If `E_(Au^(+)//Au)^(@)` is 1.69 V and `E_(Au^(3+)//Au)^(@)` is 1.40 V, then `E_(Au^(+)//Au^(3+))^(@)` will be :

Options

A. 0.9 v
B. 0.945 V
C. 1.255 V
D. None of these

Answer

To find the standard electrode potential \( E^\circ_{(Au^+ // Au^{3+})} \), we can use the given standard reduction potentials for the half-reactions involving gold ions.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- \( E^\circ_{(Au^+ // Au)} = 1.69 \, \text{V} \) (Reduction of \( Au^+ \) to \( Au \))
- \( E^\circ_{(Au^{3+} // Au)} = 1.40 \, \text{V} \) (Reduction of \( Au^{3+} \) to \( Au \))

2. **Write the Half-Reactions**:
- For \( Au^+ + e^- \rightarrow Au \) (Reduction reaction)
- For \( Au^{3+} + 3e^- \rightarrow Au \) (Reduction reaction)

3. **Reverse the Reaction for \( Au^{3+} \)**:
- To find the potential for the oxidation of \( Au \) to \( Au^{3+} \), we reverse the second half-reaction:
\[ Au \rightarrow Au^{3+} + 3e^- \]
- The potential for this oxidation reaction is the negative of the reduction potential:
\[ E^\circ_{(Au // Au^{3+})} = -E^\circ_{(Au^{3+} // Au)} = -1.40 \, \text{V} \]

4. **Combine the Half-Reactions**:
- Now we have:
- \( Au^+ + e^- \rightarrow Au \) with \( E^\circ = 1.69 \, \text{V} \)
- \( Au \rightarrow Au^{3+} + 3e^- \) with \( E^\circ = -1.40 \, \text{V} \)

- To find the overall reaction:
\[ Au^+ \rightarrow Au^{3+} + 2e^- \]

5. **Calculate the Overall Standard Electrode Potential**:
- Using the formula for the cell potential:
\[ E^\circ_{(Au^+ // Au^{3+})} = \frac{n_1 E_1 + n_2 E_2}{n_3} \]
- Where:
- \( n_1 = 1 \) (for \( Au^+ \))
- \( E_1 = 1.69 \, \text{V} \)
- \( n_2 = 3 \) (for \( Au^{3+} \))
- \( E_2 = -1.40 \, \text{V} \)
- \( n_3 = 2 \) (total electrons transferred)

- Substitute the values:
\[ E^\circ_{(Au^+ // Au^{3+})} = \frac{(1)(1.69) + (3)(-1.40)}{2} \]
\[ = \frac{1.69 - 4.20}{2} \]
\[ = \frac{-2.51}{2} = -1.255 \, \text{V} \]

6. **Final Result**:
- Therefore, the standard electrode potential \( E^\circ_{(Au^+ // Au^{3+})} = 1.255 \, \text{V} \).