Electrochemistry: Question 6

Worked solution

Step 1: Find the number of electrons from the oxidation number change

Manganese falls from $+7$ in $\text{MnO}_4^-$ to $+2$ in $\text{Mn}^{2+}$, a change of $5$. Since reduction is a gain of electrons, $5$ electrons must appear on the left-hand side:

$\text{MnO}_4^-(aq) + 5e^- \rightarrow \text{Mn}^{2+}(aq)$

Step 2: Balance the oxygen atoms with water

$\text{MnO}_4^-$ contains $4$ oxygen atoms and $\text{Mn}^{2+}$ contains none, so $4\text{H}_2\text{O}(l)$ must appear on the right:

$\text{MnO}_4^-(aq) + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l)$

Step 3: Balance the hydrogen atoms with H\textsuperscript{+}(aq)

The $4\text{H}_2\text{O}(l)$ on the right introduces $8$ hydrogen atoms, so $8\text{H}^+(aq)$ must be added to the left (acidic solution supplies the H\textsuperscript{+}):

$\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l)$

Step 4: Check both atoms and charge independently

Atoms: Mn: $1 = 1$ ✓. O: $4 = 4$ (in $4\text{H}_2\text{O}$) ✓. H: $8 = 8$ (in $4\text{H}_2\text{O}$) ✓.

Charge: left $= (-1) + 8(+1) + 5(-1) = -1 + 8 - 5 = +2$; right $= +2 + 0 = +2$. Equal ✓.

Both checks pass, so this half-equation is fully balanced.

Why the other options are wrong

• B: only $3e^-$ and $4\text{H}^+$ are used, with just $2\text{H}_2\text{O}$. The oxygen atoms do not balance ($4$ on the left against $2$ on the right).
• C: the atoms balance (same $8\text{H}^+$, $4\text{H}_2\text{O}$ as the correct equation), but only $3e^-$ are used instead of $5$, so the charge does not balance: left $= -1+8-3=+4 \neq +2$.
• D: the hydrogen atoms happen to balance ($4 = 4$), but the oxygen atoms do not ($4$ on the left against $2$ on the right, from only $2\text{H}_2\text{O}$).

Final answer

• The correctly balanced half-equation is $\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l)$, option A.