The standard electromotive force of the cell:

\(Fe | Fe^{2+}(aq) || Cd^{2+} | Cd \)is 0.0372 V

The temperature coefficient of emf is –0.125 V K^–1. Calculate the quantities \(\Delta G^0\), \(\Delta H^0\) and \(\Delta S^0\) at 25°C.

\(\Delta G^0\) = –7179.6 J, \(\Delta H^0\) = –7196.43 kJ, \(\Delta S^0\) = –24.125 kJ/K

The standard emf of the cell:

\(Fe | Fe^{2+}(aq) || Cd^{2+} | Cd \) is

\(E^0 = 0.0372 V\)

\(\left(\frac{dE}{dT}\right)_P = − 0.125 VK^{−1}\)

\(n = 2\)

 We know,

\(\Delta S^0 = nF\left(\frac{dE}{dT}\right)_P\)

or, \(\Delta S^0 = 2 × 96500 × (− 0.125)\)

or, \(\Delta S^0 = −24125 J/K\)

or, \(\Delta S^0 = − 24.125 kJ/K\)

Also,

\(\Delta G^0 = −nFE^0\)

or, \(\Delta G^0 = −2 × 96500 × 0.0372\)

or, \(\Delta G^0 = −7179.6 J\)

Now, we know

\(\Delta H^0 = \Delta G^0 + T\Delta S^0\)

or, \(\Delta H^0 = −7179.6 +298 ( − 24125)\)

or, \(\Delta H^0 = −7196429.6\)

or, \(\Delta H^0 ≈ −7196.43 kJ\)