Some standard electrode potentials at 298 K are given below :\ $\mathrm{Pb}^{2+} / \mathrm{Pb}$ $-0

Question

Some standard electrode potentials at 298 K are given below :\ $\mathrm{Pb}^{2+} / \mathrm{Pb}$ $-0.13 \mathrm{~V}$\ $\mathrm{Ni}^{2+} / \mathrm{Ni}$ $-0.24 \mathrm{~V}$\ $\mathrm{Cd}^{2+} / \mathrm{Cd}$ $-0.40 \mathrm{~V}$\ $\mathrm{Fe}^{2+} / \mathrm{Fe}$ $-0.44 \mathrm{~V}$\ To a solution containing 0.001 M of $X^{2+}$ and 0.1 M of $Y^{2+}$, the metal rods $X$ and $Y$ are inserted (at 298 K ) and connected by a conducting wire. This resulted in dissolution of $X$. The correct combination(s) of $X$ and $Y$, respectively, is(are)\ (Given: Gas constant, $R=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$,\ Faraday constant, $F=96500 \mathrm{C} \mathrm{mol}^{-1}$ )\

Professor PC Thomas Classes · Answer

$(\mathrm{a}, \mathrm{b}, \mathrm{c}): \mathrm{Pb}^{2+} / \mathrm{Pb}=-0.13 \mathrm{~V}$\ $\mathrm{Ni}^{2+} / \mathrm{Ni}=-0.24 \mathrm{~V}$\ $\mathrm{Cd}^{2+} / \mathrm{Cd}=-0.40 \mathrm{~V}$\ $\mathrm{Fe}^{2+} / \mathrm{Fe}=-0.44 \mathrm{~V}$\ As $X$ is getting dissolved that means reactions taking place are\ At anode : $X \longrightarrow X^{2+}+2 e^{-}$\                            ( 0.001 M )\ At cathode $: Y^{2+}+2 e^{-} \longrightarrow Y$\                                  (0.1 M)\ $X+Y^{2+} \longrightarrow X^{2+}+Y$\ $E_{	ext {cell }}=E_{	ext {cell }}^{\circ}-\frac{0.059}{2} \log \frac{\left[X^{2+}ight]}{\left[Y^{2+}ight]}$\ $E_{	ext {cell }}=E_{	ext {cell }}^{\circ}+0.06$\ For $\mathrm{Cd} / \mathrm{Ni}$ cell, $E_{	ext {cell }}^{\circ}=E_{	ext {red }}^{\mathrm{o}}-E_{	ext {oxid }}^{\circ}$\ $=-0.24-(-0.40)=+0.16 \mathrm{~V}$\ $E_{	ext {cell }}=0.22 \mathrm{~V}$\ For $\mathrm{Cd} / \mathrm{Fe}$ cell, $E_{	ext {cell }}^{\circ}=-0.04, E_{	ext {cell }}=0.02$\ For $\mathrm{Ni} / \mathrm{Pb}$ cell, $E_{	ext {cell }}^{\mathrm{o}}=0.11 \mathrm{~V}, E_{	ext {cell }}=0.17 \mathrm{~V}$\ For $\mathrm{Ni} /$ Fe cell, $E_{	ext {cell }}^{\circ}=-0.2 \mathrm{~V}, E_{	ext {cell }}=-0.14 \mathrm{~V}$\