Step-by-step Solution:

1. Understanding the Reaction: The problem states that metal X dissolves. This means X undergoes oxidation, losing electrons. Therefore, X acts as the anode in the electrochemical cell. The half-reaction at the anode is: $X(s) \rightarrow X^{2+}(aq) + 2e^-$ Consequently, the other metal, Y, must act as the cathode, where reduction occurs. The half-reaction at the cathode is: $Y^{2+}(aq) + 2e^- \rightarrow Y(s)$

2. Condition for Spontaneity: For a reaction to be spontaneous, the cell potential ($E_{cell}$) must be positive. The cell potential is calculated as: $E_{cell} = E_{cathode} - E_{anode}$ In this setup, Y is the cathode and X is the anode. So, the condition for the dissolution of X is: $E_{cell} = E_{Y^{2+}/Y} - E_{X^{2+}/X} > 0$ This implies that the reduction potential of the Y electrode must be greater than the reduction potential of the X electrode: $E_{Y^{2+}/Y} > E_{X^{2+}/X}$

3. Applying the Nernst Equation: Since the concentrations of the ions are not 1 M, we must use the Nernst equation to find the actual electrode potentials at 298 K. The Nernst equation for a half-cell $M^{n+}(aq) + ne^- \rightarrow M(s)$ is: $E_{M^{n+}/M} = E^o_{M^{n+}/M} - \frac{RT}{nF} \ln \frac{1}{[M^{n+}]} = E^o_{M^{n+}/M} + \frac{2.303RT}{nF} \log[M^{n+}]$ At 298 K, $\frac{2.303RT}{F} \approx 0.0591$ V. For the given reactions, $n=2$. So, the equation becomes: $E_{M^{2+}/M} = E^o_{M^{2+}/M} + \frac{0.0591}{2} \log[M^{2+}]$

4. Calculating Electrode Potentials for X and Y:

   • For the X electrode (anode), $[X^{2+}] = 0.001 \text{ M} = 10^{-3} \text{ M}$. $E_{X^{2+}/X} = E^o_{X^{2+}/X} + \frac{0.0591}{2} \log(10^{-3}) = E^o_{X^{2+}/X} - 3 \times \frac{0.0591}{2} = E^o_{X^{2+}/X} - 0.08865 \text{ V}$
   • For the Y electrode (cathode), $[Y^{2+}] = 0.1 \text{ M} = 10^{-1} \text{ M}$. $E_{Y^{2+}/Y} = E^o_{Y^{2+}/Y} + \frac{0.0591}{2} \log(10^{-1}) = E^o_{Y^{2+}/Y} - 1 \times \frac{0.0591}{2} = E^o_{Y^{2+}/Y} - 0.02955 \text{ V}$

5. Deriving the Condition on Standard Potentials ($E^o$): Substitute the expressions for the electrode potentials back into the spontaneity condition ($E_{Y^{2+}/Y} > E_{X^{2+}/X}$): $E^o_{Y^{2+}/Y} - 0.02955 > E^o_{X^{2+}/X} - 0.08865$ Rearranging the terms, we get: $E^o_{Y^{2+}/Y} - E^o_{X^{2+}/X} > 0.02955 - 0.08865$ $E^o_{Y^{2+}/Y} - E^o_{X^{2+}/X} > -0.0591$ Or equivalently: $E^o_{Y^{2+}/Y} > E^o_{X^{2+}/X} - 0.0591$

6. Evaluating Each Option: We check which pair (X, Y) satisfies the condition $E^o_{Y^{2+}/Y} > E^o_{X^{2+}/X} - 0.0591$. The given standard potentials are: $E^o_{Pb^{2+}/Pb} = -0.13$ V $E^o_{Ni^{2+}/Ni} = -0.24$ V $E^o_{Cd^{2+}/Cd} = -0.40$ V $E^o_{Fe^{2+}/Fe} = -0.44$ V

   • A: X = Cd, Y = Ni $E^o_X = -0.40$ V, $E^o_Y = -0.24$ V Is $-0.24 > -0.40 - 0.0591$? Is $-0.24 > -0.4591$? Yes. Option A is correct.

   • B: X = Cd, Y = Fe $E^o_X = -0.40$ V, $E^o_Y = -0.44$ V Is $-0.44 > -0.40 - 0.0591$? Is $-0.44 > -0.4591$? Yes. Option B is correct.

   • C: X = Ni, Y = Pb $E^o_X = -0.24$ V, $E^o_Y = -0.13$ V Is $-0.13 > -0.24 - 0.0591$? Is $-0.13 > -0.2991$? Yes. Option C is correct.

   • D: X = Ni, Y = Fe $E^o_X = -0.24$ V, $E^o_Y = -0.44$ V Is $-0.44 > -0.24 - 0.0591$? Is $-0.44 > -0.2991$? No. Option D is incorrect.

Conclusion

The correct combinations of X and Y are given in options A, B, and C.