{KBrO}_3\).
• A is wrong: in \(\text{KBrO}\), Br is +1, so each Br would lose only 1 electron; 0.00500 mol of \(\text{BrO}^-\) would then supply only 0.00500 mol of electrons, not the 0.0250 mol gained.
• B is wrong: in \(\text{KBrO}_2\), Br is +3, supplying \(3 \times 0.00500 = 0.0150\) mol of electrons, not 0.0250 mol.
• D is wrong: in \(\text{KBrO}_4\), Br is +7, supplying \(7 \times 0.00500 = 0.0350\) mol of electrons, more than the 0.0250 mol gained.