The standard e.m.f. for the cell reaction.
`2Cu^(+)(aq)rarrCu(s)+Cu^(2+)(aq)` is +0.36 V at 298K. The equilibrium constant of the reaction is

Options

A. `5xx10^(6)`
B. `1.4xx10^(12)`
C. `7.4xx10^(12)`
D. `1.3xx10^(6)`

Explanation

To find the equilibrium constant (K) for the given cell reaction, we can use the Nernst equation. The reaction provided is:

\[ 2Cu^+(aq) \rightleftharpoons Cu(s) + Cu^{2+}(aq) \]

The standard EMF (E°) for this cell reaction is given as +0.36 V at 298 K.

### Step-by-Step Solution:

1. **Identify the Nernst Equation**:
The Nernst equation relates the standard EMF of a cell to the equilibrium constant:
\[ E = E^\circ - \frac{0.0591}{n} \log K \]
At equilibrium, the cell potential (E) is 0. Therefore, we can set up the equation as:
\[ 0 = E^\circ - \frac{0.0591}{n} \log K \]

2. **Determine the number of electrons transferred (n)**:
In the given reaction, each \( Cu^+ \) ion is reduced to \( Cu \) and each \( Cu^+ \) ion oxidizes to \( Cu^{2+} \). Since 2 moles of \( Cu^+ \) produce 1 mole of \( Cu^{2+} \) and 1 mole of \( Cu \), the total number of electrons transferred (n) is 1.

3. **Rearranging the Nernst Equation**:
Rearranging the Nernst equation gives:
\[ \frac{0.0591}{n} \log K = E^\circ \]
Plugging in the values:
\[ \log K = \frac{E^\circ \cdot n}{0.0591} \]

4. **Substituting the known values**:
Substitute \( E^\circ = 0.36 \, V \) and \( n = 1 \):
\[ \log K = \frac{0.36}{0.0591} \]

5. **Calculating log K**:
Performing the calculation:
\[ \log K = 6.09 \]

6. **Finding K**:
To find K, we take the antilog:
\[ K = 10^{6.09} \]
Calculating this gives:
\[ K \approx 1.3 \times 10^6 \]

### Final Answer:
The equilibrium constant \( K \) for the reaction is approximately \( 1.3 \times 10^6 \).