Question

Calculate the standard free energy change for the following redox reaction:2Fe^2+ (aq) + H2O2 (aq) + 2H+ (aq) -> 2Fe^3+ (aq) + 2H2O (l) Given the following reduction potentials at standard conditions:Fe3+ (aq) + e- -> Fe2+ (aq)      E° = 0.77 VH2O2 (aq) + 2H+ (aq) + 2e- -> 2H2O (l)       E° = 1.78 V

Answer 1

To calculate the standard free energy change  G  for the redox reaction, we first need to determine the overall cell potential  Ecell  for the reaction. The given half-reactions are:Fe  aq  + e -> Fe  aq       E = 0.77 V  Reduction HO  aq  + 2H  aq  + 2e -> 2HO  l        E = 1.78 V  Reduction Since the first reaction is a reduction, the second reaction must be reversed to make it an oxidation reaction:HO  aq  + 2H  aq  + 2e -> 2HO  l        E = 1.78 V  Reduction 2HO  l  -> HO  aq  + 2H  aq  + 2e       E = -1.78 V  Oxidation Now, we can add the two half-reactions to get the overall redox reaction:Fe  aq  + e -> Fe  aq       E = 0.77 V2HO  l  -> HO  aq  + 2H  aq  + 2e       E = -1.78 V--------------------------------------------------------------2Fe  aq  + HO  aq  + 2H  aq  -> 2Fe  aq  + 2HO  l Now, we can calculate the overall cell potential  Ecell  by adding the half-cell potentials:Ecell = E reduction  + E oxidation  = 0.77 V +  -1.78 V  = -1.01 VFinally, we can calculate the standard free energy change  G  using the formula:G = -nFEcellwhere n is the number of moles of electrons transferred  in this case, n = 2 , F is the Faraday constant  96,485 C/mol , and Ecell is the overall cell potential.G = -2 * 96,485 C/mol *  -1.01 V G = 194,658 J/molSo, the standard free energy change for the given redox reaction is 194,658 J/mol.