Adding powdered Pb and Fe to a solution containing 1 M in each of ${\mathrm{Pb}}^{2+}$ and ${\mathrm{Fe}}^{2+}$ ions would result into the formation of  Given: $E^{\circ }\left({\mathrm{Pb}}^{2+}\mid \mathrm{Pb}\right)=-0.126\mathrm{V}$and $\left.E^{\circ }\left({\mathrm{Fe}}^{2+}\mid \mathrm{Fe}\right)=-0.44\mathrm{V}\right)$.