Question

What is the standard free energy change of the following redox reaction at 298 K and 1 bar?2Fe2+(aq) + H2O2(aq) + 2H+(aq) → 2Fe3+(aq) + 2H2O(l)Given the following standard reduction potentials:Fe3+/Fe2+ = +0.771 VH2O2/O2 = +0.695 V

Answer 1

To calculate the standard free energy change  G  of the redox reaction, we first need to determine the overall cell potential  Ecell  of the reaction using the given standard reduction potentials.The balanced half-reactions are:Fe3+ aq  + e-  Fe2+ aq   E = +0.771 V  Reduction H2O2 aq  + 2H+ aq  + 2e-  2H2O l   E = +0.695 V  Reduction We need to reverse the first half-reaction to make it an oxidation reaction:Fe2+ aq   Fe3+ aq  + e-  E = -0.771 V  Oxidation Now, we can add the two half-reactions to obtain the overall redox reaction:2Fe2+ aq  + H2O2 aq  + 2H+ aq   2Fe3+ aq  + 2H2O l Next, we can calculate the overall cell potential  Ecell  by adding the standard reduction potentials of the two half-reactions:Ecell = E Oxidation  + E Reduction Ecell = -0.771 V + 0.695 VEcell = -0.076 VNow, we can calculate the standard free energy change  G  using the following formula:G = -nFEcellwhere n is the number of moles of electrons transferred  in this case, n = 2 , F is Faraday's constant  96,485 C/mol , and Ecell is the overall cell potential.G = -2 * 96,485 C/mol *  -0.076 V G = 14,660 J/molSo, the standard free energy change  G  of the redox reaction at 298 K and 1 bar is 14,660 J/mol.