For the following cell reaction,
`Ag|Ag^(+)|AgCl|Cl^(Theta)|Cl_(2),Pt`
`DeltaG_(f)^(o)(AgCl)=-109kJ//mol`
`DeltaG_(f)^(o)(Cl^(Theta))=-129kJ//mol`
`DeltaG_(f)^(o)(Ag^(+))=78kJ//mol`
`E^(o)` of the ell is:-

Options

A. `-0.60V`
B. `0.60V`
C. `6.0V`
D. None of these

Answer

To solve the problem, we need to calculate the standard cell potential \( E^\circ \) for the given electrochemical cell reaction using the provided Gibbs free energy values. The steps are as follows:

### Step 1: Write the Reaction
The cell reaction can be represented as:
\[ \text{AgCl (s)} + \text{e}^- \rightarrow \text{Ag (s)} + \text{Cl}^- \]

### Step 2: Identify the Gibbs Free Energy Values
We have the following Gibbs free energy values:
- \( \Delta G_f^\circ (\text{AgCl}) = -109 \, \text{kJ/mol} \)
- \( \Delta G_f^\circ (\text{Cl}^-) = -129 \, \text{kJ/mol} \)
- \( \Delta G_f^\circ (\text{Ag}^+) = 78 \, \text{kJ/mol} \)

### Step 3: Calculate the Gibbs Free Energy Change for the Reaction
Using the formula:
\[ \Delta G^\circ_{\text{reaction}} = \sum \Delta G_f^\circ (\text{products}) - \sum \Delta G_f^\circ (\text{reactants}) \]

Substituting the values:
\[ \Delta G^\circ_{\text{reaction}} = [\Delta G_f^\circ (\text{Ag}) + \Delta G_f^\circ (\text{Cl}^-)] - [\Delta G_f^\circ (\text{AgCl})] \]
Since \( \Delta G_f^\circ (\text{Ag}) = 0 \) (as it is in its standard state), we have:
\[ \Delta G^\circ_{\text{reaction}} = [0 + (-129)] - [-109] \]
\[ \Delta G^\circ_{\text{reaction}} = -129 + 109 \]
\[ \Delta G^\circ_{\text{reaction}} = -20 \, \text{kJ/mol} \]

### Step 4: Convert Gibbs Free Energy to Joules
Convert \( \Delta G^\circ \) from kJ to J:
\[ \Delta G^\circ_{\text{reaction}} = -20 \times 1000 = -20000 \, \text{J/mol} \]

### Step 5: Use the Relationship Between Gibbs Free Energy and Cell Potential
The relationship is given by:
\[ \Delta G^\circ = -nFE^\circ \]
Where:
- \( n \) = number of moles of electrons transferred (1 for this reaction)
- \( F \) = Faraday's constant \( = 96500 \, \text{C/mol} \)

Rearranging the equation to solve for \( E^\circ \):
\[ E^\circ = -\frac{\Delta G^\circ}{nF} \]
Substituting the known values:
\[ E^\circ = -\frac{-20000}{1 \times 96500} \]
\[ E^\circ = \frac{20000}{96500} \]

### Step 6: Calculate the Standard Cell Potential
Calculating the value:
\[ E^\circ \approx 0.207 \, \text{V} \]

### Step 7: Final Answer
The standard cell potential \( E^\circ \) is approximately:
\[ E^\circ \approx 0.207 \, \text{V} \]