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(B) The standard reduction potential of $Pb^{2+}|Pb$ is $E^{\circ} = -0.126 \ V$. The standard reduction potential of $Fe^{2+}|Fe$ is $E^{\circ} = -0.

Vedclass · Accepted Answer

More lead and $Fe^{2+}$ ions are formed

Vedclass · Answer

(B) The standard reduction potential of $Pb^{2+}|Pb$ is $E^{\circ} = -0.126 \ V$. The standard reduction potential of $Fe^{2+}|Fe$ is $E^{\circ} = -0.44 \ V$. Since the reduction potential of $Pb^{2+}|Pb$ is greater (less negative) than that of $Fe^{2+}|Fe$,$Pb^{2+}$ ions will be reduced to $Pb$ metal,and $Fe$ metal will be oxidized to $Fe^{2+}$ ions. The spontaneous reaction is: $Fe(s) + Pb^{2+}(aq) \rightarrow Fe^{2+}(aq) + Pb(s)$. Therefore,the addition of powdered $Pb$ and $Fe$ to the solution results in the formation of more $Pb$ and $Fe^{2+}$ ions.

Adding powdered lead and iron to a solution that is $1.0 \ M$ in both $Pb^{2+}$ and $Fe^{2+}$ ions would result in a reaction in which:

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