Question

The standard e.m.f. of the cell,
$Cd(s)\left|CdC{l}_2(aq)(0\cdot 1M)\right||AgCl(s)|Ag(s)$
in which the cell reaction is
$Cd(s)+2AgCl(s)⟶2Ag(s)+C{d}^{2+}(aq)+2C{l}^{-}(aq)$
is $0.6915V$ at $0^{\circ }C$ and $0.6753V$ at $25^{\circ }C$. The enthalpy change of the reaction at $25^{\circ }C$ is

• A. −176 kJ
• B. −234.7 kJ
• C. +123.5 kJ
• D. −167.26 kJ

Answer 1

Answer: (D)

The cell reaction is
$C{d}_{\left(s\right)}+2AgC{l}_{\left(s\right)}\to 2A{g}_{\left(s\right)}+C{d}_{\left(aq\right)}^{2+}+2C{l}_{\left(aq\right)}^{-}$
$E_1=0.6915V$ at $0^{\circ }C$
$E_2=0.6753V$ at $25^{\circ }C$
Now, $\frac{\partial E_{\text{cell}}}{\partial T}=\frac{E_2-E_1}{T_2-T_1}$
$=\frac{0.6753-0.6915}{298-273}$
$=-6.48\times 10^{-4}$
$\Delta S=nF\left[\frac{\partial E_{\text{cell}}}{\partial T}\right]$
Now, we put the value
$\Delta S=2\times 96500\left(-6.48\times 10^{-4}\right)$
$=-125.064$
We know that,
$\Delta G=-nF{E}_{\text{cell}}$
$=-2\times 96500\times 0.6753$
$=-1.303\times 10^5$
As, $\Delta G=\Delta H-T\Delta S$
For calculating $\underset{(25^{\circ }C)}{\mathrm{\Delta H}}=\Delta G+T\Delta S$
$=-1.303\times 10^5+298(-125.064KJ)$
$\Delta H=-1.6726\times 10^{5}J$
$=-167.26KJ$