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What is the Gibbs free energy change for the reaction of Zinc with Copper sulfate solution in which Zinc gets oxidized to Zinc sulfate and Copper ions get reduced to Copper metal, given the standard reduction potentials for each half-reaction?

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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 redox reaction using the standard reduction potentials for each half-reaction.The half-reactions are:1. Oxidation of Zinc: Zn s   Zn aq  + 2eStandard reduction potential  E  = -0.76 V2. Reduction of Copper ions: Cu aq  + 2e  Cu s Standard reduction potential  E  = +0.34 VSince the Zinc is being oxidized, we need to reverse the sign of its standard reduction potential. The overall cell potential  E  is the sum of the standard reduction potentials for the two half-reactions:E = E - E =  +0.34 V  -  -0.76 V  = +1.10 VNow, we can calculate the Gibbs free energy change  G  using the formula:G = -nFEwhere n is the number of moles of electrons transferred  in this case, 2 moles of electrons , F is the Faraday constant  approximately 96,485 C/mol , and E is the overall cell potential.G = -2 mol  96,485 C/mol  1.10 VG = -212,267 J/molThe Gibbs free energy change for the reaction of Zinc with Copper sulfate solution is -212,267 J/mol.

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