Question

A solution is 1 molar in each of $NaCl,CdC{l}_2,ZnC{l}_2$ and $PbC{l}_2$. To this Sn metal is added, which of the following is true?
Given $E^{\circ }\left(P{b}^{2+}/Pb=-0.126V\right)$
$E^{\circ }{}_{S{n}^{2+}/Sn}=-0.136V,E_{C{d}^{2+}/Cd}^{\circ }=-0.40V$
$E_{Z{n}^{2+}/Zn}^{\circ }=-0.763V,E_{N{a}^{+}/Na}^{\circ }=-2.71V$

• A. Sn can reduce $N{a}^{+}$to $Na$
• B. Sn can reduce $Z{n}^{2+}$ to $Zn$
• C. Sn can reduce $C{d}^{2+}$ to $Cd$
• D. Sn can reduce $P{b}^{2+}$ to $Pb$

Answer 1

Answer: (D)

S.R.P or (oxdising power) $Pb>Sn>Cd>Zn>Na$
Higher value of S.R.P $\to$ more is oxidising power and less is reducing power
So, $(Sn)$ can oxidise $Cd,Zn$ and $Na$ but not $Pb$
i.e., $Sn$ can reduce $P{b}^{2+}$ to $Pb$