(1)

${\mathrm{H}}_2\mathrm{O}\to 2{\mathrm{H}}^{+}+\frac{1}{2}{\mathrm{O}}_2+2{\mathrm{e}}^{-}$

$\therefore$ $\text{No. of Faradays required for one mol}=2\mathrm{F}\left(\mathrm{A}\text{-}\mathrm{II}\right)$

          ${\mathrm{MnO}}_4^{-}+5{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{2+}$

$\therefore$   $\text{No. of Faradays required for one mol}=5\mathrm{F}\left(\mathrm{B}\text{-}\mathrm{IV}\right)$

         ${\mathrm{Ca}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Ca}$

$\therefore$   $\text{No. of Faradays required for 1.5 mol}=1.5\times 2=3\mathrm{F}\left(\mathrm{C}\text{-}\mathrm{I}\right)$

         ${\mathrm{Fe}}^{2+}\to {\mathrm{Fe}}^{3+}+{\mathrm{e}}^{-}$

$\text{No. of Faradays required for one mol}=1\mathrm{F}\left(\mathrm{D}\text{-}\mathrm{III}\right)$

(2)

Given,

Current $\left(I\right)=9.6487$ $\text{A}, \text{Time} \left(t\right)=100$ $\text{seconds}$

1F = 96487 C

Reaction : ${\mathrm{Cu}}_{\left(\mathrm{aq}\right)}^{2+}+2{\mathrm{e}}^{-}\to {\mathrm{Cu}}_{\left(\mathrm{s}\right)}$

Here, 2 electrons are exchanged, so valency factor $\left(n\right)=2$.

Applying Faraday’s first law, $W=ZIt$      $(\because Z=\frac{\text{Equivalent weight}}{F}=\frac{63}{2\times 96487})$

So, $W=\frac{63\times 9.6487\times 100}{2\times 96487}$$=0.315$ $\text{g}$

(2)

During electrolysis of dilute sulphuric acid the following reaction takes place at anode:

$2H_2{O}_{\left(l\right)}\to O_{2\left(g\right)}+4H_{\left(aq\right)}^{+}+4e^{-}; E_{cell}^{\circ }=+1.23$ $V$

i.e., $O_{2\left(g\right)}$ will be liberated at anode.

(1)

$\begin{array}{ccccc}C{a}_{\left(aq\right)}^{2+}& +& 2e^{-}& \to & C{a}_{\left(s\right)}\\ & & 2\mathrm{F}& & (1 \mathrm{mole}=40\mathrm{g})\\ & & 1\mathrm{F}& & 20\mathrm{g}\end{array}$

Thus, one Faraday is required to produce 20 g of calcium from molten $CaC{l}_2$.

(2)

During the electrolysis of molten sodium chloride, 

At cathode : $2N{a}^{+}+2e^{-}⟶2Na$

At anode :  $2C{l}^{-}⟶C{l}_2+2e^{-}$

$\begin{array}{c}\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ \mathrm{Net} \mathrm{reaction} :2N{a}^{+}+2C{l}^{-}⟶2Na+C{l}_2\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\end{array}$

According to Faraday’s first law of electrolysis, $W=Z\times I\times t$

$W=\frac{E}{96500}\times I\times t$

No. of moles of $C{l}_2$ gas $\times$ Mol. wt. of $C{l}_2$ gas

$=\frac{\text{Eq. wt. of }C{l}_2\text{ gas}\times I\times t}{96500}$

$0.10\times 71=\frac{35.5\times 3\times t}{96500}$

$t=\frac{0.10\times 71\times 96500}{35.5\times 3}=6433.33$ $\text{sec}$

$t=\frac{6433.33}{60}$ $\text{min}=107.22$ $\text{min}\approx 110$ $\text{min}$

(3)

$Q=I\times t$

$Q=1\times 60=60$ $C$

Now, $1.6\times {10}^{-19} C\equiv 1$ $\text{electron}$

$\therefore$  $60 C\equiv \frac{60}{1.6\times {10}^{-19}}=3.75\times {10}^{20}$ $\text{electrons}$

(3)

The oxidation reaction is

$\underset{0.1 \mathrm{mol}}{\stackrel{+6}{\mathrm{M}}{\mathrm{nO}}_4^{2-}} \to \underset{0.1 \mathrm{mol}}{\stackrel{+7}{\mathrm{M}}{\mathrm{nO}}_4^{-}}+{\mathrm{e}}^{-}$

$Q=0.1\times F=0.1\times 96500$ $C=9650$ $C$

(4)

According to Faraday’s second law,

$\frac{W_{Ag}}{E_{Ag}}=\frac{W_{O_2}}{E_{O_2}}$  or  $\frac{W_{Ag}}{108}=\frac{\frac{5600}{22400}\times 32}{8}$

or   $\frac{W_{Ag}}{108}=\frac{8}{8} \Rightarrow W_{Ag}=108$ $g$