$$\begin{gathered} {\mathrm{Cu}}^{+2} + \mathrm{KI} \to {\mathrm{CuI}}_2 + {\mathrm{K}}^{\oplus } \\ 2{\mathrm{CuI}}_2 \to {\mathrm{Cu}}_2{\mathrm{I}}_2 + {\mathrm{I}}_2 \end{gathered}$$

whereas with Cu²⁺ + Kcl ------      CuCl2 + k⁺

${\mathrm{H}}_2{\mathrm{O}}_2 \stackrel{ }{\to }{\mathrm{O}}_2 + 2{\mathrm{H}}^{\oplus } +\hspace{0.17em}2{\mathrm{e}}^{-}$

For 1 mol H₂O₂ required charge = 2 mol e^-

                                                = 2 F

                                                = 2 X 96500

                                                = 193000 C

Number of equivalents of NaClO₄ = Number of Faraday or, $\frac{1000}{15.13}=66\mathrm{F}$

     [Since equivalent wt. of NaClO₄ = $\frac{122.5}{8}=15.13$]

4.

In mercury cell, the cell potential is approximately $1.35$V and remains constant during its life.

Zinc is more reactive than iron, it loses electron more readily as compared to iron. In galvanized iron object, zinc acts as anode and does not allow the iron to lose  electrons

In both galvanic and electrolytic cells, oxidation takes place at the anode and electrons flow from the anode to the cathode. and reduction takes place at cathode.

(b) 

$$\begin{gathered} ∆\mathrm{G} = ∆\mathrm{G}°+\mathrm{RT} \mathcal{l}\mathrm{n} \mathrm{Q} \\ ∆\mathrm{G} = ∆\mathrm{G}° + \mathrm{RT} \mathcal{ln} \frac{{\mathrm{C}}_2}{\mathrm{C}1} \\ ∆\mathrm{G} \mathrm{is} \mathrm{function} \mathrm{of} \mathcal{ln} \frac{{\mathrm{C}}_2}{\mathrm{C}1} \end{gathered}$$

$$\begin{gathered} {\mathrm{H}}_2 + 2{\mathrm{H}}^{+} \stackrel{ }{\to }2{\mathrm{H}}^{+} + {\mathrm{H}}_2 \\ {\mathrm{P}}_1 {\mathrm{C}}_1 {\mathrm{C}}_2 {\mathrm{P}}_2 \\ \mathrm{E} = {\mathrm{E}}^{\circ }-\frac{0.059}{2}\log \frac{({\mathrm{C}}_1{)}^2 ({\mathrm{P}}_2)}{{\mathrm{P}}_1({\mathrm{C}}_2{)}^2} \\ \mathrm{E} = 0 \frac{0.059}{2}\log \frac{\left({\mathrm{C}}_1{)}^2\right({\mathrm{P}}_2)}{{\mathrm{P}}_1({\mathrm{C}}_2{)}^2} \\ \mathrm{E}= 0\frac{0.059}{2}\log \frac{{\mathrm{C}}_1^2{\mathrm{P}}_2}{{\mathrm{P}}_1{\mathrm{C}}_2^2} \\ \mathrm{E}= 0 \mathrm{if} \frac{{\mathrm{C}}_1^2{\mathrm{P}}_2}{{\mathrm{P}}_1{\mathrm{C}}_2^2} = 0 \\ {\mathrm{C}}_1^2{\mathrm{P}}_2 = {\mathrm{C}}_2^2{\mathrm{P}}_1 \end{gathered}$$

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}^{\circ } = ({\mathrm{E}}_{\mathrm{SRP}}^{\circ }{)}_{\mathrm{c}} - ({\mathrm{E}}_{\mathrm{SRP}}^{\circ }{)}_{\mathrm{a}} \\ = 1.50 -(-0.25) = 1.75 \mathrm{V} \\ {\mathrm{E}}_{\mathrm{cell} }= 1.75 \frac{-0.059}{6} \log \frac{(1{)}^3}{(1{)}^2} \\ = 1.75 \mathrm{V} \end{gathered}$$

 $\frac{{\mathrm{W}}_1}{{\mathrm{W}}_2}=\frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}$

$$\begin{gathered} \mathrm{For} \mathrm{concentration} \mathrm{cell} -{\mathrm{E}}_{\mathrm{cell}}^{\circ }=0 \\ {\mathrm{E}}_{\mathrm{cell}}=\frac{-0.0591}{2}\log \frac{[{\mathrm{H}}^{\oplus }{]}^2\mathrm{anode}}{[{\mathrm{H}}^{\oplus }{]}^2\mathrm{cathode}} \\ =\frac{-0.0591}{2}\log \frac{({10}^{-6}{)}^2}{({10}^{-4}{)}^2} \\ =(-4) \times \frac{-0.0591}{2}=0.118 \mathrm{V} \end{gathered}$$