Chemistry - Redox Reactions Question with Solution | TestHub

A species with the key element in an intermediate oxidation state (or with both an oxidizable and a reducible part) can go either way. $\mathrm{HN}{\mathrm{O}}_2$ (N = +3, between -3 and +5) can be oxidized to $\mathrm{HN}{\mathrm{O}}_3$ or reduced to $NO/{\mathrm{N}}_2\mathrm{O}$. ${\mathrm{H}}_2\mathrm{S}$: ${\mathrm{S}}^{2-}$ (-2) is oxidized, while its ${\mathrm{H}}^{+}$ can be reduced to ${\mathrm{H}}_2$ by active metals. ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_3$ (S = +4) is oxidized to ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ or reduced to S. In ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$, sulphur is at its maximum +6, so it can only be reduced. Hence (a), (c), (d).