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Calculate the Gibbs free energy change for the electrochemical reaction where Fe(s) is oxidized to Fe3+(aq) and Br2(l) is reduced to Br-(aq) with the help of the following half-reactions: Fe3+(aq) + e- ⇌ Fe2+(aq)              E° = 0.771 VBr2(l) + 2e- ⇌ 2Br-(aq)           E° = 1.087 V

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To calculate the Gibbs free energy change for the electrochemical reaction, we first need to determine the balanced overall redox reaction and the standard cell potential  Ecell  for the reaction.The given half-reactions are:Fe3+ aq  + e-  Fe2+ aq               E = 0.771 VBr2 l  + 2e-  2Br- aq            E = 1.087 VFirst, we need to balance the electrons in both half-reactions. Since the second reaction involves 2 electrons, we need to multiply the first reaction by 2 to balance the electrons:2 Fe3+ aq  + e-  Fe2+ aq                2 E  = 2 0.771 V 2Fe3+ aq  + 2e-  2Fe2+ aq               E = 1.542 VNow, we can combine the two half-reactions:2Fe3+ aq  + 2e- + Br2 l  + 2e-  2Fe2+ aq  + 2Br- aq The electrons cancel out:2Fe3+ aq  + Br2 l   2Fe2+ aq  + 2Br- aq Now, we can calculate the standard cell potential  Ecell  by adding the standard potentials of the two half-reactions:Ecell = E reduction  + E oxidation Ecell = 1.087 V + 1.542 VEcell = 2.629 VFinally, we can calculate the Gibbs free energy change  G  using the formula:G = -nFEcellwhere n is the number of moles of electrons transferred  in this case, 2 , F is the Faraday constant  96,485 C/mol , and Ecell is the standard cell potential.G = -2 * 96,485 C/mol * 2.629 VG = -507,726 J/molSo, the Gibbs free energy change for the electrochemical reaction is -507,726 J/mol.

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