The solution of `CuSO_(4)` in which `Cu` ord is dipped is diluted to 10 times, the reduction electrode potential will `:`
`a.` Decrease by `0.03V" "b.` Decrease by `0.059V`
`c.` Increase by `0.03 V" "d.` Increase by `-.059V`

Explanation

To solve the problem of how the reduction electrode potential changes when a solution of CuSO₄ is diluted to 10 times its original concentration, we can use the Nernst equation. The Nernst equation relates the reduction potential of an electrochemical reaction to the concentrations of the reactants and products involved.

### Step-by-Step Solution:

1. **Identify the Reaction:**
The reduction half-reaction for copper ions is:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]

2. **Write the Nernst Equation:**
The Nernst equation for the above reaction is given by:
\[ E = E^\circ - \frac{RT}{nF} \ln \left( \frac{1}{[\text{Cu}^{2+}]^2} \right) \]
Where:
- \( E \) = reduction potential
- \( E^\circ \) = standard reduction potential
- \( R \) = universal gas constant (8.314 J/(mol·K))
- \( T \) = temperature in Kelvin
- \( n \) = number of electrons transferred (which is 2 for Cu)
- \( F \) = Faraday's constant (96485 C/mol)
- \( [\text{Cu}^{2+}] \) = concentration of copper ions

3. **Effect of Dilution:**
When the solution is diluted to 10 times, the concentration of \(\text{Cu}^{2+}\) ions decreases to:
\[ [\text{Cu}^{2+}]_{\text{diluted}} = \frac{[\text{Cu}^{2+}]_{\text{initial}}}{10} \]

4. **Substituting into the Nernst Equation:**
The new potential after dilution becomes:
\[ E_{\text{diluted}} = E^\circ - \frac{RT}{nF} \ln \left( \frac{1}{\left(\frac{[\text{Cu}^{2+}]_{\text{initial}}}{10}\right)^2} \right) \]
Simplifying this gives:
\[ E_{\text{diluted}} = E^\circ - \frac{RT}{nF} \ln \left( \frac{10^2}{[\text{Cu}^{2+}]_{\text{initial}}^2} \right) \]
\[ E_{\text{diluted}} = E^\circ - \frac{RT}{nF} \left( 2 \ln(10) \right) + \frac{RT}{nF} \ln \left( [\text{Cu}^{2+}]_{\text{initial}}^2 \right) \]

5. **Calculating the Change in Potential:**
The change in potential due to dilution can be approximated using the fact that at room temperature (298 K), the term \(\frac{RT}{nF}\) can be approximated as:
\[ \frac{RT}{nF} \approx 0.059 \text{ V} \text{ (for } n = 1\text{)} \]
Since we have \(n = 2\) for copper, we will use:
\[ \Delta E \approx -0.059 \text{ V} \times 2 = -0.118 \text{ V} \]
However, we need to consider that the potential increases when the concentration decreases. Thus, the effective change in potential is:
\[ \Delta E = 0.059 \text{ V} \text{ (for each log unit increase)} \]

6. **Final Answer:**
Since the concentration of \(\text{Cu}^{2+}\) decreases by a factor of 10, the reduction potential will increase by approximately:
\[ \Delta E \approx 0.059 \text{ V} \]
Therefore, the correct option is:
**d. Increase by 0.059 V.**