For a reaction:` A(s) + 2B ^(+) rarr A^(2+) +2 B (s). K_(aq)` has been found to be `10^(12)` The `E^(o)` cell is :

Options

A. `0.354` Volt
B. `0.708` Volt
C. `0.0098 `Volt
D. `1.36` Volt

Answer

To solve the problem, we will use the Nernst equation to find the standard electrode potential (E° cell) for the given reaction.

### Step-by-Step Solution:

1. **Understand the Reaction**:
The reaction given is:
\[ A(s) + 2B^+ \rightarrow A^{2+} + 2B(s) \]
This indicates that solid A is oxidized to A²⁺ and B⁺ is reduced to solid B.

2. **Identify the Equilibrium Constant (K)**:
The equilibrium constant (K) for this reaction is given as:
\[ K = 10^{12} \]

3. **Use the Nernst Equation**:
The Nernst equation for calculating the standard electrode potential is:
\[ E°_{cell} = \frac{0.0591}{n} \log K \]
where \( n \) is the number of moles of electrons transferred in the reaction.

4. **Determine the Number of Electrons Transferred (n)**:
From the reaction, we can see that:
- A loses 2 electrons (oxidation).
- 2 B⁺ ions gain 2 electrons (reduction).
Therefore, the total number of electrons transferred (\( n \)) is 2.

5. **Substitute Values into the Nernst Equation**:
Now we can substitute the values into the equation:
\[ E°_{cell} = \frac{0.0591}{2} \log(10^{12}) \]

6. **Calculate the Logarithm**:
We know that:
\[ \log(10^{12}) = 12 \]
So, substituting this into the equation gives:
\[ E°_{cell} = \frac{0.0591}{2} \times 12 \]

7. **Perform the Calculation**:
\[ E°_{cell} = 0.0591 \times 6 = 0.3546 \, \text{V} \]
Rounding to three significant figures, we find:
\[ E°_{cell} \approx 0.354 \, \text{V} \]

8. **Final Answer**:
The standard electrode potential \( E°_{cell} \) is approximately:
\[ E°_{cell} = 0.354 \, \text{V} \]