Calculate the standard emf of a cell that uses the \(\mathrm{Mg} / \mathrm{Mg}^{2+}\) and \(\mathrm{Cu} / \mathrm{Cu}^{2+}\) half-cell reactions at \(25^{\circ} \mathrm{C} .\) Write the equation for the cell reaction that occurs under standard-state conditions.

The given half-cell reactions are Mg/Mg2+ and Cu/Cu2+. The standard reduction potentials for these reactions from the standard reduction potential table are as follows: \n\n\(Mg^{2+} + 2e^- → Mg \, E° = -2.37 \, V\)\n\(Cu^{2+} + 2e^- → Cu \, E° = +0.34 \, V\)

An oxidation-reduction (redox) reaction consists of two half-reactions, one for oxidation and one for reduction. In this case, magnesium (Mg) undergoes oxidation to form Mg2+, and copper ions (Cu2+) undergo reduction to form copper (Cu). Therefore, the half-reactions can be written as follows:\n\nOxidation: \(Mg → Mg^{2+} + 2e^-\) \nReduction: \(Cu^{2+} + 2e^- → Cu\) \n\nThese can be added to give the overall cell reaction:\n\n\(Mg + Cu^{2+} → Mg^{2+} + Cu\) \n\nThe standard cell potential (\(E^{°}_{cell}\)) is the difference between the standard reduction potentials of the cathode and anode. Since the Cu2+/Cu half-cell acts as the cathode (where reduction occurs) and the Mg/Mg2+ half-cell acts as the anode (where oxidation occurs), \(E^{°}_{cell}\) can be calculated as follows:\n\n\(E^{°}_{cell} = E^{°}_{cathode} - E^{°}_{anode} = 0.34V - (-2.37V) = 2.71V\)

The standard emf of the cell is calculated to be 2.71V. The cell reaction that occurs under standard-state conditions is \(Mg + Cu^{2+} → Mg^{2+} + Cu\).