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 $0.18V$ and $0.28V$ respectively?
The cell $:(Pt)H_2/H^{+}||KCl/H{g}_{2}C{l}_2/Hg$
(Given $:1F=96500$ coulombs)

• A. $\Delta S^{\circ }=2017J{K}^{-1}$ and $\Delta H^{\circ }=630.54kJ$
• B. $\Delta S^{\circ }=-4532J{K}^{-1}$ and $\Delta H^{\circ }=768.73kJ$
• C. $\Delta S^{\circ }=3425J{K}^{-1}$ and $\Delta H^{\circ }=-530.75kJ$
• D. $\Delta S^{\circ }=1930J{K}^{-1}$ and $\Delta H^{\circ }=530.75kJ$

Answer 1

Answer: (D)

Cathode: $H{g}_{2}C{l}_2(s)+2e^{-}\to 2H{g}_{(l)}+2C{l}_{(g)}$
Anode: $2H^{+}+2e^{-}\to H_2;n=2$
$\Delta G^{\circ }=-nF{E}^{\circ }$
$=-2\times 96500\times 0.18$
$=-34740J$
$\Delta S^{\circ }=nF\left(\frac{dE}{dT}\right)$
We can take $\frac{dE}{dt}=\frac{\Delta E}{\Delta T}$
$=\frac{0.28-0.18}{10}$
$=0.01V/K$
(From temperature $293K$ to $303K)$
$\Delta s^{\circ }=2\times 96500\times 0.01$
$=1930J{K}^{-1}$
$\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$
$\Delta H^{\circ }=\Delta G^{\circ }+T\Delta S^{\circ }$
$=-34740+293\times 1930$
$=530750J$
$=530.75KJ$