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

$$E_{{\text{cell}}}^ \circ = E_{{\text{cathode}}}^ \circ - E_{{\text{anode}}}^ \circ $$
It will remain unchanged.

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

$$\eqalign{ & {\text{Oxidation half cell:}} \cr & {H_2}\left( g \right) \to 2{H^ + }\left( {1M} \right) + 2{e^ - } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{P_1} \cr & {\text{Reduction half cell}} \cr & 2{H^ + }\left( {1M} \right) + 2{e^ - } \to {H_2}\left( g \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{P_2} \cr & {\text{The net cell reaction}} \cr & {H_2}\left( g \right) \to {H_2}\left( g \right) \cr & {P_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{P_2} \cr & E_{cell}^o = 0.00\,V\,\,\,\,\,\,\,\,n = 2 \cr & \therefore \,\,{E_{cell}} = {E^o}_{cell} - \frac{{RT}}{{nF}}{\log _e}K = 0 - \frac{{RT}}{{nF}}{\log _e}\frac{{{P_2}}}{{{P_1}}} \cr & or\,\,{E_{cell}} = \frac{{RT}}{{2F}}{\log _e}\frac{{{P_2}}}{{{P_1}}} \cr} $$

$$F{e^{3 + }} + {e^ - } \to F{e^{2 + }}\Delta {G^ \circ } = $$      $$ - 1 \times F \times 0.77$$
$$S{n^{2 + }} + 2{e^ - } \to Sn\left( s \right)\Delta {G^ \circ } = $$      $$ - 2 \times F\left( { - 0.14} \right)$$
$${\text{for}}\,\,Sn\left( s \right) + 2F{e^{3 + }}\left( {aq} \right) \to $$      $$2F{e^{2 + }}\left( {aq} \right) + S{n^{2 + }}\left( {aq} \right)$$
$$\therefore {\text{ Standard potential for the given reaction}}$$
$$\eqalign{ & {\text{or}}\,\,\,E_{cell}^o = E_{\frac{{Sn}}{{S{n^{2 + }}}}}^o + E_{\frac{{F{e^{3 + }}}}{{F{e^{2 + }}}}}^o \cr & = 0.14 + 0.77 \cr & = 0.91\,V \cr} $$

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2{I^ - } \to {I_2} + 2{e^ - }\,\,\,\,\,\,\,\,{\text{(Oxidation)}} \cr & \underline {B{r_2} + 2{e^ - } \to 2B{r^ - }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} {\text{(Reduction)}} \cr & \underline {2{I^ - } + B{r_2} \to {I_2} + 2B{r^ - }\,\,} {\text{(Net cell reaction)}} \cr} $$

$$A{l^{3 + }} + 3{e^ - } \to Al;$$    $${\text{Eq}}{\text{.}}\,wt.\,{\text{of}}\,Al = \frac{{27}}{3} = 9$$
$$\eqalign{ & W = Z \times I \times t \Rightarrow Z = \frac{{{\text{Eq}}.\,wt.}}{{96500}} \cr & t = \frac{{W \times 96500}}{{Eq.\,wt. \times I}} \cr & \,\,\, = \frac{{50 \times 96500}}{{9 \times 105}} \cr & \,\,\, = 5105.82\,s\,\,{\text{or}}\,\,1.42\,h \cr} $$

The equilibrium constant is related to the standard emf of cell by the expression
$$\eqalign{ & \log K = {E^ \circ }_{cell} \times \frac{n}{{0.059}} = 0.295 \times \frac{2}{{0.059}} \cr & \log K = \frac{{590}}{{59}} = 10\,\,{\text{or}}\,K = 1 \times {10^{10}} \cr} $$

NOTE : In an electrolytic cell, electrons do not flow themselves. It is the migration of ions towards oppositely charged electrodes that indirectly constitutes the flow of electrons from cathode to anode through internal supply.

Conductivity decreases because number of ions per unit volume decreases.