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What is the Gibbs free energy change at 298 K for the reaction 2Fe2+(aq) + Cl2(g) --> 2Fe3+(aq) + 2Cl-(aq) if the standard electrode potentials for the half-reactions are E°(Fe3+(aq) + e- --> Fe2+(aq)) = -0.77 V and E°(Cl2(g) + 2e- --> 2Cl-(aq)) = 1.36 V?

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To calculate the Gibbs free energy change  G  for the reaction, we first need to determine the overall cell potential  E  for the reaction. This can be done by adding the standard electrode potentials for the two half-reactions:E Fe3+ aq  + e-  Fe2+ aq   = -0.77 VE Cl2 g  + 2e-  2Cl- aq   = 1.36 VSince the Fe3+/Fe2+ half-reaction needs to be reversed to match the overall reaction, we need to change the sign of its potential:E Fe2+ aq   Fe3+ aq  + e-  = 0.77 VNow, we can add the potentials for the two half-reactions:E overall  = E Fe2+ aq   Fe3+ aq  + e-  + E Cl2 g  + 2e-  2Cl- aq  E overall  = 0.77 V + 1.36 VE overall  = 2.13 VNext, we can use the Nernst equation to calculate the Gibbs free energy change  G :G = -nFEwhere n is the number of moles of electrons transferred in the reaction  in this case, 2  and F is the Faraday constant  approximately 96,485 C/mol .G = -2 * 96,485 C/mol * 2.13 VG = -411,032 J/molSo, the Gibbs free energy change  G  for the reaction at 298 K is -411,032 J/mol.

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