Consider the reduction potentials below: Pb2+(aq) + 2e^- <=> Pb(s) E° = -0.13 V and Mg2+(aq) + 2e^- <=> Mg(s) E° = -0.76 V. a) Write the overall Redox reaction that takes place when the above half cel

The species with the more negative standard reduction potential will be oxidized (anode), and the species with the less negative (or more positive) standard reduction potential will be reduced (cathode).
Comparing $\text{E}^\circ$ values:

• $\text{Mg}^{2+}/\text{Mg}$: $\text{E}^\circ = -0.76 \text{ V}$ (more negative, so $\text{Mg}$ will be oxidized)
• $\text{Pb}^{2+}/\text{Pb}$: $\text{E}^\circ = -0.13 \text{ V}$ (less negative, so $\text{Pb}^{2+}$ will be reduced)

Oxidation (Anode): $\text{Mg}(s) \to \text{Mg}^{2+}(aq) + 2\text{e}^-$
Reduction (Cathode): $\text{Pb}^{2+}(aq) + 2\text{e}^- \to \text{Pb}(s)$

The electrons are already balanced.

Where:

• $\text{E}^\circ_{cathode} = \text{E}^\circ(\text{Pb}^{2+}/\text{Pb}) = -0.13 \text{ V}$
• $\text{E}^\circ_{anode} = \text{E}^\circ(\text{Mg}^{2+}/\text{Mg}) = -0.76 \text{ V}$

The $\text{E}^\circ$ value of the cell is $\boxed{0.63 \text{ V}}$