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 finding the change in emf when the `Cu^(2+)//Cu` electrode is diluted 100 times, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Nernst Equation**:
The Nernst equation relates the cell potential (emf) to the standard cell potential and the concentrations of the reactants and products. The equation is given by:
\[ E_{cell} = E^0_{cell} - \frac{2.303RT}{nF} \log \left( \frac{[\text{oxidized}]}{[\text{reduced}]} \right) \]
where:
- \(E_{cell}\) = cell potential
- \(E^0_{cell}\) = standard cell potential
- \(R\) = universal gas constant (8.314 J/(mol·K))
- \(T\) = temperature in Kelvin
- \(n\) = number of moles of electrons transferred in the half-reaction
- \(F\) = Faraday's constant (96485 C/mol)

2. **Determine the Change in Concentration**:
When the `Cu^(2+)` solution is diluted 100 times, the concentration of `Cu^(2+)` decreases. If the initial concentration is \(C\), the new concentration after dilution is:
\[ [\text{Cu}^{2+}] = \frac{C}{100} \]

3. **Substituting into the Nernst Equation**:
For the copper half-reaction:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]
Here, \(n = 2\). The Nernst equation becomes:
\[ E_{cell} = E^0_{cell} - \frac{0.0591}{2} \log \left( \frac{[\text{Cu}^{2+}]}{[\text{Cu}]} \right) \]
Assuming the concentration of solid copper is constant and can be treated as 1, we have:
\[ E_{cell} = E^0_{cell} - \frac{0.0591}{2} \log \left( \frac{C/100}{1} \right) \]
This simplifies to:
\[ E_{cell} = E^0_{cell} - \frac{0.0591}{2} \log \left( \frac{C}{100} \right) \]

4. **Calculate the Logarithm**:
The logarithm can be simplified:
\[ \log \left( \frac{C}{100} \right) = \log(C) - \log(100) = \log(C) - 2 \]
Thus, substituting back:
\[ E_{cell} = E^0_{cell} - \frac{0.0591}{2} (\log(C) - 2) \]

5. **Change in emf**:
The change in emf due to dilution can be calculated as:
\[ \Delta E = E_{cell, \text{diluted}} - E_{cell, \text{initial}} = - \frac{0.0591}{2} (-2) = 0.0591 \text{ volts} \]
Converting this to millivolts:
\[ \Delta E = 0.0591 \times 1000 = 59.1 \text{ mV} \]

6. **Final Answer**:
Since the emf decreases due to dilution, we express this as:
\[ \Delta E = -59.1 \text{ mV} \]