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(D) The standard reduction potentials are $E^{\circ}_{Pb^{2+}/Pb} = -0.13 \, V$ and $E^{\circ}_{Fe^{2+}/Fe} = -0.44 \, V$. Since $E^{\circ}_{Pb^{2+}/P

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$Fe^{2+}$ concentration increases and $Pb^{2+}$ concentration decreases

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(D) The standard reduction potentials are $E^{\circ}_{Pb^{2+}/Pb} = -0.13 \, V$ and $E^{\circ}_{Fe^{2+}/Fe} = -0.44 \, V$. Since $E^{\circ}_{Pb^{2+}/Pb} > E^{\circ}_{Fe^{2+}/Fe}$,$Pb^{2+}$ ions will act as the oxidizing agent and $Fe$ metal will act as the reducing agent. The spontaneous cell reaction is: $Pb^{2+}(aq) + Fe(s) \rightarrow Fe^{2+}(aq) + Pb(s)$. In this reaction,$Pb^{2+}$ ions are reduced to $Pb$ metal,so the concentration of $Pb^{2+}$ decreases. Simultaneously,$Fe$ metal is oxidized to $Fe^{2+}$ ions,so the concentration of $Fe^{2+}$ increases.

Addition of powdered lead and iron to a solution which is $1.0 \, M$ in both $Pb^{2+}$ and $Fe^{2+}$ would result in [$E^{\circ}_{Fe^{2+}/Fe} = -0.44 \, V$ and $E^{\circ}_{Pb^{2+}/Pb} = -0.13 \, V$]

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