`DeltaG=DeltaH-TDeltaS` and ltBRgt `DeltaG=DeltaH+T[(d(DeltaG))/(dT)]_(p),` then` ((dE_(cell))/(dT))` is

Options

A. `(DeltaS)/(nF)`
B. `(nE)/(DeltaS)`
C. `-nFE_(cell)`
D. `nFE_(cell)`

Explanation

To solve the problem step-by-step, we will analyze the given equations and derive the required expression for \(\frac{dE_{cell}}{dT}\).

### Step 1: Understand the given equations
We have two equations:
1. \(\Delta G = \Delta H - T \Delta S\)
2. \(\Delta G = \Delta H + T \left(\frac{d(\Delta G)}{dT}\right)_{p}\)

### Step 2: Set the equations equal to each other
Since both expressions are equal to \(\Delta G\), we can set them equal to each other:
\[ \Delta H - T \Delta S = \Delta H + T \left(\frac{d(\Delta G)}{dT}\right)_{p} \]

### Step 3: Cancel out \(\Delta H\)
Subtract \(\Delta H\) from both sides:
\[ -T \Delta S = T \left(\frac{d(\Delta G)}{dT}\right)_{p} \]

### Step 4: Divide by T
Assuming \(T \neq 0\), divide both sides by \(T\):
\[ -\Delta S = \left(\frac{d(\Delta G)}{dT}\right)_{p} \]

### Step 5: Relate \(\Delta S\) to \(\frac{dE_{cell}}{dT}\)
From thermodynamics, we know that:
\[ \Delta S = nF \left(\frac{dE_{cell}}{dT}\right) \]
Thus, we can substitute this into our equation:
\[ -\Delta S = -nF \left(\frac{dE_{cell}}{dT}\right) \]

### Step 6: Rearranging the equation
This gives us:
\[ \left(\frac{dE_{cell}}{dT}\right) = \frac{\Delta S}{nF} \]

### Conclusion
Thus, the final expression for \(\frac{dE_{cell}}{dT}\) is:
\[ \frac{dE_{cell}}{dT} = \frac{\Delta S}{nF} \]

### Final Answer
The correct option is:
\[ \frac{\Delta S}{nF} \]

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