Celll equation : `A + 2B^(+) rarr A^(2+) + 2 B`
`A^(2+) + 2e rarr A`
`E^@ = + 0.34 V` and `log_(10) K= 15. 6 ` at `300 K` for cell reactions Find `E^@` for `B^(+) +e rarr B`
Given `[( 2.303 RT)/(nF) =0.059]` at `300 K`.

Options

A. 0.81
B. 1.26
C. `-0.54`
D. `+0.94`

Explanation

To solve the problem step by step, we will follow the provided information and apply the relevant equations.

### Step 1: Understand the Cell Reaction
The cell reaction is given as:
\[ A + 2B^+ \rightarrow A^{2+} + 2B \]
This indicates that \( A \) is being oxidized to \( A^{2+} \) and \( B^+ \) is being reduced to \( B \).

### Step 2: Identify the Half-Reactions
From the overall cell reaction, we can identify the half-reactions:
1. Oxidation (Anode):
\[ A \rightarrow A^{2+} + 2e^- \]
2. Reduction (Cathode):
\[ B^+ + e^- \rightarrow B \]

### Step 3: Use the Nernst Equation
The standard cell potential \( E^\circ_{cell} \) can be calculated using the equation:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]

### Step 4: Relate \( E^\circ_{cell} \) to Equilibrium Constant \( K \)
We are given:
\[ \log_{10} K = 15.6 \]
Using the relationship between cell potential and equilibrium constant:
\[ E^\circ_{cell} = \frac{0.059}{n} \log_{10} K \]
where \( n \) is the number of electrons transferred in the balanced equation. Here, \( n = 2 \).

### Step 5: Substitute Values into the Equation
Substituting the values into the equation:
\[ E^\circ_{cell} = \frac{0.059}{2} \times 15.6 \]

### Step 6: Calculate \( E^\circ_{cell} \)
Calculating the above expression:
1. Calculate \( \frac{0.059}{2} = 0.0295 \)
2. Now multiply by \( 15.6 \):
\[ E^\circ_{cell} = 0.0295 \times 15.6 = 0.4602 \, V \]

### Step 7: Determine the Standard Reduction Potential for \( B^+ \)
We know:
\[ E^\circ_{cell} = E^\circ_{B^+/B} - E^\circ_{A^{2+}/A} \]
Given \( E^\circ_{A^{2+}/A} = 0.34 \, V \):
\[ 0.4602 = E^\circ_{B^+/B} - 0.34 \]
Rearranging gives:
\[ E^\circ_{B^+/B} = 0.4602 + 0.34 = 0.8002 \, V \]

### Final Answer
Thus, the standard reduction potential for \( B^+ + e^- \rightarrow B \) is approximately:
\[ E^\circ_{B^+/B} = 0.80 \, V \]

---