To find the standard electrode potential \( E^\circ_{(Au^+ // Au^{3+})} \), we can use the given standard electrode potentials for the half-reactions involving gold:
1. **Given Values**:
- \( E^\circ_{(Au^+ // Au)} = 1.69 \, \text{V} \)
- \( E^\circ_{(Au^{3+} // Au)} = 1.40 \, \text{V} \)
2. **Write the Half-Reactions**:
- The reduction half-reaction for \( Au^+ \) to \( Au \):
\[
Au^+ + e^- \rightarrow Au \quad (E^\circ = 1.69 \, \text{V})
\]
- The reduction half-reaction for \( Au^{3+} \) to \( Au \):
\[
Au^{3+} + 3e^- \rightarrow Au \quad (E^\circ = 1.40 \, \text{V})
\]
3. **Reverse the Second Half-Reaction**:
- To find the potential for the reaction from \( Au^+ \) to \( Au^{3+} \), we need to reverse the second half-reaction:
\[
Au \rightarrow Au^{3+} + 3e^- \quad (E^\circ = -1.40 \, \text{V})
\]
4. **Combine the Half-Reactions**:
- Now we can add the two half-reactions together. However, we need to ensure that the number of electrons is the same. The first half-reaction involves 1 electron, while the second involves 3 electrons. To balance this, we can multiply the first half-reaction by 3:
\[
3(Au^+ + e^- \rightarrow Au) \quad (E^\circ = 3 \times 1.69 \, \text{V} = 5.07 \, \text{V})
\]
- This gives us:
\[
3Au^+ + 3e^- \rightarrow 3Au \quad (E^\circ = 5.07 \, \text{V})
\]
- Now, we can add the reversed second half-reaction:
\[
3Au \rightarrow 3Au^{3+} + 9e^- \quad (E^\circ = -1.40 \, \text{V})
\]
5. **Final Reaction**:
- The overall reaction becomes:
\[
3Au^+ \rightarrow 3Au^{3+} + 6e^-
\]
6. **Calculate the Standard Electrode Potential**:
- Now we can calculate the standard electrode potential for the overall reaction:
\[
E^\circ_{(Au^+ // Au^{3+})} = \frac{E^\circ_{(Au^+ // Au)} \times n_1 + E^\circ_{(Au^{3+} // Au)} \times n_2}{n}
\]
- Where \( n_1 = 3 \) (from \( Au^+ \)), \( n_2 = 1 \) (from \( Au^{3+} \)), and \( n = 2 \) (total electron transfer).
- Substituting the values:
\[
E^\circ_{(Au^+ // Au^{3+})} = \frac{3 \times 1.69 - 1 \times 1.40}{2}
\]
\[
= \frac{5.07 - 1.40}{2} = \frac{3.67}{2} = 1.835 \, \text{V}
\]
7. **Final Answer**:
- Therefore, the standard electrode potential \( E^\circ_{(Au^+ // Au^{3+})} \) is approximately \( 1.835 \, \text{V} \).
