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Calculate the standard free energy change (∆G°) for the following redox reaction at 25°C:2 Fe³⁺(aq) + 3 H₂(g) → 2 Fe(s) + 6 H⁺(aq)

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ago by (510 points)
To calculate the standard free energy change  G  for the redox reaction, we need to use the Nernst equation:G = -nFEwhere n is the number of moles of electrons transferred, F is the Faraday constant  96,485 C/mol , and E is the standard cell potential.First, we need to determine the half-reactions for the redox reaction:Oxidation half-reaction  H is oxidized to H :H g   2 H aq  + 2 eSince there are 3 moles of H, multiply the half-reaction by 3:3 H g   6 H aq  + 6 eReduction half-reaction  Fe is reduced to Fe :Fe aq  + 3 e  Fe s Since there are 2 moles of Fe, multiply the half-reaction by 2:2 Fe aq  + 6 e  2 Fe s Now, we need to find the standard reduction potentials  E  for each half-reaction. These values can be found in a standard reduction potential table:E H/H  = 0.00 V  by definition E Fe/Fe  = -0.036 VNext, we need to calculate the standard cell potential  E  for the overall redox reaction:E cell  = E cathode  - E anode E cell  = E Fe/Fe  - E H/H E cell  = -0.036 V - 0.00 VE cell  = -0.036 VNow, we can calculate the standard free energy change  G  using the Nernst equation:G = -nFEG = -6 mol e * 96,485 C/mol e *  -0.036 V G = 20,923.56 J/molSince the standard free energy change is usually expressed in kJ/mol, we can convert the value:G = 20.92 kJ/mol

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