The emf of the cell,
`Zn|Zn^(2+) (0.01 M)||Fe^(2+) (0.001 M)|Fe` at 298 K is 0.2905 V then the value of equilibrium constant for the cell reaction is :

Options

A. `e^(0.32//0.0295)`
B. `10^(0.32//0.0295)`
C. `10^(0.26//0.0295)`
D. `10^(0.32//0.0591)`

Answer

To find the equilibrium constant for the cell reaction given the emf of the cell, we can follow these steps:

### Step 1: Write the Cell Reaction
The cell consists of a zinc electrode and an iron electrode. The half-reactions are:
- Oxidation: \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \)
- Reduction: \( \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \)

The overall cell reaction is:
\[ \text{Zn} + \text{Fe}^{2+} \rightarrow \text{Zn}^{2+} + \text{Fe} \]

### Step 2: Identify the Standard Electrode Potentials
From standard reduction potential tables, we find:
- For zinc: \( E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \, \text{V} \) (oxidation potential is +0.76 V)
- For iron: \( E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44 \, \text{V} \)

### Step 3: Calculate the Standard Cell Potential
The standard cell potential \( E^\circ_{\text{cell}} \) is calculated as follows:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
\[ E^\circ_{\text{cell}} = (-0.44) - (-0.76) = 0.32 \, \text{V} \]

### Step 4: Use the Nernst Equation
The Nernst equation relates the cell potential to the concentrations of the reactants and products:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q \]
Where:
- \( n = 2 \) (number of electrons transferred)
- \( Q = \frac{[\text{Zn}^{2+}]}{[\text{Fe}^{2+}]} = \frac{0.01}{0.001} = 10 \)

### Step 5: Substitute Values into the Nernst Equation
Given \( E_{\text{cell}} = 0.2905 \, \text{V} \):
\[ 0.2905 = 0.32 - \frac{0.0591}{2} \log(10) \]

### Step 6: Solve for \( \log K \)
Rearranging gives:
\[ 0.2905 = 0.32 - 0.02955 \]
\[ 0.2905 = 0.29045 \]
This confirms our calculations are consistent. Now we can find \( K \):
Since at equilibrium, \( E_{\text{cell}} = 0 \):
\[ 0 = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log K \]
\[ \log K = \frac{n \cdot E^\circ_{\text{cell}}}{0.0591} \]
\[ \log K = \frac{2 \cdot 0.32}{0.0591} \]
\[ \log K = \frac{0.64}{0.0591} \approx 10.81 \]

### Step 7: Calculate \( K \)
\[ K = 10^{10.81} \approx 6.46 \times 10^{10} \]

### Final Answer
The value of the equilibrium constant \( K \) for the cell reaction is approximately \( 6.46 \times 10^{10} \).

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