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) 

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

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

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

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

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

 ${\mathrm{MnO}}_2 +\hspace{0.17em}{\mathrm{NH}}_4^{+} +\hspace{0.17em}\mathrm{e} \stackrel{ }{\to }\mathrm{MnO}\left(\mathrm{OH}\right) + {\mathrm{NH}}_3$

NH₃ furthur combines with Zn⁺² and forms $\mathrm{Zn}({\mathrm{NH}}_3{)}_4^{+2}$ 

$$\begin{gathered} \mathrm{W} = \frac{\mathrm{E}}{96500} \times \mathrm{i} \times \mathrm{t} = \frac{31.75}{96500} \times 5 \times 2 \times 3600 \\ \mathrm{W} = 11.84 \mathrm{g} \\ \% \mathrm{of} \mathrm{current} \mathrm{efficiency} = \frac{9.8}{11.84} \times 100 = 82.8\% \end{gathered}$$

In a Daniell cell, the electrode potential depends on the concentrations of Zn2+ and Cu2+ ions. When the concentration of Zn2+ increases and Cu2+ decreases, the electrode potential (cell emf) increases according to the Nernst equation: E = E° - (RT/nF) ln Q, where Q is the reaction quotient.