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

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

we will follow these steps:

### Step 1: Identify the half-reactions and their standard electrode potentials.
The given standard electrode potentials are:
- For the reduction of \(\text{Fe}^{3+}\) to \(\text{Fe}^{2+}\):
\[ E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = 0.77 \, \text{V} \]
- For the reduction of \(\text{Ag}^+\) to \(\text{Ag}\):
\[ E^\circ(\text{Ag}^+/Ag) = 0.80 \, \text{V} \]

### Step 2: Determine the overall cell potential (\(E^\circ_{\text{cell}}\)).
The overall cell potential can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

In our case:
- \(\text{Ag}^+\) is reduced at the cathode (0.80 V).
- \(\text{Fe}^{3+}\) is oxidized at the anode (0.77 V).

Thus,
\[ E^\circ_{\text{cell}} = 0.80 \, \text{V} - 0.77 \, \text{V} = 0.03 \, \text{V} \]

### Step 3: Relate the cell potential to the Gibbs free energy change (\(\Delta G^\circ\)).
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 (\(96500 \, \text{C/mol}\)),
- \(E^\circ_{\text{cell}}\) = \(0.03 \, \text{V}\).

Substituting the values:
\[ \Delta G^\circ = -1 \times 96500 \times 0.03 = -2895 \, \text{J/mol} \]

### Step 4: Relate \(\Delta G^\circ\) to the equilibrium constant (K).
The relationship between \(\Delta G^\circ\) and the equilibrium constant is given by:
\[ \Delta G^\circ = -2.303RT \log K \]
where:
- \(R\) = gas constant (\(8.314 \, \text{J/(mol K)}\)),
- \(T\) = temperature in Kelvin (standard temperature \(298 \, \text{K}\)).

Rearranging gives:
\[ \log K = -\frac{\Delta G^\circ}{2.303RT} \]

Substituting the values:
\[ \log K = -\frac{-2895}{2.303 \times 8.314 \times 298} \]

Calculating the denominator:
\[ 2.303 \times 8.314 \times 298 \approx 5730.2 \]

Now substituting:
\[ \log K = \frac{2895}{5730.2} \approx 0.505 \]

### Step 5: Calculate K from \(\log K\).
To find \(K\):
\[ K = 10^{0.505} \approx 3.16 \]

### Final Answer:
Thus, the equilibrium constant \(K\) for the reaction is approximately \(3.16\).

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