The equilibrium constant (K) for the reaction
`Fe^(2+)(aq)+Ag^(+)(aq)toFe^(3+)(aq)+Ag(s)` will be
Given `E^(@)(Fe^(3+)//Fe^(2+))=0.77V,`
`E^(@)(Ag^(+)//Ag)=0.80V`

Options

A. 100
B. 3.16
C. 10
D. 0.5

Explanation

To find the equilibrium constant (K) for the reaction:

\[ \text{Fe}^{2+}(aq) + \text{Ag}^{+}(aq) \rightarrow \text{Fe}^{3+}(aq) + \text{Ag}(s) \]

we will follow these steps:

### Step 1: Identify the half-reactions
The half-reactions for the given reaction are:
1. Reduction (cathode):
\[ \text{Ag}^{+} + e^{-} \rightarrow \text{Ag}(s) \]
2. Oxidation (anode):
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^{-} \]

### Step 2: Write the standard reduction potentials
Given:
- \( E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = 0.77 \, \text{V} \)
- \( E^\circ(\text{Ag}^{+}/\text{Ag}) = 0.80 \, \text{V} \)

### Step 3: Determine the overall cell potential
The overall cell potential (\( E^\circ_{\text{cell}} \)) can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Substituting the values:
\[ E^\circ_{\text{cell}} = 0.80 \, \text{V} - 0.77 \, \text{V} = 0.03 \, \text{V} \]

### Step 4: Relate cell potential to Gibbs free energy
The relationship between the standard cell potential and the Gibbs free energy change is given by:
\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]
where:
- \( n \) = number of moles of electrons transferred (which is 1 for this reaction)
- \( F \) = Faraday's constant \( (96500 \, \text{C/mol}) \)

### Step 5: Calculate \( \Delta G^\circ \)
Substituting the values:
\[ \Delta G^\circ = -1 \times 96500 \, \text{C/mol} \times 0.03 \, \text{V} = -2895 \, \text{J/mol} \]

### Step 6: Relate Gibbs free energy to the equilibrium constant
The relationship between Gibbs free energy and the equilibrium constant is:
\[ \Delta G^\circ = -RT \ln K \]
where:
- \( R \) = universal gas constant \( (8.314 \, \text{J/(mol K)}) \)
- \( T \) = temperature in Kelvin (standard temperature is 298 K)

### Step 7: Rearrange to find \( K \)
Rearranging the equation gives:
\[ \ln K = -\frac{\Delta G^\circ}{RT} \]
Substituting the values:
\[ \ln K = -\frac{-2895}{8.314 \times 298} \]

### Step 8: Calculate \( K \)
Calculating the right side:
\[ \ln K = \frac{2895}{2477.572} \approx 1.167 \]

Now, exponentiating to find \( K \):
\[ K = e^{1.167} \approx 3.216 \]

### Final Result
Thus, the equilibrium constant \( K \) for the reaction is approximately:
\[ K \approx 3.16 \]

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