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What is the Gibbs free energy change for the electrochemical reaction:Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)given that the standard reduction potentials are E°(Cu2+|Cu) = +0.34 V and E°(Zn2+|Zn) = -0.76 V, and the reaction is carried out at 25°C?

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To calculate the Gibbs free energy change  G  for the electrochemical reaction, we first need to determine the standard cell potential  Ecell  for the reaction. This can be done using the standard reduction potentials given:Ecell = E cathode  - E anode In this reaction, Cu2+ is reduced to Cu, so it is the cathode, and Zn is oxidized to Zn2+, so it is the anode.Ecell = E Cu2+|Cu  - E Zn2+|Zn  =  +0.34 V  -  -0.76 V  = +1.10 VNow that we have the standard cell potential, we can calculate the Gibbs free energy change using the following equation:G = -nFEcellWhere n is the number of moles of electrons transferred in the reaction  in this case, 2 moles of electrons, as both Cu2+ and Zn2+ have a charge of +2 , F is the Faraday constant  96,485 C/mol , and Ecell is the standard cell potential.G = -2 mol  96,485 C/mol  1.10 V = -212,667 J/molSince the value is usually expressed in kJ/mol, we can convert it:G = -212,667 J/mol   1 kJ/1000 J  = -212.67 kJ/molSo, the Gibbs free energy change for the given electrochemical reaction at 25C is -212.67 kJ/mol.

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