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Calculate the Gibbs free energy change for the electrochemical reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Given the standard reduction potentials: E°(Cu2+/Cu) = 0.34 V E°(Zn2+/Zn) = -0.76 V Assuming standard conditions of 1 M concentration and 1 atm pressure 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 overall cell potential  Ecell  for the reaction. The cell potential is the difference between the reduction potentials of the two half-reactions:Ecell = E cathode  - E anode In this case, the Cu2+/Cu half-reaction acts as the cathode  reduction  and the Zn2+/Zn half-reaction acts as the anode  oxidation :Ecell = E Cu2+/Cu  - E Zn2+/Zn  = 0.34 V -  -0.76 V  = 1.10 VNow that we have the 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, F is the Faraday constant  96,485 C/mol , and Ecell is the cell potential.For this reaction, the balanced redox reaction is:Zn s  + Cu2+ aq   Zn2+ aq  + Cu s The number of moles of electrons transferred  n  is 2, as one Zn atom loses 2 electrons and one Cu2+ ion gains 2 electrons.Now we can calculate G:G = -nFEcell = -2 mol * 96,485 C/mol * 1.10 V = -212,667 J/molSince it's common to express Gibbs free energy change 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 electrochemical reaction is -212.67 kJ/mol.

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