Which of the following change is possible in aqueous medium -

Options

A. `2Cu^(+)toCu^(2+)+Cu^(0)`
B. `2Cu^(0)underset(-3e^(-))(to)Cu^(+)+Cu^(+2)`
C. `2Cu^(+2)underset(+3e^(-))(to)Cu^(0)+Cu^(+)`
D. `Cu^(+2)+e^(-)toCu^(+)`

Answer

To determine which of the given changes is possible in an aqueous medium, we need to analyze the standard reduction potentials (E°) for the reactions involved. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Reactions and Their Standard Reduction Potentials
We have the following reactions with their respective standard reduction potentials:
1. \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \) (E° = +0.34 V)
2. \( \text{Cu}^{+} + e^- \rightarrow \text{Cu} \) (E° = +0.52 V)
3. \( 2\text{H}_2O \rightarrow O_2 + 4H^+ + 4e^- \) (E° = -1.23 V)
4. \( 2H^+ + 2e^- \rightarrow H_2 \) (E° = -0.828 V)

### Step 2: Calculate the Standard Cell Potential for Each Reaction
We will calculate the cell potential (E) for each proposed reaction.

#### Reaction A: \( 2\text{Cu}^{+} \rightarrow \text{Cu}^{2+} + \text{Cu} \)
- For \( \text{Cu}^{+} \rightarrow \text{Cu}^{2+} + e^- \) (reverse of reduction): E° = -0.52 V
- For \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \): E° = +0.34 V
- Total E = -0.52 V + 0.34 V = -0.18 V (not feasible)

#### Reaction B: \( \text{Cu} \rightarrow \text{Cu}^{+} + \text{Cu}^{2+} \)
- For \( \text{Cu} \rightarrow \text{Cu}^{+} + e^- \): E° = -0.52 V
- For \( \text{Cu} \rightarrow \text{Cu}^{2+} + 2e^- \): E° = -0.34 V
- Total E = -0.52 V + -0.34 V = -0.86 V (not feasible)

#### Reaction C: \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}^{+} \)
- For \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \): E° = +0.34 V
- For \( \text{Cu}^{+} + e^- \rightarrow \text{Cu} \): E° = +0.52 V
- Total E = +0.34 V - 0.52 V = -0.18 V (not feasible)

#### Reaction D: \( \text{Cu}^{2+} + e^- \rightarrow \text{Cu}^{+} \)
- For \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \): E° = +0.34 V
- For \( \text{Cu}^{+} + e^- \rightarrow \text{Cu} \): E° = +0.52 V
- Total E = +0.34 V - 0.52 V = -0.18 V (not feasible)

### Step 3: Determine Feasibility
- A reaction is feasible if the cell potential (E) is greater than 0.
- From our calculations, all reactions resulted in negative potentials, indicating that none of the reactions are feasible in an aqueous medium.

### Conclusion
None of the proposed changes are possible in an aqueous medium based on the calculated standard reduction potentials.