To calculate the standard free energy change G for the redox reaction, we can use the Nernst equation:G = -nFEwhere n is the number of moles of electrons transferred in the reaction, F is the Faraday constant 96,485 C/mol , and E is the standard cell potential.First, we need to determine the balanced redox reaction and the number of electrons transferred. The given half-reactions are:Mg2+ + 2e- -> Mg E = -2.37 V - Reduction half-reactionO2 + 4e- + 2H2O -> 4OH- E = +1.23 V - Reduction half-reactionWe need to reverse the first half-reaction to make it an oxidation half-reaction:Mg -> Mg2+ + 2e- E = +2.37 V - Oxidation half-reactionNow, we need to balance the electrons in both half-reactions. Since the oxidation half-reaction involves 2 electrons and the reduction half-reaction involves 4 electrons, we need to multiply the oxidation half-reaction by 2:2Mg -> 2Mg2+ + 4e- E = +2.37 V Now, we can add the balanced half-reactions:2Mg + O2 + 4e- + 2H2O -> 2Mg2+ + 4OH- + 4e-The electrons cancel out:2Mg + O2 + 2H2O -> 2Mg2+ + 4OH-Now, we can combine Mg2+ and OH- to form MgO:2Mg + O2 + 2H2O -> 2MgO + 2H2OFinally, we can cancel out the water molecules:2Mg + O2 -> 2MgONow that we have the balanced redox reaction, we can calculate the standard cell potential E by adding the potentials of the half-reactions:E = E oxidation + E reduction E = +2.37 V + +1.23 V E = +3.60 VSince the balanced redox reaction involves 4 moles of electrons n = 4 , we can now calculate the standard free energy change G :G = -nFEG = - 4 mol 96,485 C/mol 3.60 V G = -1,387,032 J/molSo, the standard free energy change for the redox reaction at 298 K is -1,387,032 J/mol.