Question

A solution is 1 molar in each of $NaCl , CdCl _{2}, ZnCl _{2}$ and $PbCl _{2}$. To this Sn metal is added, which of the following is true?
Given $E ^{\circ}\left( Pb ^{2+} / Pb =-0.126 \,V \right)$
$E ^{\circ}{ }_{ Sn ^{2+} / Sn }=-0.136 V , E _{ Cd ^{2+} / Cd }^{\circ}=-0.40\, V$
$E _{ Zn ^{2+} / Zn }^{\circ}=-0.763 V , E _{ Na ^{+} / Na }^{\circ}=-2.71 \,V$

• A. Sn can reduce $Na ^{+}$to $Na$
• B. Sn can reduce $Zn ^{2+}$ to $Zn$
• C. Sn can reduce $Cd ^{2+}$ to $Cd$
• D. Sn can reduce $Pb ^{2+}$ to $Pb$

Answer 1

Answer: (D)

S.R.P or (oxdising power) $Pb > Sn > Cd > Zn > Na$
Higher value of S.R.P $\rightarrow$ 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 $Pb ^{2+}$ to $Pb$