Question

What is the cell entropy change of the following cell ?
$Pt\left(s\right)|H_2(g,1atm)|C{H}_{3}COOH,HCl||KCl\left(aq\right)|H{g}_{2}C{l}_2\left(s\right)|Hg$
EMF of the cell is found to be $0.0045V$ at 298 K and temperature coefficient is $3.4\times {10}^{-4}V{K}^{-1}$

• A. $105kJ{K}^{-1}$
• B. $205.7J{K}^{-1}$
• C. $5.45J{K}^{-1}$
• D. $65.61J{K}^{-1}$

Answer 1

The correct option is D $65.61J{K}^{-1}$

Reactions involved in the cell are
$H_2\to 2H^{+}+2e^{-}$
$H{g}^{2+}+2e^{-}\to 2Hg$

Thus, 2 electron is involved in the cell reaction.
We know,
Entropy change,
$\Delta S=nF{\left(\frac{\partial E}{\partial T}\right)}_P$
$\left(\frac{\partial E}{\partial T}\right)$ is temperature coefficient of cell reaction.

$\Delta S=2\times 96485C\times 3.4\times {10}^{-4}V{K}^{-1}=65.61J{K}^{-1}$