If the temperature coefficient `((delE)/(delT))` is zero for a cell reaction then out of `DeltaS, DeltaH`, and `DeltaG` , the `…………………..` is zero.

Explanation

To solve the question, we need to analyze the relationship between the temperature coefficient of the cell reaction and the thermodynamic quantities: ΔS (change in entropy), ΔH (change in enthalpy), and ΔG (change in Gibbs free energy).

### Step-by-Step Solution:

1. **Understanding the Temperature Coefficient**:
- The temperature coefficient \(\left(\frac{\partial E}{\partial T}\right)\) being zero implies that the cell potential (E) does not change with temperature.

2. **Relating ΔS to the Temperature Coefficient**:
- The change in entropy (ΔS) can be expressed in terms of the temperature coefficient:
\[ \Delta S = nF \left(\frac{\partial E}{\partial T}\right) \]
- Here, \(n\) is the number of moles of electrons transferred, and \(F\) is Faraday's constant.

3. **Substituting the Given Condition**:
- Since \(\frac{\partial E}{\partial T} = 0\), we substitute this into the equation for ΔS:
\[ \Delta S = nF \cdot 0 = 0 \]
- Therefore, we conclude that ΔS is zero.

4. **Analyzing ΔH**:
- The change in enthalpy (ΔH) is given by the equation:
\[ \Delta H = -nF E_{\text{cell}} - T\left(\frac{\partial E}{\partial T}\right) \]
- Since \(\frac{\partial E}{\partial T} = 0\), this simplifies to:
\[ \Delta H = -nF E_{\text{cell}} \]
- This means ΔH is not necessarily zero, as \(E_{\text{cell}}\) is generally not zero.

5. **Analyzing ΔG**:
- The change in Gibbs free energy (ΔG) is related to ΔH and ΔS by the equation:
\[ \Delta G = \Delta H - T\Delta S \]
- Substituting ΔS = 0 into this equation gives:
\[ \Delta G = \Delta H - T \cdot 0 = \Delta H \]
- Since ΔH is not zero, ΔG is also not zero.

6. **Final Conclusion**:
- Out of ΔS, ΔH, and ΔG, the only quantity that is zero when \(\frac{\partial E}{\partial T} = 0\) is ΔS.

### Answer:
The correct answer is that **ΔS is zero**.

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