The cell potential `(E_(cell))` of a reaction is related as `/_\G=-nF E_(cell)`, where `/_\G `represents max. useful electrical work
n=no. of moles of electrons exchanged during the section
for reversible cell reaction `d(/_\G)=(/_\_rV)dp-(/_\_rS),dT`
at constant pressure `d(/_\G)=-(/_\_rS).dT`
:' At constant pressure `/_\G=/_\H-T./_\S`
:. `/_\G=H+T(d(/_\G))/((dT)_P)`
`((dE_(cell))/(dT))_P` is known as temperture coefficient of the e.m.f of the cell.
The temperature coefficient of the e.m.f of cell, `((dE)/(dT))_P` si given by:

Options

A. `(nF)/(/_\S)`
B. `(/_\S)/(nF)`
C. `(/_\S)/(nFT)`
D. #REF!

Answer

To find the temperature coefficient of the e.m.f. of the cell, we start with the relationship between the Gibbs free energy change (ΔG) and the cell potential (E_cell). The equation is given by:

\[ \Delta G = -nFE_{cell} \]

where:
- ΔG is the Gibbs free energy change,
- n is the number of moles of electrons exchanged,
- F is Faraday's constant,
- E_cell is the cell potential.

### Step 1: Differentiate ΔG with respect to temperature
At constant pressure, we know that:

\[ d\Delta G = -\Delta S dT \]

where ΔS is the entropy change. This means that:

\[ \frac{d\Delta G}{dT} = -\Delta S \]

### Step 2: Differentiate E_cell with respect to temperature
From the relationship between ΔG and E_cell, we can differentiate E_cell:

\[ dE_{cell} = -\frac{d\Delta G}{nF} \]

Substituting the expression for dΔG from Step 1 into this equation gives:

\[ dE_{cell} = -\left(-\Delta S dT\right) \frac{1}{nF} = \frac{\Delta S}{nF} dT \]

### Step 3: Find the temperature coefficient
Now, we can find the temperature coefficient of the e.m.f. of the cell, which is defined as:

\[ \left(\frac{dE_{cell}}{dT}\right)_{P} = \frac{\Delta S}{nF} \]

### Conclusion
Thus, the temperature coefficient of the e.m.f. of the cell is given by:

\[ \left(\frac{dE_{cell}}{dT}\right)_{P} = \frac{\Delta S}{nF} \]

### Final Answer
The temperature coefficient of the e.m.f. of the cell is \( \frac{\Delta S}{nF} \).

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