Calculate the useful work of the reaction Ag(s)+`1//2Cl_2(g)to AgCl(s)`
Given `E_(C l^(-)//C l^(-))^(@)=+1.36 V, E_(Ag|AgCl|Cl^(-))^(@)=0.22 V`
If `P_(Cl_(2))`=1 atm and T=298 K

Options

A. 110 kJ/mol
B. `220 "kJ"//"mol"`
C. `55 "kJ"//"mol"`
D. `1000 "kJ"//"mol"`

Explanation

To calculate the useful work of the reaction \( \text{Ag}(s) + \frac{1}{2} \text{Cl}_2(g) \rightarrow \text{AgCl}(s) \), we will follow these steps:

### Step 1: Identify the half-reactions
The overall reaction can be broken down into two half-reactions:

1. **Reduction at the cathode**:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag}(s) \]
This half-reaction corresponds to the reduction of silver ions to solid silver.

2. **Oxidation at the anode**:
\[ \frac{1}{2} \text{Cl}_2(g) + e^- \rightarrow \text{Cl}^- \]
This half-reaction corresponds to the oxidation of chlorine gas to chloride ions.

### Step 2: Write the standard electrode potentials
From the problem, we have the following standard electrode potentials:
- \( E^\circ_{\text{Cl}^-/\text{Cl}_2} = +1.36 \, \text{V} \)
- \( E^\circ_{\text{Ag}^+/Ag} = +0.22 \, \text{V} \)

### Step 3: Determine the standard cell potential \( E^\circ_{\text{cell}} \)
The standard cell potential is calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
In our case:
- Cathode (Ag): \( E^\circ_{\text{cathode}} = 0.22 \, \text{V} \)
- Anode (Cl): \( E^\circ_{\text{anode}} = 1.36 \, \text{V} \)

Thus,
\[ E^\circ_{\text{cell}} = 0.22 \, \text{V} - 1.36 \, \text{V} = -1.14 \, \text{V} \]

### Step 4: Calculate the Gibbs free energy change \( \Delta G \)
The Gibbs free energy change can be calculated using the formula:
\[ \Delta G = -nFE^\circ_{\text{cell}} \]
Where:
- \( n \) = number of moles of electrons transferred (which is 1 for this reaction)
- \( F \) = Faraday's constant \( = 96500 \, \text{C/mol} \)
- \( E^\circ_{\text{cell}} = -1.14 \, \text{V} \)

Substituting the values:
\[ \Delta G = -1 \times 96500 \, \text{C/mol} \times (-1.14 \, \text{V}) \]
\[ \Delta G = 96500 \times 1.14 = 110,010 \, \text{J/mol} = 110.01 \, \text{kJ/mol} \]

### Step 5: Conclusion
The useful work done by the reaction is approximately \( 110 \, \text{kJ/mol} \).