Equivalent weight of  Pb(NO₃)₂ in the equation

$2\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\to 2\mathrm{PbO}+4{\mathrm{NO}}_2+{\mathrm{O}}_2$ is

$2\mathrm{Pb}{\left(\stackrel{+5}{\mathrm{N}}{\stackrel{-2}{\mathrm{O}}}_3\right)}_2 \stackrel{∆}{→} 2\mathrm{PbO}+4\stackrel{+4}{\mathrm{N}}{\mathrm{O}}_2 +{\stackrel{0}{\mathrm{O}}}_2$

'N' atom Undergoes reduction

'O' atom Undergoes Oxidation

Total increase in Ox. state of 'N' per 2 moles of lead nitrate = 20-16=4

EW of Pb(NO₃)₂ = 2M/4=M/2

Alternate method:

                 Total decrease in Ox. state of Oxygen per 2 moles of lead nitrate = [-20-(-24)]=4

EW of Pb(NO₃)₂ = 2M/4=M/2