The logarithm of the equilibriium constant of the cell reaction corresponding to the cell
`X(s)|x^(2+)(aq)||Y^(+)(aq)|Y(s)` with standard cell potential `E_(cell)^(@)=1.2V` given by

Options

A. 40.2
B. 47.2
C. 12.5
D. 21.5

Answer

To solve the problem of finding the logarithm of the equilibrium constant (Kc) for the given cell reaction, we will follow these steps:

### Step 1: Identify the half-reactions
The cell reaction is given as:
\[ X(s) | X^{2+}(aq) || Y^{+}(aq) | Y(s) \]

The half-reactions can be identified as:
1. Oxidation: \( X \rightarrow X^{2+} + 2e^- \)
2. Reduction: \( 2Y^{+} + 2e^- \rightarrow 2Y \)

### Step 2: Write the balanced overall cell reaction
Combining the half-reactions, we get the overall balanced cell reaction:
\[ X + 2Y^{+} \rightarrow X^{2+} + 2Y \]

### Step 3: Determine the number of electrons transferred (n)
From the balanced equation, we can see that 2 electrons are transferred in the reaction:
\[ n = 2 \]

### Step 4: Use the Nernst equation
The Nernst equation for the standard cell potential (\( E^\circ_{cell} \)) is given by:
\[ E^\circ_{cell} = \frac{0.0591}{n} \log K_c \]

### Step 5: Substitute the known values into the equation
We know:
- \( E^\circ_{cell} = 1.2 \, V \)
- \( n = 2 \)

Substituting these values into the Nernst equation gives:
\[ 1.2 = \frac{0.0591}{2} \log K_c \]

### Step 6: Rearrange the equation to solve for \( \log K_c \)
Rearranging the equation to isolate \( \log K_c \):
\[ \log K_c = \frac{1.2 \times 2}{0.0591} \]

### Step 7: Calculate \( \log K_c \)
Calculating the right-hand side:
\[ \log K_c = \frac{2.4}{0.0591} \approx 40.5 \]

### Step 8: Conclusion
Thus, the logarithm of the equilibrium constant \( K_c \) is approximately:
\[ \log K_c \approx 40.5 \]

### Final Answer
The required option is \( 40.5 \).

---