The standard reduction potential of $Mg^{2+}/Mg, Pb^{2+}/Pb$ and $Sn^{2+}/Sn$ are -2.36, -0.13 and -0.14 V respectively. The reaction $X+Y^{2+}→X^{2+}+Y$ will be spontaneous when:

$X=Mg, Y=Pb$

The correct answer is Option (1) → $X=Mg, Y=Pb$.

To determine which reaction will be spontaneous, we need to understand how thestandard reduction potential \((E^o)\) values influence the spontaneity of a redox reaction.

The general redox reaction is:

\(X + Y^{2+} \rightarrow X^{2+} + Y\)

For the reaction to be spontaneous, the species \(X \)must be a stronger reducing agent than \(Y\). This means \(X\) must lose electrons and be oxidized, while \(Y^{2+}\) gains electrons and is reduced. In other words, the species \(X\) should have a more negative reduction potential than \(Y^{2+}/Y\).

Given data:

\( \text{Mg}^{2+}/\text{Mg} = -2.36 \, V \)

\( \text{Pb}^{2+}/\text{Pb} = -0.13 \, V \)

\( \text{Sn}^{2+}/\text{Sn} = -0.14 \, V \)

A redox reaction is spontaneous if the difference in standard reduction potential between the two half-reactions \((E^o)\) gives a positive \(E^o\) cell when the more negative reduction potential is written as oxidation.

To predict spontaneity:

The species with a more negative \(E^o\) will act as the reducing agent (it will be oxidized).

The species with a more positive \(E^o\)  will be reduced.

Now, let's calculate the cell potential for each option:

Option 1: \( X = \text{Mg}, Y = \text{Pb} \)

Oxidation half-reaction \((Mg \longrightarrow Mg^{2+})\): \( E^o = +2.36 \, V \)

Reduction half-reaction \((Pb^{2+} \longrightarrow Pb)\): \( E^o = -0.13 \, V \)

\(E^o_{cell} = 2.36 \, V - (-0.13 \, V) = 2.36 + 0.13 = +2.23 \, V\)

This reaction is spontaneous.

Option 2: \( X = \text{Pb}, Y = \text{Sn} \)

Oxidation half-reaction \((Pb \longrightarrow Pb^{2+})\): \(E^o = +0.13 \, V \)

Reduction half-reaction \((Sn^{2+} \longrightarrow Sn)\): \( E^o = -0.14 \, V \)

\(E^o_{cell} = 0.13 \, V - (-0.14 \, V) = +0.13 + 0.14 = +0.27 \, V\)

This reaction is spontaneous.

Option 3: \( X = \text{Pb}, Y = \text{Mg} \)

Oxidation half-reaction \((Pb \longrightarrow Pb^{2+})\): \( E^o = +0.13 \, V \)

Reduction half-reaction \((Mg^{2+} \longrightarrow Mg)\): \( E^o = -2.36 \, V \)

\(E^o_{cell} = 0.13 \, V - (-2.36 \, V) = 0.13 + 2.36 = +2.49 \, V\)

This reaction is spontaneous.

Option 4: \( X = \text{Sn}, Y = \text{Mg} \)

Oxidation half-reaction \((Sn \longrightarrow Sn^{2+})\): \( E^o = +0.14 \, V \)

Reduction half-reaction \((Mg^{2+} \longrightarrow Mg)\): \( E^o = -2.36 \, V \)

\(E^o_{cell} = 0.14 \, V - (-2.36 \, V) = 0.14 + 2.36 = +2.50 \, V\)

This reaction is spontaneous

Conclusion:

The following reactions are spontaneous:

1. \( X = \text{Mg}, Y = \text{Pb} \)

2. \( X = \text{Pb}, Y = \text{Sn} \)

3. \( X = \text{Pb}, Y = \text{Mg} \)

4. \( X = \text{Sn}, Y = \text{Mg} \)

Therefore, the correct answer is Option 1: X = Mg, Y = Pb.