(a) Calculate the emf of the cell in which the following reaction takes place.

Ni (s) 2Ag⁺ (0.002 M) Ni²⁺ (0.160 M) + 2Ag (s)

Given that, E°_cell = 1.05 V

(b) Define electrochemical cell. What happens if external potential applied becomes greater than ${\mathrm{E}}_{\mathrm{cell}}^{\circ }$ of electrochemical cell?

(c) Calculate the potential of hydrogen electrode, which is in contact with a solution whose pH is 10.

OR

The resistance of a conductivity cell, when filled with 0.05 M solution of an electrolyte x, is 100 at 40°C. The same conductivity cell when filled with 0.01 M solution of electrolyte y, has a resistance of 50. The conductivity of 0.05 M solution of electrolyte x is 1.0 x 10^-4 S cm ^-1.

Calculate

(a) Cell constant.

(b) Conductivity of 0.01 M y solution

(c) Molar conductivity of 0.01 M y solution

(a) 
$\begin{array}{l}{\mathrm{E}}^{\circ }={\mathrm{E}}_{\text{cell }}^{\circ }-\frac{0.0591}{\mathrm{n}}\log \frac{\left[{\mathrm{Ni}}^{2+}\right]}{{\left[{\mathrm{Ag}}^{+}\right]}^2}\\ =1.05\mathrm{V}-\frac{0.0591}{2}\log \frac{(0.160)}{(0.002{)}^2}\\ =1.05-\frac{0.0591}{2}\log \left(4\times {10}^4\right)\\ =1.05-\frac{0.0591}{2}(4.6021)\\ =1.05-0.14=0.91\mathrm{V}\\ {\mathrm{E}}_{\text{cell }}=0.91\mathrm{V}\end{array}$

(b) Electrochemical cell is a device through which chemical energy changes to electrical energy. Current will flows in opposite direction as the chemical reaction of the cell is reversed. Opposing emf will be more than the cell.
(c) For Hydrogen electrode, ${\mathrm{H}}^{+}+{\mathrm{e}}^{-}⟶\frac{1}{2}{\mathrm{H}}_2$
Applying Nernst equation, 
$\begin{array}{l}{\mathrm{E}}_{\left({\mathrm{H}}^{+}/1/2{\mathrm{H}}_2\right)}={\mathrm{E}}_{\left({\mathrm{H}}^{+}/1/2{\mathrm{H}}_2\right)}^{\circ }-\frac{0.0591}{\mathrm{n}}\log \frac{1}{\left[{\mathrm{H}}^{+}\right]}\\ =0-\frac{0.0591}{1}\log \frac{1}{\left({10}^{-10}\right)}\left[\mathrm{pH}=10;\left[{\mathrm{H}}^{+}\right]={10}^{-10}\mathrm{M}\right]\\ =0-\frac{0.0591}{1}\times (10\log 10)=-0.591\mathrm{V}\end{array}$
${\mathrm{E}}_{\left({\mathrm{H}}^{+}/1/2{\mathrm{H}}_2\right)}=-0.591\mathrm{V}$

OR

(a) 
$\text{ Cell constant, }{G}^{\ast }=\text{ resistance }\left(R\right)\times \text{ conductivity }\left(k\right)$
$=100\times 1.0\times {10}^{-4}={10}^{-2}{cm}^{-1}$
(b)  Conductivity of solution y,
$\mathrm{\kappa }=\frac{\text{ Cell constant }}{\text{ Resistance }}=\frac{{10}^{-2}}{50}=2\times {10}^{-4}\mathrm{S}{\mathrm{cm}}^{-1}$

(c)  Molar conductivity of solution y 
${\mathrm{\Lambda }}_{\mathrm{m}}=\frac{\kappa \times 1000}{\text{ Molarity }}=\frac{2\times {10}^{-4}\times 1000}{0.01}=20\mathrm{S}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$