$NH_4HS(s) \rightleftharpoons NH_3 (g) + H_2S(g)$
If more $NH_4HS$ is added to the equilibrium
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The correct statement for product $B$ is. It is
$\begin{array}{*{20}{c}}
O \\
{||} \\
{C{H_3} - C{H_2} - C - C{H_2} - C{H_2}}
\end{array}$
$\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\
{C{H_3} - C{H_2} - C{H_2} - C - N{H_2}}
\end{array}$
${B^{3 + }}_{(aq.)} + 3{e^ - } \to B(s)$ ; $E = 0.6\ Volt$
Using above information find the ion which will be deposited first at cathode if solution containing $A_{(aq.)}^{2 + }$ & $B_{(aq.)}^{ 3+ }$ is electrolysed