STATEMENT-1: If `((dE_(cell))/(dT))_Pgt0,` For a cell reaction then `/_\S` is positive.
STATEMENT-2: `/_\S = nFT((dE)/(dT))_p`

Options

A. If both the statements are TRUE and STATEMENTS-2 is the correct explantion of STATEMENTS-11
B. If both the statements are TRUE but STATEMENTS-2 is NOT the correct explanation of STATEMENTS-13
C. If STATEMENTS-1 is TRUE and STATEMENTS-2 is FALSE
D. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE

Answer

To solve the problem, we need to analyze both statements and their interrelation step by step.

### Step 1: Understanding Statement 1
Statement 1 claims that if \(\left(\frac{dE_{cell}}{dT}\right)_P > 0\) for a cell reaction, then \(\Delta S\) is positive.

- The term \(\frac{dE_{cell}}{dT}\) represents the temperature coefficient of the cell potential.
- A positive value for this coefficient indicates that as temperature increases, the cell potential also increases.

### Step 2: Relating \(\Delta S\) to \(\frac{dE_{cell}}{dT}\)
From thermodynamics, we know the relationship between the change in Gibbs free energy (\( \Delta G \)), cell potential (\( E_{cell} \)), and entropy (\( \Delta S \)) can be expressed as:
\[ \Delta G = -nFE_{cell} \]
At constant pressure and temperature, the change in Gibbs free energy is also related to entropy by:
\[ \Delta G = \Delta H - T\Delta S \]
Differentiating with respect to temperature at constant pressure gives:
\[ \left(\frac{dG}{dT}\right)_P = -nF\left(\frac{dE_{cell}}{dT}\right)_P \]
This implies:
\[ \Delta S = -\left(\frac{dG}{dT}\right)_P = nF\left(\frac{dE_{cell}}{dT}\right)_P \]
Thus, if \(\left(\frac{dE_{cell}}{dT}\right)_P > 0\), it follows that \(\Delta S > 0\).

### Step 3: Analyzing Statement 2
Statement 2 states that:
\[ \Delta S = nFT\left(\frac{dE}{dT}\right)_P \]
This statement is incorrect because the correct relationship derived earlier does not include the temperature \(T\) multiplying the term \(\left(\frac{dE_{cell}}{dT}\right)_P\). The correct expression is:
\[ \Delta S = nF\left(\frac{dE_{cell}}{dT}\right)_P \]

### Conclusion
- **Statement 1** is correct: If \(\left(\frac{dE_{cell}}{dT}\right)_P > 0\), then \(\Delta S\) is indeed positive.
- **Statement 2** is incorrect as it misrepresents the relationship by including an extra factor of \(T\).

### Final Answer
- Statement 1 is correct.
- Statement 2 is incorrect.