Question

What will be the $ \Delta H^{\circ} $ and $ \Delta S^{\circ} $ values for the given cell having $ E^{\circ} $ at $ 20^{\circ} C $ and $ 30^{\circ} C $ to be $ 0.18\, V $ and $ 0.28\, V $ respectively?
The cell $ : (Pt) H_2/H^ + || KCl/Hg_2Cl_2/Hg $
(Given $ : 1 \,F = 96500 $ coulombs)

• A. $ \Delta S^{\circ} = 2017\, JK^{-1} $ and $ \Delta H^{\circ} = 630 .54\, kJ $
• B. $ \Delta S^{\circ} = - 4532 \,JK^{-1} $ and $ \Delta H^{\circ} = 768.73\, kJ $
• C. $ \Delta S^{\circ} = 3425 \,JK^{-1} $ and $ \Delta H^{\circ} = -530.75\, kJ $
• D. $ \Delta S^{\circ} = 1930 \,JK^{-1} $ and $ \Delta H^{\circ} = 530.75\, kJ $

Answer 1

Answer: (D)

Cathode: $Hg_{2}Cl_{2}(s)+2e^{-} \to 2Hg_{(l)}+2Cl_{(g)}$
Anode: $2H^{+}+2e^{-} \to H_{2} ; n=2$
$\Delta\,G^{\circ}=-nFE^{\circ}$
$=-2 \times 96500\times 0.18$
$=-34740\,J$
$\Delta\, S^{\circ} = nF \left(\frac{dE}{dT}\right)$
We can take $\frac{dE}{dt} =\frac {\Delta\, E}{\Delta T}$
$=\frac{0.28-0.18}{10}$
$=0.01V/K$
(From temperature $293\,K$ to $303\,K)$
$\Delta\, s^{\circ}=2\times 96500\times 0.01$
$=1930\,J\,K^{-1}$
$\Delta\,G^{\circ}=\Delta\, H^{\circ}-T \Delta \, S^{\circ}$
$\Delta \,H^{\circ}=\Delta\, G^{\circ}+T \Delta\,S^{\circ}$
$=-34740+293 \times 1930$
$=530750\,J$
$=530.75\,KJ$