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

Answer

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

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

### Step 1: Identify Oxidation and Reduction
In the reaction \( X + Y^{2+} \rightarrow X^{2+} + Y \):
- \( X \) is oxidized to \( X^{2+} \) (loses electrons).
- \( Y^{2+} \) is reduced to \( Y \) (gains electrons).

### Step 2: Determine the Standard Cell Potential
The standard cell potential \( E^\circ_{cell} \) can be calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
where:
- \( E^\circ_{cathode} \) is the standard reduction potential of the species being reduced.
- \( E^\circ_{anode} \) is the standard reduction potential of the species being oxidized.

### Step 3: Evaluate Each Option
We will evaluate each option to see if \( E^\circ_{cell} > 0 \) for spontaneity.

#### Option A: \( X = Ni, Y = Fe \)
- \( E^\circ_{cathode} = E^\circ_{Fe} = -0.44 \, V \)
- \( E^\circ_{anode} = E^\circ_{Ni} = -0.23 \, V \)
\[ E^\circ_{cell} = -0.44 - (-0.23) = -0.44 + 0.23 = -0.21 \, V \quad (\text{Not spontaneous}) \]

#### Option B: \( X = Ni, Y = Zn \)
- \( E^\circ_{cathode} = E^\circ_{Zn} = -0.76 \, V \)
- \( E^\circ_{anode} = E^\circ_{Ni} = -0.23 \, V \)
\[ E^\circ_{cell} = -0.76 - (-0.23) = -0.76 + 0.23 = -0.53 \, V \quad (\text{Not spontaneous}) \]

#### Option C: \( X = Fe, Y = Zn \)
- \( E^\circ_{cathode} = E^\circ_{Zn} = -0.76 \, V \)
- \( E^\circ_{anode} = E^\circ_{Fe} = -0.44 \, V \)
\[ E^\circ_{cell} = -0.76 - (-0.44) = -0.76 + 0.44 = -0.32 \, V \quad (\text{Not spontaneous}) \]

#### Option D: \( X = Zn, Y = Ni \)
- \( E^\circ_{cathode} = E^\circ_{Ni} = -0.23 \, V \)
- \( E^\circ_{anode} = E^\circ_{Zn} = -0.76 \, V \)
\[ E^\circ_{cell} = -0.23 - (-0.76) = -0.23 + 0.76 = 0.53 \, V \quad (\text{Spontaneous}) \]

### Conclusion
The reaction \( X + Y^{2+} \rightarrow X^{2+} + Y \) will be spontaneous when \( X = Zn \) and \( Y = Ni \). Therefore, the correct answer is:

**Option D: \( X = Zn, Y = Ni \)**