Question

Some standard electrode potentials at $298K$ are given below:
$P{b}^{2+}/Pb-0.13V$
$N{i}^{2+}Ni-0.24V$
$C{d}^{2+}/Cd-0.40V$
$F{e}^{2+}/Fe-0.44V$
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)

(A) $Cd+N{i}^{+2}⟶C{d}^{+2}+Ni$
$E_{cell}=0.40+(-24)-\frac{0.0591}{2}\log \frac{0.001}{0.1}$
$=0.16+\frac{0.0591}{2}\times 2$
$=6.64+0.551=0.71(+ve)$
(B) $E_{\text{cell }}=0.40+(-0.44)-\frac{0.591}{2}\log \frac{0.01}{0.1}$
$=-0.04+\frac{0.591}{2}\times 2$
$=-0.04+0.06=0.02(+ve)$
(C) $E_{\text{cell }}=0.24+(-0.13)+\frac{.0591}{2}\times 2$
$=0.11+0.06=0.33(+ve)$
(D) $E_{\text{cell }}=0.24+(-0.44)+\frac{0.0591}{2}\times 2$
$=-0.20+0.06=-0.14(-ve)$