The standard potential of the cell formed by combining the `Cl_(2)//Cl^(-)(aq)` half - cell with the standard hydrogen electrode is `+1.36V and ((delE^(@))/(delT))_(P)=-1.2xx10^(-3)VK^(-1)`. What is the value of `DeltaS_("reaction")^(@)` for reaction
`H_(2(g))+Cl_(2(g))rarr2H_((aq))^(+)+2Cl_((aq))^(-)`

Options

A. `-1xx10^(2)JK^(-1)`
B. `2.3xx10^(2)JK^(-1)`
C. `-2.3xx10^(2)JK^(-1)`
D. `1JK^(-1)`

Answer

To solve the problem, we need to find the value of ΔS (entropy change) for the given reaction using the provided data. Let's break down the steps:

### Step 1: Identify the Reaction and Determine N
The reaction given is:
\[ H_{2(g)} + Cl_{2(g)} \rightarrow 2H_{(aq)}^{+} + 2Cl_{(aq)}^{-} \]

In this reaction:
- Chlorine gas \( Cl_2 \) is reduced to \( 2Cl^{-} \), which involves the transfer of 2 electrons.
- Hydrogen gas \( H_2 \) is oxidized to \( 2H^{+} \), which also involves the transfer of 2 electrons.

Thus, the total number of electrons transferred (N) in the overall reaction is:
\[ N = 2 + 2 = 2 \]

### Step 2: Use the Formula for ΔS
The formula to calculate the standard entropy change (ΔS) for the reaction is given by:
\[ \Delta S^{\circ}_{\text{reaction}} = N \cdot F \cdot \left( \frac{dE^{\circ}}{dT} \right)_{P} \]

Where:
- \( N \) is the number of moles of electrons transferred (which we found to be 2).
- \( F \) is Faraday's constant, approximately \( 96500 \, C/mol \).
- \( \left( \frac{dE^{\circ}}{dT} \right)_{P} \) is given as \( -1.2 \times 10^{-3} \, V/K \).

### Step 3: Substitute the Values into the Formula
Now, substituting the values into the formula:
\[ \Delta S^{\circ}_{\text{reaction}} = 2 \cdot 96500 \, C/mol \cdot (-1.2 \times 10^{-3} \, V/K) \]

### Step 4: Calculate ΔS
Calculating the value:
\[ \Delta S^{\circ}_{\text{reaction}} = 2 \cdot 96500 \cdot (-1.2 \times 10^{-3}) \]
\[ = 2 \cdot 96500 \cdot -0.0012 \]
\[ = -231.6 \, J/K \]

### Step 5: Final Result
Thus, the value of \( \Delta S^{\circ}_{\text{reaction}} \) is approximately:
\[ \Delta S^{\circ}_{\text{reaction}} \approx -2.3 \times 10^{2} \, J/K \]

### Conclusion
The final answer is:
\[ \Delta S^{\circ}_{\text{reaction}} = -2.3 \times 10^{2} \, J/K \]

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