Question

Some standard electrode potentials at $298K$ are given below:
$P{b}^{2+}/Pb-0.13VN{i}^{2+}/Ni-0.24VC{d}^{2+}/Cd-0.40VF{e}^{2+}/Fe-0.44V$
To a solution containing $0.001M$ of $X^{2+}$ and $0.1M$ of $Y^{2+}$, the metal rods $X$ and $Y$ are inserted (at $298K$) 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, $R=8.314J{K}^{-1}mo{l}^{-1}$,
Faraday constant, $F=96500Cmo{l}^{-1}$)

• A. $Cd$ and $Ni$
• B. $Cd$ and $Fe$
• C. $Ni$ and $Pb$
• D. $Ni$ and $Fe$

Answer 1

Answer: (A), (B), (C)

$Ni$ and $Pb$

$\left(A\right)Cd+N{i}^{+2}\underset{}{\overset{}{\to }}C{d}^{+2}+Ni{E}_{cell}=0.40+(-24)-\frac{0.0591}{2}log\frac{0.001}{0.1}=0.16+\frac{0.0591}{2}\times 2=.16+0.0591=0.21(+ve)$

$\left(B\right)Fe+C{d}^{+2}\underset{}{\overset{}{\to }}F{e}^{+2}+Cd{E}_{cell}=0.40+(-0.40)-\frac{0.0591}{2}log\frac{0.001}{0.1}=0.4+\frac{0.0591}{2}\times 2=-0.044+0.061=0.02(+ve)$

$\left(C\right)Ni+P{b}^{+2}\underset{}{\overset{}{\to }}N{i}^{+2}+Pb{E}_{cell}=0.24+(-0.13)+\frac{0.0591}{2}\times 2=0.11+0.06=0.17(+ve)$

$\left(D\right)Fe+N{i}^{+2}\underset{}{\overset{}{\to }}F{e}^{+2}+Ni{E}_{cell}=0.24+(-0.44)+\frac{0.0591}{2}\times 2=-0.20+0.06=-0.14(-ve)$
For spontaneous Reaction The $E_{cell}$ must be positive value.
Thus correct Answer is Option (A),(B),(C)