Adding powdered lead and iron to a solution that $\text{ is }1\mathrm{M}\text{ in both }{\mathrm{Pb}}^{2+}\text{ and }{\mathrm{Fe}}^{2+}\text{ ions. }$would results to a reaction in which $\left({\mathrm{E}}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}^0=\right.$ $\left.-0.44\mathrm{V},{\mathrm{E}}_{{\mathrm{Pb}}^{2+}/\mathrm{Pb}}^0=-0.13\mathrm{V}\right)$

As the value of standard electrode potential for Pb²⁺is more than that of Fe²⁺

i.e.E^∘ Pb²⁺/Pb > E^∘  Fe²⁺/Fe

So,Fe will get oxidised and Pb²⁺ will get reduced. 

Hence, we will have more of Pb and Fe²⁺ions.