Question

Some standard electrode potentials at $298\, K$ are given below:
$Pb ^{2+} / Pb -0.13 \,V$
$Ni ^{2+} Ni -0.24 \,V$
$Cd ^{2+} / Cd -0.40 \,V$
$Fe ^{2+} / Fe -0.44\, V$
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, $R =8.314 \,J \,K ^{-1}\, mol ^{-1}$,
Faraday constant, $F =96500\, C \,mol ^{-1}$ )

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

Answer 1

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

(A) $Cd + Ni ^{+2} \longrightarrow Cd ^{+2}+ Ni$
$E _{ cell }=0.40+(-24)-\frac{0.0591}{2} \log \frac{0.001}{0.1}$
$=0.16+\frac{0.0591}{2} \times 2$
$=6.64+0.551=0.71(+ ve )$
(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) $E _{\text {cell }}=0.24+(-0.44)+\frac{0.0591}{2} \times 2$
$=-0.20+0.06=-0.14(- ve )$