Question

The temperature coefficient ${\left(\frac{\partial E}{\partial T}\right)}_P$ for a given cell is $1.5\times {10}^{-4}J{K}^{-1}$ at 300 K. The change in entropy during cell reaction is:
$Pb\left(s\right)+HgC{l}_2\left(aq\right)\to PbC{l}_2\left(aq\right)+Hg\left(l\right)$

• A. 28.95 JK${}^{-1}$
• B. 14.47 JK${}^{-1}$
• C. 57.9 JK${}^{-1}$
• D. 21.70 JK${}^{-1}$

Answer 1

The correct option is A 28.95 JK${}^{-1}$

As we know,

${\left(\frac{\partial E}{\partial T}\right)}_P=\frac{\Delta S}{nF}$

$\therefore \Delta S={\left(\frac{\partial E}{\partial T}\right)}_P\times nF$

$=1.5\times {10}^{-4}\times 2\times 96500=28.95J{K}^{-1}$ (as n is 2)