Some standard electrode potentials at $298 \mathrm{K}$ are given below:

$\begin{array}{ll}{\mathrm{Pb}}^{2+}/\mathrm{Pb}& -0.13 \mathrm{V}\end{array}$

$\begin{array}{ll}{\mathrm{Ni}}^{2+}/\mathrm{Ni}& -0.24 \mathrm{V}\end{array}$

$\begin{array}{ll}{\mathrm{Cd}}^{2+}/\mathrm{Cd}& -0.40 \mathrm{V}\end{array}$

$\begin{array}{ll}{\mathrm{Fe}}^{2+}/\mathrm{Fe}& -0.44 \mathrm{V}\end{array}$

To a solution containing $0.001 \mathrm{M}$ of ${\mathbf{X}}^{2+}$ and $0.1 \mathrm{M}$ of ${\mathbf{Y}}^{2+}$, the metal rods $\mathbf{X}$ and $\mathbf{Y}$ are inserted $($at $298 \mathrm{K})$ and connected by a conducting wire. This resulted in dissolution of $\mathbf{X}$. The correct combination(s) of $\mathbf{X}$ and $\mathbf{Y}$, respectively, is(are)

(Given: Gas constant, $\mathrm{R}=8.314 \mathrm{J}{ \mathrm{K}}^{-1}{ \mathrm{mol}}^{-1}$, Faraday constant, $\mathrm{F}=96500 \mathrm{C} {\mathrm{mol}}^{-1}$)

• Correct answer
  A

  $\mathrm{Cd}$ and $\mathrm{Ni}$
• Correct answer
  B

  $\mathrm{Cd}$ and $\mathrm{Fe}$
• Correct answer
  C

  $\mathrm{Ni}$ and $\mathrm{Pb}$
• D

  $\mathrm{Ni}$ and $\mathrm{Fe}$

Hence, the correct answer is options: 'A', 'B', 'C'.