Consider the following electrode potentials
`{:(,Mg^(+2)+2e^(-)toMg,,,E^(0)=-2.37V),(,V^(+2)+2e^(-)toV,,,E^(0)=-1.18V),(,Cu^(+2)e^(-)toCu^(+),,,E^(0)=0.15V):}`
Which of the following reactions will proceed from left to right spontaneously ?

Options

A. `Mg^(+2)+VtoMg+V^(2+)`
B. `Mg^(+2)+2Cu^(+)toMg+2Cu^(+2)`
C. `V^(+2)+2Cu^(+)toV+2Cu^(+2)`
D. `V+2Cu^(+2)toV^(2)+2Cu^(+)`

Answer

To determine which of the given reactions will proceed spontaneously from left to right, we need to analyze the standard reduction potentials (E°) for the half-reactions involved. The general rule is that a reaction will be spontaneous if the overall cell potential (E°cell) is positive.

### Step-by-Step Solution:

1. **Identify the Half-Reactions and Their Potentials**:
- \( \text{Mg}^{2+} + 2e^- \rightarrow \text{Mg} \) (E° = -2.37 V)
- \( \text{V}^{2+} + 2e^- \rightarrow \text{V} \) (E° = -1.18 V)
- \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \) (E° = 0.15 V)

2. **Understand the Concept of Spontaneity**:
- A reaction is spontaneous if the reduction potential of the species being reduced is higher than that of the species being oxidized. If the overall E°cell is positive, the reaction is spontaneous.

3. **Evaluate Each Reaction**:
- **Reaction 1**: \( \text{Mg} + \text{V}^{2+} \rightarrow \text{Mg}^{2+} + \text{V} \)
- Here, Mg is oxidized (E° = -2.37 V) and V is reduced (E° = -1.18 V).
- E°cell = E°(reduction) - E°(oxidation) = (-1.18) - (-2.37) = 1.19 V (positive, spontaneous).

- **Reaction 2**: \( \text{Mg}^{2+} + 2\text{Cu}^{+} \rightarrow \text{Mg} + 2\text{Cu}^{2+} \)
- Mg is reduced (E° = -2.37 V) and Cu is oxidized (E° = 0.15 V).
- E°cell = E°(reduction) - E°(oxidation) = (-2.37) - (0.15) = -2.52 V (negative, non-spontaneous).

- **Reaction 3**: \( \text{V}^{2+} + \text{Cu}^{+} \rightarrow \text{V} + \text{Cu}^{2+} \)
- V is reduced (E° = -1.18 V) and Cu is oxidized (E° = 0.15 V).
- E°cell = E°(reduction) - E°(oxidation) = (-1.18) - (0.15) = -1.33 V (negative, non-spontaneous).

- **Reaction 4**: \( \text{V} + \text{Cu}^{2+} \rightarrow \text{V}^{2+} + \text{Cu} \)
- Cu is reduced (E° = 0.15 V) and V is oxidized (E° = -1.18 V).
- E°cell = E°(reduction) - E°(oxidation) = (0.15) - (-1.18) = 1.33 V (positive, spontaneous).

4. **Conclusion**:
- The reactions that proceed spontaneously from left to right are Reaction 1 and Reaction 4.

### Final Answer:
- The reactions that will proceed spontaneously from left to right are **Reaction 1** and **Reaction 4**.