Question

An electric current is passed through an aqueous solution of lead(II) chloride. The half-reactions are:
Cathode- ${\text{Pb}}^{x+}\left(\text{aq}\right)+x{e}^{-}\to \text{Pb(s)}$
Anode - ${\text{Cl}}^{-y}\left(\text{aq}\right)\to \text{Cl(s)}+y{e}^{-}$
Find the value of x + y.

• A. 3

Answer 1

The correct option is A 3
We know oxidation occurs at anode and reduction occurs at cathode.
In ${\text{PbCl}}_2$, ions are ${\text{Pb}}^{2+}$ and ${\text{Cl}}^{-}$.
So, ${\text{Pb}}^{2+}$ will gain electrons (i.e., reduction) and ${\text{Cl}}^{-}$ will lose electron (i.e., oxidation).

At anode:
${\text{Cl}}^{-}\left(\text{aq}\right)\to \text{Cl(s)}+e^{-}$ (oxidation)

At cathode:
${\text{Pb}}^{2+}\left(\text{aq}\right)+2e^{-}\to \text{Pb(s)}$ (reduction)

On comparing these equations with the equations given in the question, we get x =2 and y = 1
So x + y = 3