5fe2+ + Mno4- +8H+ -----> 5Fe3+ +Mn2+ +4H2O determine Eo determmine delta G init

Question

5fe2+ + Mno4- +8H+     -----> 5Fe3+ +Mn2+ +4H2O

determine Eo
determmine delta G initial
and equilibrium constant K

I know how to balance it back to initital but then after that ihave no idea!
Fe2+ + MnO4- --> Fe3+ + Mn2+


Fe2+ + MnO4- --> Fe3+ + Mn2+

Explanation / Answer

                     5(Fe2+-> Fe3+ +e)               E_half = -0.77   MnO4 + 8H+ +5e -> Mn2+ +4H2O               E_half = 1.51 5fe2+ + Mno4- +8H+ -----> 5Fe3+ +Mn2+ + 4H2O you multiply the first equation by 5 to get the correct electronstoichiometric coefficient. Electrons do no appear in the overallreaction. Note that you DO NOT multiply the half cell reaction 1 by5. E0 = 1.51 - 0.77 = 0.74 V delG0 = -nFE = -5 mol e- *96485 C/mol e- * 0.74 V = -356994.5 J/mol= -357 kJ/mol delG0 = -RTlnK = -nFE0 K = exp(nFE0/(RT)) = exp(5*96485*0.74/8.314/298) = 3.78

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