Answer: C. Find the oxidation state of each element on both sides and look for the one that has decreased (reduction = gain of electrons = decrease in oxidation number).
In \(\text{VO}^{2+}\): O is \(-2\); the overall charge is \(+2\), so V is \(+4\). In \(\text{V}^{3+}\) on the right: V is \(+3\). V has gone from \(+4\) to \(+3\) — it has gained one electron per ion. V is reduced.
Zinc goes from 0 (metal) to \(+2\) in \(\text{Zn}^{2+}\) — Zn is oxidised, not reduced.
Hydrogen stays at \(+1\) in \(\text{H}^+\) and in \(\text{H}_2\text{O}\). Oxygen stays at \(-2\) on both sides. Neither H nor O changes.
Only vanadium has been reduced — option C.
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