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Calculate the Gibbs free energy change for the electrochemical reaction where zinc metal (Zn) is oxidized by aqueous hydrogen ions (H+) to form aqueous zinc ions (Zn2+) and hydrogen gas (H2) at 298 K, given that the standard reduction potential of Zn2+/Zn is -0.76 V and the standard hydrogen electrode potential is 0 V.

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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  E_cell  for the reaction. The cell potential can be calculated using the standard reduction potentials of the two half-reactions involved.The overall reaction can be written as:Zn s  + 2H+ aq   Zn2+ aq  + H2 g The half-reactions are:1. Oxidation of Zn: Zn s   Zn2+ aq  + 2e-  E = -0.76 V 2. Reduction of H+: 2H+ aq  + 2e-  H2 g   E = 0 V Now, we can calculate the overall cell potential  E_cell  by subtracting the standard reduction potential of the oxidation half-reaction from the standard reduction potential of the reduction half-reaction:E_cell = E reduction  - E oxidation E_cell = 0 V -  -0.76 V E_cell = 0.76 VNow that we have the cell potential, we can calculate the Gibbs free energy change  G  using the following equation:G = -nFE_cellwhere n is the number of moles of electrons transferred in the reaction  in this case, n = 2 , F is the Faraday constant  approximately 96,485 C/mol , and E_cell is the cell potential.G = - 2 mol  96,485 C/mol  0.76 V G = -146,327.6 J/molSince we usually express Gibbs free energy change in kJ/mol, we can convert the value:G = -146.33 kJ/molSo, the Gibbs free energy change for the electrochemical reaction where zinc metal is oxidized by aqueous hydrogen ions to form aqueous zinc ions and hydrogen gas at 298 K is -146.33 kJ/mol.

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