The `E^(@)` for the following cell is +0.34 V. `In(s)|In (OH)_3(aq)"||"SbO_2^(-)(aq)"|"Sb(s)`.
Using `E^(@)=-1.0 V` for the `In (OH)_3| In`, couple, calculate `E^(@)` for the `SbO_2^(-)|Sb `half-reaction:

Options

A. -1.34
B. 0.66
C. 0.82
D. -0.66

Answer

To solve the problem, we need to calculate the standard electrode potential (E°) for the half-reaction of SbO2⁻ to Sb using the given information about the cell and the standard electrode potential for the indium half-reaction.

### Step-by-Step Solution:

1. **Understand the Cell Representation**:
The cell is represented as:
\[ \text{In(s)} | \text{In(OH)}_3(aq) || \text{SbO}_2^{-}(aq) | \text{Sb(s)} \]
Here, In is oxidized at the anode and SbO2⁻ is reduced at the cathode.

2. **Identify the Half-Reactions**:
- At the anode (oxidation):
\[ \text{In(s)} \rightarrow \text{In(OH)}_3(aq) + 3e^- \]
- At the cathode (reduction):
\[ \text{SbO}_2^{-}(aq) + 3e^- \rightarrow \text{Sb(s)} \]

3. **Use the Given Standard Electrode Potentials**:
- The standard cell potential (E°cell) is given as +0.34 V.
- The standard electrode potential for the indium half-reaction (E°anode) is given as -1.0 V.

4. **Apply the Formula for Cell Potential**:
The relationship between the cell potential and the half-cell potentials is given by:
\[ E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} \]
Here, we need to find \(E°_{\text{cathode}}\) (which is the potential for the SbO2⁻ to Sb half-reaction).

5. **Substitute the Known Values**:
\[ 0.34 \, \text{V} = E°_{\text{cathode}} - (-1.0 \, \text{V}) \]
This simplifies to:
\[ 0.34 \, \text{V} = E°_{\text{cathode}} + 1.0 \, \text{V} \]

6. **Rearrange to Solve for E°cathode**:
\[ E°_{\text{cathode}} = 0.34 \, \text{V} - 1.0 \, \text{V} \]
\[ E°_{\text{cathode}} = -0.66 \, \text{V} \]

7. **Final Result**:
The standard electrode potential for the half-reaction of SbO2⁻ to Sb is:
\[ E° = -0.66 \, \text{V} \]

### Summary:
The calculated standard electrode potential for the half-reaction \( \text{SbO}_2^{-} \rightarrow \text{Sb} \) is **-0.66 V**.