Question

Calculate the standard free energy change for the following redox reaction at 25°C:2Fe(s) + 3/2 O2(g) -> Fe2O3(s)Given the following half-reactions and their standard reduction potentials at 25°C:Fe2O3(s) + 3H2O(l) + 2e- -> 2Fe(OH)3(s)     E°= -0.33VO2(g) + 4H+(aq) + 4e- -> 2H2O(l)             E°= +1.23VFe(OH)3(s) -> Fe3+(aq) + 3OH-(aq)            E°= -0.25V

Answer 1

First, we need to find the balanced redox reaction. We can do this by combining the given half-reactions:1. Fe2O3 s  + 3H2O l  + 2e- -> 2Fe OH 3 s      E= -0.33V2. O2 g  + 4H+ aq  + 4e- -> 2H2O l              E= +1.23V3. Fe OH 3 s  -> Fe3+ aq  + 3OH- aq             E= -0.25VFirst, reverse reaction 1 to obtain the Fe2O3 s  on the product side:1. 2Fe OH 3 s  -> Fe2O3 s  + 3H2O l  + 2e-     E= +0.33VNow, multiply reaction 3 by 2 to balance the number of Fe atoms:3. 2Fe OH 3 s  -> 2Fe3+ aq  + 6OH- aq             E= -0.50VAdd reaction 1 and 3:4. Fe2O3 s  + 6OH- aq  -> 2Fe3+ aq  + 3H2O l  + 2e-     E= -0.17VNow, multiply reaction 2 by 3/2 to balance the number of O2 molecules:2. 3/2 O2 g  + 6H+ aq  + 6e- -> 3H2O l              E= +1.845VAdd reaction 4 and the modified reaction 2:2Fe s  + 3/2 O2 g  -> Fe2O3 s      E= +1.675VNow we can calculate the standard free energy change  G  using the formula:G = -nFEwhere n is the number 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 * 1.675G = -323,298 J/molThe standard free energy change for the reaction 2Fe s  + 3/2 O2 g  -> Fe2O3 s  at 25C is -323,298 J/mol.