If `Cu^(2+)//Cu` electrode is dilutes 100 times, the change in emf is

Options

A. Increase of 59mV
B. Decrease of 59mV
C. Increase of 29.5mV
D. Decrease of 29.5mV

Answer

To solve the problem of determining the change in emf when the concentration of the `Cu^(2+)` ion is diluted 100 times, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The half-reaction for the copper electrode is:
\[ Cu^{2+} + 2e^- \leftrightarrow Cu \]
Here, copper ions are being reduced to solid copper.

2. **Use the Nernst Equation**:
The Nernst equation relates the emf of the cell to the standard emf and the concentrations of the reactants and products:
\[ E = E^0 - \frac{2.303RT}{nF} \log \left( \frac{[Cu^{2+}]}{[Cu]} \right) \]
where:
- \( E \) is the cell potential,
- \( E^0 \) is the standard cell potential,
- \( R \) is the universal gas constant (8.314 J/(mol·K)),
- \( T \) is the temperature in Kelvin,
- \( n \) is the number of moles of electrons transferred (which is 2 for copper),
- \( F \) is Faraday's constant (96485 C/mol).

3. **Determine the Change in Concentration**:
If the concentration of \( Cu^{2+} \) is diluted 100 times, we can express this as:
\[ [Cu^{2+}]_{new} = \frac{[Cu^{2+}]_{initial}}{100} \]

4. **Calculate the Logarithmic Term**:
The change in emf will be due to the change in the concentration of \( Cu^{2+} \):
\[ E_{new} = E^0 - \frac{2.303RT}{nF} \log \left( \frac{[Cu^{2+}]_{new}}{[Cu]} \right) \]
Substituting the new concentration:
\[ E_{new} = E^0 - \frac{2.303RT}{nF} \log \left( \frac{[Cu^{2+}]_{initial}/100}{[Cu]} \right) \]
This simplifies to:
\[ E_{new} = E^0 - \frac{2.303RT}{nF} \left( \log \left( [Cu^{2+}]_{initial} \right) - \log(100) - \log \left( [Cu] \right) \right) \]

5. **Calculate the Change in EMF**:
The change in emf (\( \Delta E \)) is:
\[ \Delta E = E_{new} - E^0 = -\frac{2.303RT}{nF} \left( -\log(100) \right) \]
Since \( \log(100) = 2 \):
\[ \Delta E = \frac{2.303RT}{nF} \cdot 2 \]

6. **Substitute Values**:
Using \( R = 8.314 \, J/(mol·K) \), \( T = 298 \, K \), \( n = 2 \), and \( F = 96485 \, C/mol \):
\[ \Delta E = \frac{2.303 \times 8.314 \times 298}{2 \times 96485} \cdot 2 \]
\[ \Delta E \approx 0.0591 \, V \quad \text{(since we multiply by 2)} \]

7. **Convert to Millivolts**:
To express this in millivolts:
\[ \Delta E \approx 59.1 \, mV \]

8. **Final Answer**:
The change in emf when the \( Cu^{2+} \) concentration is diluted 100 times is approximately:
\[ \Delta E \approx -59.1 \, mV \]