0 votes
19 views
ago in Chemical thermodynamics by (550 points)
Calculate the standard free energy change for the following redox reaction at 298 K:Mg(s) + 2H+(aq) -> Mg2+(aq) + H2(g)Given: Eºred for Mg2+/Mg = -2.37 V & Eºred for H+/H2 = 0 V.

1 Answer

0 votes
ago by (410 points)
To calculate the standard free energy change  G  for the redox reaction, we first need to determine the standard cell potential  Ecell  for the reaction. The cell potential can be found using the Nernst equation:Ecell = Ered cathode  - Ered anode In this reaction, Mg is being oxidized  losing electrons  and H+ is being reduced  gaining electrons . Therefore, Mg is the anode and H+ is the cathode. We are given the standard reduction potentials for both half-reactions:Ered Mg2+/Mg  = -2.37 V  anode Ered H+/H2  = 0 V  cathode Now we can calculate the standard cell potential:Ecell = 0 V -  -2.37 V  = 2.37 VNext, we can use the relationship between the standard cell potential and the standard free energy change:G = -nFEcellwhere n is the number of moles of electrons transferred in the reaction  in this case, 2 moles  and F is the Faraday constant  96,485 C/mol .G = - 2 mol  96,485 C/mol  2.37 V  = -457,331.9 J/molSince the standard free energy change is typically expressed in kJ/mol, we can convert the value:G = -457.33 kJ/molSo, the standard free energy change for the given redox reaction at 298 K is -457.33 kJ/mol.

Related questions

Welcome to Sarvan Science Q&A, where you can ask questions and receive answers from other members of the community.
...