No explanation is given for this question. Let's discuss the answer together.

$${\text{For}}\,{M^ + } + {X^ - } \to M + X,\,\,E_{cell}^{^{\text{o}}} = 0.44 - 0.33 = 0.11V$$           is positive, hence reaction is spontaneous.

Key Idea This problem is based on Faraday's first law of electrolysis which states that when an electric current is passed through an electrolytic solution, the amount of substance $$(w),$$  deposited at the electrode, is proportional to the electric charge $$(q)$$  passed through the electrolytic solution.
The formula used in the problem is
$$w = \frac{{Eit}}{{96500}}\,\,\,...{\text{(i)}}$$
where, $$E = $$  gram-equivalent mass of $$C{l^ - }$$
$$i =$$  current, $$t =$$  time, $$it= q$$
Given, $$w = 0.10\,mol = \left( {0.10 \times 71} \right)g,$$
$$i = 3A,E = 35.5$$
The following reactions occured,
\[\begin{align} & \text{At cathode :}\,2{{H}_{2}}O\xrightarrow{+2{{e}^{-}}}{{H}_{2}}+2O{{H}^{-}} \\ & \text{At anode : }\underset{35.5\,g}{\mathop{2C{{l}^{-}}}}\,\xrightarrow{-2{{e}^{-}}}\underset{71\,g}{\mathop{C{{l}_{2}}}}\, \\ \end{align}\]
Putting all values in expression (i) we get
\[\begin{align} & \left( 0.10\times 71 \right)=\frac{35.5}{96500}\times 3\times t\,\,\text{or}\,\,t=6433\,s \\ & \,\text{or}\,\,t=107.22\,\min \\ & \simeq \,110\,\min \,\,\left[ 1\,s=\frac{1}{60}\min \right] \\ & \\ \end{align}\]

$${\text{When}}\,Q = {K_C},E = 0$$

$$\eqalign{ & 2A{g^ + } + {H_2} \to 2{H^ + } + 2Ag \cr & E = {E^ \circ } - \frac{{0.0591}}{2}\log \frac{{{{\left[ {{H^ + }} \right]}^2}}}{{{P_{{H_2}}} \times {{\left[ {A{g^ + }} \right]}^2}}} \cr & 0.222 = 0.7995 - \frac{{0.0591}}{2}\log \frac{1}{{{{\left[ {A{g^ + }} \right]}^2}}} \cr & \left[ {A{g^ + }} \right] = {10^{ - 9.8}} \cr & {K_{sp}} = \left[ {A{g^ + }} \right]\left[ {C{l^ - }} \right] \cr & = \left( {{{10}^{ - 9.8}}} \right) \times \left( 1 \right) \cr & = {10^{ - 9.8}} \cr} $$

In eclectrochemical series, $$Al$$  is placed above $$Zn$$  and all other are present below $$Zn.$$  So, aluminium displaces zinc from $$ZnC{l_2}$$  solution. Hence, it cannot keep in contact with $$Al.$$

Cathode and anode are $$\left( - \right)ve$$  terminals in electrolytic and galvanic cell.

$$\eqalign{ & {\text{At anode :}}\,2{H_{2\left( g \right)}} + 4OH_{\left( {aq} \right)}^ - \to 4{H_2}{O_{\left( l \right)}} + 4{e^ - } \cr & \underline {{\text{At cathode :}}\,{O_{2\left( g \right)}} + 2{H_2}{O_{\left( l \right)}} + 4{e^ - } \to 4OH_{\left( {aq} \right)}^ - } \cr & \underline {{\text{Overall}}\,\,{\text{reaction}}\,{\text{:}}\,2{H_{2\left( g \right)}} + {O_{2\left( g \right)}} \to 2{H_2}{O_{\left( l \right)}}\,\,\,\,\,\,\,} \cr} $$

$$2H{I^{ - 1}} + {H_2}\mathop {S{O_4}}\limits^{ + 6} \to I_2^0 + \mathop {S{O_2}}\limits^{ + 4} + 2{H_2}O$$         in this reaction oxidation number of $$S$$ is decreasing from $$+ 6$$  to $$+4$$  hence undergoing reduction and for $$HI$$ oxidation Number of $$I$$ is increasing from $$-1$$ to $$0$$ hence underegoing oxidation therefore $${H_2}S{O_4}$$  is acting as oxidising agent.

\[\begin{matrix} A \\ +0.5C \\ \end{matrix}\,\,\,\begin{matrix} B \\ -3.0V \\ \end{matrix}\,\,\,\begin{matrix} C \\ -1.2V \\ \end{matrix}\]
NOTE : The higher the negative value of reduction potential, the more is the reducing power.
Hence $$B > C > A.$$