JAMB Chemistry 2012 Paper

Concept:Oxidation involves loss of electrons, shown by an increase in the oxidation number of an element.Explanation:The balanced reaction is$2\mathrm{Na_2O_2} + 2\mathrm{H_2O} \rightarrow 4\mathrm{NaOH} + \mathrm{O_2}$In $\mathrm{Na_2O_2}$, oxygen has an oxidation number of $-1$.In the products, oxygen appears in $\mathrm{O_2}$ with oxidation number $0$, and in $\mathrm{NaOH}$ with oxidation number $-2$.The change from $-1$ to $0$ means oxygen loses electrons, so it is oxidized.The oxygen in water remains at $-2$ throughout, so water is not oxidized.Thus, the substance that is oxidized is $\mathrm{Na_2O_2}$.Answer:A. $\mathrm{Na_2O_{2(s)}}$