If `DeltaG=DeltaH-TDeltaS and DeltaG = DeltaH+T[(d(DeltaG))/(dT)]` , then variation EMF of a cell E, with temperature T, is given by

Options

A. `(DeltaH)/(nF)`
B. `(DeltaG)/(nF)`
C. `(DeltaS)/(nF)`
D. `(-DeltaS)/(nF)`

Answer

To derive the variation of EMF of a cell (E) with temperature (T) using the given equations, we can follow these steps:

### Step-by-Step Solution:

1. **Start with the Gibbs Free Energy Equations**:
We have two equations:
- \( \Delta G = \Delta H - T \Delta S \) (Equation 1)
- \( \Delta G = \Delta H + T \left( \frac{d(\Delta G)}{dT} \right) \) (Equation 2)

2. **Set the Two Equations Equal**:
Since both equations equal \( \Delta G \), we can set them equal to each other:
\[ \Delta H - T \Delta S = \Delta H + T \left( \frac{d(\Delta G)}{dT} \right) \]

3. **Cancel Out \( \Delta H \)**:
Subtract \( \Delta H \) from both sides:
\[ -T \Delta S = T \left( \frac{d(\Delta G)}{dT} \right) \]

4. **Divide by T**:
Assuming \( T \neq 0 \), we can divide both sides by \( T \):
\[ -\Delta S = \frac{d(\Delta G)}{dT} \]

5. **Relate to EMF**:
The change in Gibbs free energy (\( \Delta G \)) is related to the EMF of the cell (\( E \)) by the equation:
\[ \Delta G = -nFE \]
where \( n \) is the number of moles of electrons transferred and \( F \) is Faraday's constant.

6. **Differentiate with Respect to Temperature**:
Taking the derivative with respect to temperature:
\[ \frac{d(\Delta G)}{dT} = -nF \frac{dE}{dT} \]

7. **Substitute Back**:
Substitute \( \frac{d(\Delta G)}{dT} \) from step 4 into this equation:
\[ -\Delta S = -nF \frac{dE}{dT} \]

8. **Rearranging the Equation**:
Rearranging gives us:
\[ \frac{dE}{dT} = \frac{\Delta S}{nF} \]

### Final Result:
Thus, the variation of EMF of the cell with temperature is given by:
\[ \frac{dE}{dT} = \frac{\Delta S}{nF} \]