If the solution of the `CuSO_(4)` in which copper rod is immersed is diluted to 10 times, the electrode potential :

Options

A. increases by 0.030 volt
B. decreases by 0.030 volt
C. increases by 0.059 volt
D. decreases by 0.0059 volt

Answer

To solve the problem of how the electrode potential changes when the solution of CuSO₄ is diluted 10 times, we can use the Nernst equation. Here’s a step-by-step solution:

### Step 1: Understand the Nernst Equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{RT}{nF} \ln Q \]
Where:
- \( E \) = electrode potential
- \( E^\circ \) = standard electrode potential
- \( R \) = universal gas constant (8.314 J/(mol·K))
- \( T \) = temperature in Kelvin (298 K)
- \( n \) = number of moles of electrons transferred in the half-reaction
- \( F \) = Faraday's constant (96485 C/mol)
- \( Q \) = reaction quotient

### Step 2: Calculate the Nernst Equation at Initial Concentration
For the copper half-reaction:
\[ \text{Cu}^{2+} + 2e^- \leftrightarrow \text{Cu (s)} \]

The reaction quotient \( Q \) is given by:
\[ Q = \frac{1}{[\text{Cu}^{2+}]} \]

Thus, the Nernst equation becomes:
\[ E = E^\circ - \frac{0.059}{2} \log [\text{Cu}^{2+}] \]

### Step 3: Determine the Effect of Dilution
When the solution is diluted 10 times, the concentration of \( \text{Cu}^{2+} \) becomes:
\[ [\text{Cu}^{2+}]_{\text{new}} = \frac{[\text{Cu}^{2+}]_{\text{initial}}}{10} \]

### Step 4: Substitute the New Concentration into the Nernst Equation
Now, substituting the new concentration into the Nernst equation:
\[ E' = E^\circ - \frac{0.059}{2} \log \left( \frac{[\text{Cu}^{2+}]_{\text{initial}}}{10} \right) \]

Using the logarithmic property:
\[ \log \left( \frac{a}{b} \right) = \log a - \log b \]
We can rewrite it as:
\[ E' = E^\circ - \frac{0.059}{2} \left( \log [\text{Cu}^{2+}]_{\text{initial}} - \log 10 \right) \]

### Step 5: Simplify the Equation
Since \( \log 10 = 1 \):
\[ E' = E^\circ - \frac{0.059}{2} \log [\text{Cu}^{2+}]_{\text{initial}} + \frac{0.059}{2} \]

### Step 6: Calculate the Change in Electrode Potential
The change in electrode potential \( \Delta E \) can be calculated as:
\[ \Delta E = E' - E \]
Substituting the values:
\[ \Delta E = \left( E^\circ - \frac{0.059}{2} \log [\text{Cu}^{2+}]_{\text{initial}} + \frac{0.059}{2} \right) - \left( E^\circ - \frac{0.059}{2} \log [\text{Cu}^{2+}]_{\text{initial}} \right) \]

This simplifies to:
\[ \Delta E = \frac{0.059}{2} \]

### Step 7: Determine the Sign of Change
Since we have diluted the solution, the electrode potential decreases:
\[ \Delta E = -\frac{0.059}{2} = -0.0295 \text{ V} \]

Thus, the electrode potential decreases by approximately \( 0.03 \text{ V} \).

### Final Answer
The electrode potential decreases by \( 0.03 \text{ V} \) when the solution of CuSO₄ is diluted 10 times.

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