The Standard reduction potential values, `E^(@)(Bi^(3+)//Bi)` and `E^(@)(Cu^(2+)//Cu)` are 0.226 V and 0.344 V respectively. A mxiture of salt of bismut and copper at unit concentration each is electrolysed at `25^(@)C` to what value can `[Cu^(2+)]` be brought down before bismuth starts to deposit, in electrolysis.

Answer

To solve the problem, we need to determine the concentration of \( [Cu^{2+}] \) before bismuth starts to deposit during electrolysis. We will use the Nernst equation and the given standard reduction potentials.

### Step-by-Step Solution:

1. **Identify the Standard Reduction Potentials**:
- For \( Bi^{3+} + 3e^- \rightarrow Bi \), \( E^\circ = 0.226 \, V \)
- For \( Cu^{2+} + 2e^- \rightarrow Cu \), \( E^\circ = 0.344 \, V \)

2. **Write the Nernst Equation**:
The Nernst equation is given by:
\[ E = E^\circ + \frac{0.059}{n} \log \left( \frac{1}{[Red]} \right) \]
where \( n \) is the number of moles of electrons transferred.

3. **Calculate the Cell Potential for Copper**:
For the reduction of copper:
\[ E_{Cu} = E^\circ_{Cu} + \frac{0.059}{2} \log [Cu^{2+}] \]
Substituting the values:
\[ E_{Cu} = 0.344 + \frac{0.059}{2} \log [Cu^{2+}] \]

4. **Calculate the Cell Potential for Bismuth**:
For the reduction of bismuth:
\[ E_{Bi} = E^\circ_{Bi} + \frac{0.059}{3} \log [Bi^{3+}] \]
Substituting the values:
\[ E_{Bi} = 0.226 + \frac{0.059}{3} \log [Bi^{3+}] \]

5. **Determine the Condition for Bismuth Deposition**:
Bismuth will start to deposit when the potential of copper equals the potential of bismuth:
\[ E_{Cu} = E_{Bi} \]
Therefore:
\[ 0.344 + \frac{0.059}{2} \log [Cu^{2+}] = 0.226 + \frac{0.059}{3} \log [Bi^{3+}] \]

6. **Assume Initial Concentrations**:
Initially, both \( [Cu^{2+}] \) and \( [Bi^{3+}] \) are 1 M. Thus, \( \log [Bi^{3+}] = 0 \).

7. **Set Up the Equation**:
\[ 0.344 + \frac{0.059}{2} \log [Cu^{2+}] = 0.226 \]
Rearranging gives:
\[ \frac{0.059}{2} \log [Cu^{2+}] = 0.226 - 0.344 \]
\[ \frac{0.059}{2} \log [Cu^{2+}] = -0.118 \]

8. **Solve for \( [Cu^{2+}] \)**:
\[ \log [Cu^{2+}] = \frac{-0.118 \times 2}{0.059} = -4 \]
Therefore:
\[ [Cu^{2+}] = 10^{-4} \, M \]

### Final Answer:
The concentration of \( [Cu^{2+}] \) can be brought down to \( 10^{-4} \, M \) before bismuth starts to deposit.