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For a reversible galvanic cell with $n$ electrons transferred, the entropy change of the cell reaction is related to the temperature coefficient of the EMF by

A $Δ_{r} S = n F (\frac{\partial E}{\partial T})_{P}$
B $Δ_{r} H = n F (\frac{\partial E}{\partial T})_{P}$
C $Δ_{r} S = - n F (\frac{\partial E}{\partial T})_{P}$
D $Δ_{r} G = n F (\frac{\partial E}{\partial T})_{P}$

Answer: A

From

Δ_{r} G = - n FE

and the thermodynamic relation

(\frac{\partial Δ G}{\partial T})_{P} = - Δ S

, we get

- n F (\frac{\partial E}{\partial T})_{P} = - Δ_{r} S

, so

Δ_{r} S = n F (\frac{\partial E}{\partial T})_{P}

. A cell whose EMF rises with temperature has a positive entropy change.

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