Atom in highest oxidation state can oxidize iodide to liberate ${\mathrm{I}}_2$ which is volumetrically measured by iodometric titration using hypo solution.
$2{\mathrm{I}}^{-}\to {\mathrm{I}}_2$
$\mathrm{P}{\mathrm{b}}^{+2}\to$ Lowest oxidation state can not oxidise iodide to ${\mathrm{I}}_2$ .