Explanation

(d)
The cell reaction is
Cd (s) + 2AgCl (s) → 2 Ag (s) + ${Cd}^{2 +}$ (aq) + ${2Cl}^{-}$ (aq)
$E_{1}$ = 0.6915 V at 0℃
$E_{2}$ = 0.6753 V at 25℃
Now, $\frac{\partial\ E_{cell}}{\partial\ T}$ = $\frac{E_{2} - \ E_{1}}{T_{2} - \ T_{1}}$
= $\frac{0.6753 - 0.6915}{298 - 273}$
= - 6.48 × $10^{- 4}$
△S =nF $\left\lbrack \frac{\partial\ E_{cell}}{\partial\ T} \right\rbrack$
Now, we put the value
△S = 2 × 96500 (- 6.48 × $10^{- 4}$ ) = - 125.064
We know that,
△G $= \ - nFE_{cell}$
= - 2 × 96500 × 0.6753
= - 1.303 × $10^{5}$
As, △G = △H - T△S
For calculating △H = △G + T△S
= - 1.303 × $10^{5}$ + 298 (- 125.064 kJ)
△H = - 1.6726 × $10^{5}$ J
= - 167.26 kJ