Question

The standard e.m.f. of the cell,
$Cd ( s )\left| CdCl _{2}( aq )(0 \cdot 1 M )\right|| AgCl ( s )| Ag ( s )$
in which the cell reaction is
$Cd ( s )+2 AgCl ( s ) \longrightarrow 2 Ag ( s )+ Cd ^{2+}( aq )+2 Cl ^{-}( aq )$
is $0.6915 V$ at $0^{\circ} C$ and $0.6753 V$ 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
$Cd_{\left(s\right)}+2AgCl_{\left(s\right)} \rightarrow 2Ag_{\left(s\right)}+Cd_{\left(aq\right)}^{2+}+2Cl_{\left(aq\right)}^{-}$
$E_{1}=0.6915 V$ at $0^{\circ}C $
$E_{2}=0.6753 V$ at $25^{\circ}C $
Now, $\frac{\partial E_{\text{cell}}}{\partial T}=\frac{E_{2}-E_{1}}{T_{2}-T_{1}}$
$=\frac{0.6753-0.6915}{298-273}$
$=-6.48\times10^{-4}$
$\Delta S=nF \left[\frac{\partial E_{\text{cell}}}{\partial T}\right]$
Now, we put the value
$\Delta S=2\times96500\left(-6.48\times10^{-4}\right)$
$=-125.064$
We know that,
$\Delta G=-nFE_{\text{cell}} $
$=-2\times96500\times0.6753$
$=-1.303\times10^{5}$
As, $\Delta G=\Delta H -T \Delta S$
For calculating $\underset{{(25^{\circ}C)}} {{\Delta\,H}}=\Delta\,G+T \Delta\,S$
$=-1.303\times 10^{5}+298(-125.064 KJ)$
$\Delta\,H = -1.6726 \times 10^{5}\,J$
$=-167.26\,KJ$