Question

Select the incorrect relation among the following.

• A. $\Delta S = (\frac{\partial E}{\partial T})_P \times nF$
• B. $-\Delta S = (\frac{\partial E}{\partial T})_P \times nF$
• C. $(\frac{\partial E}{\partial T})_P = (\frac{\partial \Delta S}{\partial T})$
• D. $(\frac{\partial E}{\partial T})_P = \frac{\Delta H + nEF}{nFT}$

Answer 1

Answer: (B)

B. $-\Delta S = (\frac{\partial E}{\partial T})_P \times nF$

Answer 2

The question asks to identify the incorrect relation among the given options, which are related to the thermodynamics of electrochemical cells.

We start with the fundamental thermodynamic relations:

1. The change in Gibbs free energy for a reversible electrochemical cell is given by: $\Delta G = -nFE$ where $n$ is the number of moles of electrons transferred, $F$ is Faraday's constant, and $E$ is the cell potential.

2. The general thermodynamic relation between Gibbs free energy change, enthalpy change, and entropy change is: $\Delta G = \Delta H - T\Delta S$

3. The temperature dependence of Gibbs free energy at constant pressure is related to entropy change: $(\frac{\partial \Delta G}{\partial T})_P = -\Delta S$

Now, let's derive the correct relations and compare them with the given options:

Derivation of $\Delta S$ in terms of $E$ and $T$: Differentiate equation (1) with respect to temperature at constant pressure: $(\frac{\partial \Delta G}{\partial T})_P = -nF(\frac{\partial E}{\partial T})_P$

Equating this with equation (3): $-nF(\frac{\partial E}{\partial T})_P = -\Delta S$ $\Delta S = nF(\frac{\partial E}{\partial T})_P$

Checking Option A: Option A states: $\Delta S = (\frac{\partial E}{\partial T})_P \times nF$. This matches our derived relation. Therefore, Option A is a correct relation.

Checking Option B: Option B states: $-\Delta S = (\frac{\partial E}{\partial T})_P \times nF$. This is the negative of the correct relation derived above. Therefore, Option B is an incorrect relation.

Derivation of $(\frac{\partial E}{\partial T})_P$ in terms of $\Delta H$ and $E$ (Gibbs-Helmholtz equation for cell potential): Substitute $\Delta G = -nFE$ into the Gibbs-Helmholtz equation: $\Delta G = \Delta H + T(\frac{\partial \Delta G}{\partial T})_P$. $-nFE = \Delta H + T(\frac{\partial (-nFE)}{\partial T})_P$ $-nFE = \Delta H - nFT(\frac{\partial E}{\partial T})_P$ Rearranging the terms to solve for $(\frac{\partial E}{\partial T})_P$: $nFT(\frac{\partial E}{\partial T})_P = \Delta H + nFE$ $(\frac{\partial E}{\partial T})_P = \frac{\Delta H + nFE}{nFT}$

Checking Option D: Option D states: $(\frac{\partial E}{\partial T})_P = \frac{\Delta H + nEF}{nFT}$. This matches our derived relation. Therefore, Option D is a correct relation.

Checking Option C: Option C states: $(\frac{\partial E}{\partial T})_P = (\frac{\partial \Delta S}{\partial T})$. From our derived correct relation, we know $(\frac{\partial E}{\partial T})_P = \frac{\Delta S}{nF}$. So, Option C implies $\frac{\Delta S}{nF} = (\frac{\partial \Delta S}{\partial T})$. This relation is generally not true for thermodynamic processes. For this equality to hold, $\Delta S$ would have to follow a specific exponential function of $T$ (i.e., $\Delta S = A e^{T/nF}$), which is not a general thermodynamic relationship. Therefore, Option C is an incorrect relation.

Conclusion: Both Option B and Option C are incorrect relations. However, in typical multiple-choice questions asking to select "the incorrect relation", usually only one option is intended to be incorrect. Option B is a direct sign error of a fundamental and widely used relation for the temperature coefficient of cell potential and entropy change. This type of error is a common distractor in such questions. Option C, while also incorrect, relates derivatives in a less direct manner. Given the standard format of such questions, a direct sign error (Option B) is often the intended incorrect option when multiple options appear incorrect.