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Calculate the standard free energy change for the reaction of HCl with MnO2 to form MnCl2 and H2O at a temperature of 298 K, given that the standard reduction potentials are E°(MnO2/MnCl2)=-0.13 V and E°(H2O/HCl)=1.23 V.

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ago by (350 points)
To calculate the standard free energy change  G  for the reaction, we first need to determine the overall standard cell potential  E  for the reaction. This can be done using the given standard reduction potentials.The balanced redox reaction is:MnO2 + 4HCl -> MnCl2 + 2H2O + Cl2The reduction half-reaction is:MnO2 + 4H+ + 2e- -> Mn2+ + 2H2O  E = -0.13 V The oxidation half-reaction is:2HCl -> Cl2 + 2H+ + 2e-  E = 1.23 V Now, we can determine the overall standard cell potential  E  for the reaction by adding the standard reduction potentials:E cell  = E reduction  + E oxidation E cell  =  -0.13 V  +  1.23 V E cell  = 1.10 VNow that we have the standard cell potential, we can calculate the standard free energy change  G  using the following equation:G = -nFEwhere n is the number of moles of electrons transferred  in this case, 2 , F is Faraday's constant  96,485 C/mol , and E is the standard cell potential.G = -2 * 96,485 C/mol * 1.10 VG = -212,267 J/molSince we typically express free energy changes in kJ/mol, we can convert this value:G = -212.27 kJ/molSo, the standard free energy change for the reaction of HCl with MnO2 to form MnCl2 and H2O at 298 K is -212.27 kJ/mol.

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