# Question #cdd5a

In the first reaction, the oxidation state of Cl in HCL is -1 and it changes to 0 in the free form of $C {l}_{2}$. Hence it lost electrons and so was reduced.
Similarly the oxidation state of Mn in $M n {O}_{2}$ is 4+ and this becomes 2+ in $M n C {l}_{2}$ so the Mn gained electrons and was reduced.
By definition, HCL is then the oxidizing agent and $M n {O}_{2}$ is the reducing agent.
A similar analysis may be applied in the second reaction to show that since $Z n \to Z {n}^{2 +} \mathmr{and} {H}^{\pm} > {H}_{2}$, it means that Zn was oxidized since it lost electrons, and H was reduced since it gained electrons. (Cl is a spectator ion here as its oxidation state does not change).