$\mathrm{x}(\mathrm{s}) \longrightarrow \mathrm{x}^{+2}(0.001 \mathrm{M})+2 \mathrm{e}^{-}($anode $)$
$\mathrm{y}^{+2}(0.1 \mathrm{M})+2 \mathrm{e}^{-} \longrightarrow \mathrm{y}$ (s) (cathode)

$\mathrm{E}_{\text {oell }}=\mathrm{E}^{\circ}{ }_{\text {cell }}-\frac{0.06}{2} \log \frac{\mathrm{x}^{+2}}{\mathrm{y}^{+2}}$
$E_{\text {cell }}=E^{\circ}{ }_{\text {cell }}+0.06$
(A) Cd and $\mathrm{Ni} \mathrm{E}_{\text {cell }}^{\circ}=+0.4-0.24$. $\mathrm{E}_{\text {cell }}=0.22$
(B) Cd and $\mathrm{Fe} \mathrm{E}^{\circ}{ }_{\text {cell }}=-0.04$; $\mathrm{E}_{\text {cell }}=0.02$
(C) Ni and Pb $\mathrm{E}_{\text {cell }}^{\circ}=0.11$; $\mathrm{E}_{\text {cell }}=0.17$
(D) Ni and $\mathrm{Fe} \mathrm{E}_{\text {cell }}^{\circ}=-02$; $\mathrm{E}_{\text {cell }}=-0.14$
since in $(A)(B)(C) E_{\text {cell }}$ is positive hence answer is $(A)(B)(C)$.