The solution of `CuSO_(4)` in which copper rod is immersed is diluted to 10 times. The reduction electrode potential

Options

A. Increases by `30mV`
B. Decreases by `30mV`
C. Increases by `59mV`
D. Decrease by `59mV`

Explanation

To solve the problem of finding the reduction electrode potential after diluting a copper sulfate solution by 10 times, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Setup**:
- We have a copper half-cell where copper ions (\(Cu^{2+}\)) are reduced to solid copper (\(Cu\)).
- The standard reduction potential (\(E^\circ\)) for the half-reaction \(Cu^{2+} + 2e^- \rightarrow Cu\) is given as \(0.34 \, V\).

2. **Determine the Effect of Dilution**:
- When the solution is diluted 10 times, the concentration of \(Cu^{2+}\) ions decreases.
- If the original concentration of \(Cu^{2+}\) is \(C\), after dilution, the new concentration becomes \(\frac{C}{10}\).

3. **Use the Nernst Equation**:
- The Nernst equation for the half-cell can be expressed as:
\[ E = E^\circ - \frac{0.059}{n} \log \left( \frac{1}{[Cu^{2+}]} \right) \]
- Here, \(n = 2\) (the number of electrons transferred in the half-reaction).

4. **Calculate the Initial and New Electrode Potentials**:
- **Initial Electrode Potential (\(E_{cell1}\))**:
\[ E_{cell1} = E^\circ - \frac{0.059}{2} \log \left( \frac{1}{C} \right) \]
- **New Electrode Potential after Dilution (\(E_{cell2}\))**:
\[ E_{cell2} = E^\circ - \frac{0.059}{2} \log \left( \frac{1}{\frac{C}{10}} \right) \]
- Simplifying \(E_{cell2}\):
\[ E_{cell2} = E^\circ - \frac{0.059}{2} \log \left( \frac{10}{C} \right) \]
\[ E_{cell2} = E^\circ - \frac{0.059}{2} \left( \log(10) - \log(C) \right) \]
\[ E_{cell2} = E^\circ - \frac{0.059}{2} \left( 1 - \log(C) \right) \]

5. **Find the Change in Electrode Potential**:
- To find the change in potential (\(\Delta E\)):
\[ \Delta E = E_{cell2} - E_{cell1} \]
- Substituting the expressions for \(E_{cell1}\) and \(E_{cell2}\):
\[ \Delta E = \left( E^\circ - \frac{0.059}{2} \left( 1 - \log(C) \right) \right) - \left( E^\circ - \frac{0.059}{2} \log(C) \right) \]
- Simplifying:
\[ \Delta E = - \frac{0.059}{2} \cdot 1 \]
\[ \Delta E = -0.0295 \, V \, \text{or} \, -29.5 \, mV \]

6. **Conclusion**:
- The reduction electrode potential decreases by approximately \(30 \, mV\) (rounded from \(29.5 \, mV\)) due to the dilution.

### Final Answer:
The reduction electrode potential decreases by \(30 \, mV\).

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