`Cu^(2+)+2e^(-) rarr Cu.` On increasing `[Cu^(2+)]`, electrode potential

Options

A. Increases
B. Decreases
C. No change
D. First increases, then decreases

Explanation

To analyze the effect of increasing the concentration of \( \text{Cu}^{2+} \) on the electrode potential of the half-reaction \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \), we can follow these steps:

### Step 1: Write the Nernst Equation
The Nernst equation relates the standard electrode potential to the concentration of the reactants and products. For the half-reaction given, the Nernst equation can be expressed as:

\[ E = E^\circ - \frac{0.0591}{n} \log Q \]

where:
- \( E \) is the electrode potential,
- \( E^\circ \) is the standard electrode potential,
- \( n \) is the number of electrons transferred in the reaction,
- \( Q \) is the reaction quotient.

### Step 2: Determine the Reaction Quotient \( Q \)
For the reaction \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \), the reaction quotient \( Q \) is defined as:

\[ Q = \frac{[\text{Products}]}{[\text{Reactants}]} = \frac{[\text{Cu}]}{[\text{Cu}^{2+}]} \]

Since the activity of a solid (copper) is considered to be 1, we can simplify \( Q \) to:

\[ Q = \frac{1}{[\text{Cu}^{2+}]} \]

### Step 3: Substitute \( Q \) into the Nernst Equation
Substituting the expression for \( Q \) into the Nernst equation gives:

\[ E = E^\circ - \frac{0.0591}{n} \log \left(\frac{1}{[\text{Cu}^{2+}]}\right) \]

This can be rewritten as:

\[ E = E^\circ + \frac{0.0591}{n} \log [\text{Cu}^{2+}] \]

### Step 4: Determine the Value of \( n \)
In this half-reaction, \( n = 2 \) because 2 electrons are involved in the reduction of \( \text{Cu}^{2+} \) to \( \text{Cu} \).

### Step 5: Analyze the Effect of Increasing \( [\text{Cu}^{2+}] \)
When the concentration of \( \text{Cu}^{2+} \) increases, the term \( \log [\text{Cu}^{2+}] \) also increases. Since this term is added to \( E^\circ \), it follows that:

\[ E \text{ increases as } [\text{Cu}^{2+}] \text{ increases.} \]

### Conclusion
Thus, increasing the concentration of \( \text{Cu}^{2+} \) leads to an increase in the electrode potential.