The standard emf of the cell \(Cd(s)|CdCl_2(0.1M)||AgCl(s) |Ag(s)\) in which the cell reaction is \(Cd(s) + 2AgCl(s) \rightarrow 2Ag(s) + Cd^{2+}(aq) + 2Cl^–(aq)\) is 0.6915 V at 0°C and 0.6753 V at 25° C. \(\Delta H\) of the reaction is

–167 kJ

The correct answer is option 4. –167 kJ

We know,

\(\Delta G = −nFE_{cell}\)

\(\Delta G = −2 × 96500 × 0.6753 \) at \(25^oC\)

\(\Delta G = −130333J\)

\(\Delta S = nF\left(\frac{\delta E}{\delta T}\right)\) \(\left[\left[\frac{\delta E}{\delta T}\right] \text{is called temperature coefficient of cell}\right]\)

\(\left(\frac{\delta E}{\delta T}\right)_p = \frac{E_2 − E_1}{T_2 − T_1}\)

\(\left(\frac{\delta E}{\delta T}\right)_p = \frac{0.6753 − 0.6915}{298 − 273}\)

\(\left(\frac{\delta E}{\delta T}\right)_p = −6.48 × 10^{−4}\)

∴\(\Delta S = 2 × 96500 × (−6.48 × 10^{−4}) = − 125.064 J mol^{-1}\)

We know,

\(\Delta G = \Delta H − T\Delta S\)

or, \(\Delta H = \Delta G + T\Delta S\)

or, \(\Delta H = −130333 + 298 × (−125.064)\)

or, \(\Delta H = −167.6 kJ\)

As the reaction is an exothermic reaction so the sign is negative. Hence,  the magnitude of \(\Delta H\) is (4) 167.6 kJ.