The standard reduction potential for `Zn^(2+)//Zn, Ni^(2+)//Ni` and `Fe^(2+)//Fe` are `-0.76, -0.23` and `0.44V` respectively. The reaction `X + Y^(2) rarr X^(2+) + Y` will be spontaneous when:

Options

A. `X = Ni, Y = Zn`
B. `X = Fe, Y = Zn`
C. `X = Zn, Y = Ni`
D. `X = Ni, Y = Fe`

Explanation

To determine when the reaction \( X + Y^{2+} \rightarrow X^{2+} + Y \) will be spontaneous, we need to analyze the standard reduction potentials of the involved species. The standard reduction potentials for the half-reactions are given as follows:

- \( \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \) has \( E^\circ = -0.76 \, \text{V} \)
- \( \text{Ni}^{2+} + 2e^- \rightarrow \text{Ni} \) has \( E^\circ = -0.23 \, \text{V} \)
- \( \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \) has \( E^\circ = 0.44 \, \text{V} \)

### Step 1: Identify the reducing and oxidizing agents
To determine the spontaneity of the reaction, we need to identify which species will be oxidized and which will be reduced:
- The species with the most negative standard reduction potential will act as a good reducing agent (it will be oxidized).
- The species with the most positive standard reduction potential will act as a good oxidizing agent (it will be reduced).

### Step 2: Determine the reducing agent (X)
Among the given standard reduction potentials, zinc (\( \text{Zn} \)) has the most negative potential (\( -0.76 \, \text{V} \)). Therefore, zinc will be oxidized and will be our \( X \).

### Step 3: Determine the oxidizing agent (Y)
The species with the highest standard reduction potential is iron (\( \text{Fe}^{2+} \)) with \( 0.44 \, \text{V} \). However, since we are looking for the oxidizing agent that will be reduced in this reaction, we need to check the other options. Nickel (\( \text{Ni}^{2+} \)) has a potential of \( -0.23 \, \text{V} \), which is more positive than zinc and less positive than iron. Therefore, we can choose nickel as our \( Y \).

### Step 4: Write the half-reactions
- Oxidation half-reaction (for \( X \)):
\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \quad (E^\circ = +0.76 \, \text{V}) \]
- Reduction half-reaction (for \( Y \)):
\[ \text{Ni}^{2+} + 2e^- \rightarrow \text{Ni} \quad (E^\circ = -0.23 \, \text{V}) \]

### Step 5: Calculate the cell potential \( E^\circ_{\text{cell}} \)
The standard cell potential can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Here, the cathode is where reduction occurs (nickel) and the anode is where oxidation occurs (zinc):
\[ E^\circ_{\text{cell}} = (-0.23 \, \text{V}) - (-0.76 \, \text{V}) = 0.53 \, \text{V} \]

### Step 6: Determine spontaneity
Since \( E^\circ_{\text{cell}} \) is positive (\( 0.53 \, \text{V} \)), the reaction is spontaneous.

### Conclusion
Thus, the reaction \( X + Y^{2+} \rightarrow X^{2+} + Y \) will be spontaneous when:
- \( X = \text{Zn} \)
- \( Y = \text{Ni} \)