To calculate the standard free energy change G for the redox reaction, we need to first determine the overall cell potential Ecell and then use the following equation:G = -nFEcellwhere n is the number of moles of electrons transferred, F is Faraday's constant 96,485 C/mol , and Ecell is the cell potential.First, we need to identify the half-reactions:Oxidation half-reaction: H2O2 aq O2 g + 2H+ aq + 2e-Reduction half-reaction: Fe3+ aq + e- Fe2+ aq Now, we need to balance the half-reactions to find the overall redox reaction:Oxidation half-reaction multiplied by 2 : 2H2O2 aq 2O2 g + 4H+ aq + 4e-Reduction half-reaction multiplied by 4 : 4Fe3+ aq + 4e- 4Fe2+ aq Adding the two balanced half-reactions, we get:2H2O2 aq + 4Fe3+ aq 2O2 g + 4H+ aq + 4Fe2+ aq However, this is not the same as the given balanced chemical equation. To make it match, we need to multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2:Oxidation half-reaction multiplied by 3 : 3H2O2 aq 3O2 g + 6H+ aq + 6e-Reduction half-reaction multiplied by 2 : 2Fe3+ aq + 2e- 2Fe2+ aq Now, adding the two balanced half-reactions, we get the given balanced chemical equation:2Fe3+ aq + 3H2O2 aq 2Fe2+ aq + 3O2 g + 6H+ aq Now we can find the overall cell potential Ecell using the standard reduction potentials given:Ecell = Ecathode - EanodeSince Fe3+ is being reduced, it is the cathode, and H2O2 is the anode:Ecell = +0.77 V - +0.70 V = +0.07 VNow we can calculate the standard free energy change G using the equation:G = -nFEcellThe number of moles of electrons transferred n is 6, as seen in the balanced equation:G = - 6 mol 96,485 C/mol +0.07 V G = -40,560.9 J/molSince it is more common to express G in kJ/mol, we can convert the value:G = -40.56 kJ/molSo, the standard free energy change for the given redox reaction is -40.56 kJ/mol.