Given, molar conductance at $$0.1$$ $$M$$ concentration,
$${\lambda _c} = 9.54\,{\Omega ^{ - 1}}\,c{m^2}\,mo{l^{ - 1}}$$
Molar conductance at infinite dilution,
$$\lambda _c^\infty = 238\,{\Omega ^{ - 1}}\,c{m^2}\,mo{l^{ - 1}}$$
We know that, degree of ionisation,
$$\eqalign{ & \alpha = \frac{{{\lambda _c}}}{{\lambda _c^\infty }} \times 100 \cr & = \frac{{9.54}}{{238}} \times 100 \cr & = 4.008\% \cr} $$

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According to convention, the standard hydrogen electrode is assigned a zero potential at all temperatures.

The salt used to make ‘salt—bridge’ must be such that the ionic mobility of cation and anion are of comparable order so that they can keep the anode and cathode half cells neutral at all times. $$KN{O_3}$$  is used because velocities of $${K^ + }$$ and $$NO_3^ - $$ ions are nearly same

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In $${H_2} - {O_2}$$   fuel cell, the combustion of $${H_2}$$  occurs to create potential difference between the two electrodes

$$ArN{O_2} + 6{H^ + } + 6{e^ - } \to $$     $$ArN{H_2} + 2{H_2}O$$
$$W = \frac{{E.wt \times Q}}{{96500}};$$     $$20.50 = \frac{M}{6} \times \frac{{2 \times 96500}}{{96500}} \times \frac{{50}}{{100}}$$
$$\therefore \,\,M = 123.0\,g$$

Equivalent $$wt.$$  of $$Ag = \frac{{108}}{1} = 108$$
Equivalent $$wt.$$  of $$Cu = \frac{{63.5}}{2} = 31.75$$
$$\frac{{{\text{Eq}}.\,wt\,{\text{of}}\,\,Ag}}{{{\text{Eq}}.\,wt.\,{\text{of}}\,\,Cu}}$$     $$ = \frac{{{\text{Mass of }}Ag{\text{ deposited}}}}{{{\text{Mass of }}Cu{\text{ deposited}}}}$$
$$\frac{{108}}{{31.75}} = \frac{{15.28}}{W} \Rightarrow W$$     $$ = \frac{{15.28 \times 31.75}}{{108}} = 4.49\,g$$

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