$\mathrm{H}_{2} \mathrm{~S}$ acts only as a reducing agent while $\mathrm{SO}_{2}$ can act both as a…
$\mathrm{H}_{2} \mathrm{~S}$ acts only as a reducing agent while $\mathrm{SO}_{2}$ can act both as a reducing and oxidizing agent because
oxygen is more $-$ ve in $\mathrm{SO}_{2}$
hydrogen in $\mathrm{H}_{2} \mathrm{~S}$ is more $+$ ve than oxygen
$\mathrm{S}$ in $\mathrm{SO}_{2}$ has one oxidation state
$\mathrm{S}$ in $\mathrm{H}_{2} \mathrm{~S}$ has $-2$ oxidation state
Solution
$\mathrm{H}_{2} \mathrm{~S}$, the oxidation state of $\mathrm{S}$ is $-2 .$ So it cannot accept more electrons because on accepting 2 electrons $S$ accquires a noble gas configuration. So, it can acts only as a reducing agent by loosing electron. On the other hand, the oxidation state of $\mathrm{S}$ in $\mathrm{SO}_{2}$ is $+4$ which is an intermediate oxidation state of sulphur so it can reduce as well oxidise.
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