Consider the following table of standard reduction potentials.
`{:("Reaction",E^@(V)),(A^(3+)+2e^(-) rarrA^+,1.36),(B^(2+)+2e^(-)rarrB,0.72),(C^(2+)+2e^(-)rarrC,-0.28),(D^(+)+e^(-)rarrD,-1.42):}`
Which of the following reaction will occur ?
(i) `B^(2+) +C rarr B +C^(2+)`
(ii) `C^(2+)+ArarrC+A^(2+)`

Answer

To determine which of the given reactions will occur based on the standard reduction potentials, we will analyze each reaction step by step.

### Given Data:
- **Standard Reduction Potentials (E° values)**:
1. \( A^{3+} + 2e^- \rightarrow A^+ \) (E° = 1.36 V)
2. \( B^{2+} + 2e^- \rightarrow B \) (E° = 0.72 V)
3. \( C^{2+} + 2e^- \rightarrow C \) (E° = -0.28 V)
4. \( D^{+} + e^- \rightarrow D \) (E° = -1.42 V)

### Reactions to Analyze:
1. \( B^{2+} + C \rightarrow B + C^{2+} \)
2. \( C^{2+} + A \rightarrow C + A^{2+} \)

### Step 1: Analyze Reaction (i) \( B^{2+} + C \rightarrow B + C^{2+} \)

- **Identify Oxidation and Reduction**:
- \( B^{2+} \) is being reduced to \( B \) (gaining electrons).
- \( C \) is being oxidized to \( C^{2+} \) (losing electrons).

- **Reduction Potentials**:
- For \( B^{2+} + 2e^- \rightarrow B \): E° = 0.72 V (reduction potential)
- For \( C \rightarrow C^{2+} + 2e^- \): The oxidation potential is the negative of the reduction potential of \( C^{2+} \rightarrow C \), which is -(-0.28 V) = 0.28 V.

- **Comparison**:
- Since \( E°_{B^{2+}/B} = 0.72 \, V \) (reduction) is greater than \( E°_{C/C^{2+}} = 0.28 \, V \) (oxidation), the reaction \( B^{2+} + C \rightarrow B + C^{2+} \) is favorable and will occur.

### Step 2: Analyze Reaction (ii) \( C^{2+} + A \rightarrow C + A^{2+} \)

- **Identify Oxidation and Reduction**:
- \( C^{2+} \) is being reduced to \( C \) (gaining electrons).
- \( A \) is being oxidized to \( A^{2+} \) (losing electrons).

- **Reduction Potentials**:
- For \( C^{2+} + 2e^- \rightarrow C \): E° = -0.28 V (reduction potential)
- For \( A \rightarrow A^{2+} + 2e^- \): The reduction potential is E° = 1.36 V.

- **Comparison**:
- Since \( E°_{A/A^{2+}} = 1.36 \, V \) (reduction) is greater than \( E°_{C^{2+}/C} = -0.28 \, V \) (reduction), the reaction \( C^{2+} + A \rightarrow C + A^{2+} \) is not favorable because \( A \) will not oxidize when \( C^{2+} \) is present.

### Conclusion:
- The reaction that will occur is **(i) \( B^{2+} + C \rightarrow B + C^{2+} \)**.