a. Write the cell reaction and calculate the e.m.f. of the following cell at 298K :
$\mathrm{Sn}\left(\mathrm{s}\right)\left|{\mathrm{Sn}}^{2+}(0.004\mathrm{M})\parallel {\mathrm{H}}^{+}(0.020\mathrm{M})\right|{\mathrm{H}}_2\left(\mathrm{g}\right)\left(1\mathrm{bar}\right)\mid \mathrm{Pt}\left(\mathrm{s}\right)$
(Given: $\left.{\mathrm{E}}_{{\mathrm{Sn}}^{2+}/\mathrm{Sn}}^0=-0\cdot 14\mathrm{V}\right)$
b. Give reasons:
(i) On the basis of E⁰ values, O₂ gas should be liberated at anode but it is Cl₂ gas, which is liberated in the electrolysis of aqueous NaCl.
(ii) Conductivity of CH₃COOH decreases on dilution.

OR

(a) For the reaction
$\begin{array}{l}2\mathrm{AgCl}\left(\mathrm{s}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)\left(1\mathrm{atm}\right)⟶2\mathrm{Ag}\left(\mathrm{s}\right)+2{\mathrm{H}}^{+}(0\cdot 1\mathrm{M})+2{\mathrm{Cl}}^{-}(0\cdot 1\mathrm{M})\\ {\mathrm{\Delta G}}^0=-43600\mathrm{J}\text{ at }{25}^{\circ }\mathrm{C}\end{array}$
Calculate the e.m.f. of the cell.
$\left(\log {10}^{-\mathrm{n}}=-\mathrm{n}\right)$
(b) Define fuel cell and write its two advantages.

a.
The cell reaction is
$\mathrm{Sn}\left(\mathrm{s}\right)+2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\to {\mathrm{Sn}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)$
$\begin{array}{l}{\mathrm{E}}_{\text{cell }}^{\circ }={\mathrm{E}}_{\mathrm{SHE}}^{\circ }-{\mathrm{E}}_{\mathrm{Sn}\mid {\mathrm{Sn}}^{2+}}^0\\ {\mathrm{E}}_{\text{cell }}^{\circ }=0.0\mathrm{V}-(-0.14\mathrm{V})=0.14\mathrm{V}\\ {\mathrm{E}}_{\text{cell }}={\mathrm{E}}_{\text{cell }}^{\circ }-\frac{0.0592}{\mathrm{n}}\log \frac{\left[{\mathrm{Sn}}^{2+}\right]\times {\mathrm{P}}_{\mathrm{H}2}}{{\left[{\mathrm{H}}^{+}\right]}^2}\\ {\mathrm{E}}_{\text{cell }}=0.14\mathrm{V}-\frac{0.0592}{2}\log \frac{0.004\mathrm{M}\times 0.987\mathrm{atm}}{[0.020{]}^2}\\ \end{array}$
Note: 1bar=0.987 atm
${\mathrm{E}}_{\text{cell }}=0.111\mathrm{V}$

b. (i) Overvoltage of O₂
(ii) The number of ions per unit volume decreases.

OR

a) $\begin{array}{l}{\mathrm{\Delta G}}^0=-{\mathrm{nFE}}^{\circ }\\ -43600=-2\times 96500\times {\mathrm{E}}^{\circ }\end{array}$
$\begin{array}{l}{\mathrm{E}}^{\circ }=0.226\mathrm{V}\\ \mathrm{E}={\mathrm{E}}^{\circ }-0.059/2\log \left({\left[{\mathrm{H}}^{+}\right]}^2{\left[{\mathrm{Cl}}^2\right]}^2/\left[{\mathrm{H}}_2\right]\right)\\ =0.226-0.059/2\log \left[{\left({\mathrm{O}}^{-1}1\right)}^2\times (0.1{)}^2\right]/1\\ =0.226-0.059/2\log {10}^{-4}\\ =0.226+0.118=0.344\mathrm{V}\end{array}$
b. Cells that convert the energy of combustion of fuels directly into electrical energy are
called fuel cells.
Advantages: (i) High efficiency.
(ii) Non polluting