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.
Answer & explanation
Correct answer: option 1
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, \(\Delta H^0 ≈ −7196.43 kJ\)