Question #be447

1 Answer
Al E.
Nov 17, 2017

This is a beast of a problem, but possible.

By (i) balancing the half reactions in acid, (ii) adding equivalent hydroxide on both sides of the equation to neutralize acid, (iii) balancing the electrons used, and (iv) adding the equations together!

#H_2O+2Fe_3O_4 to 3Fe_2O_3+2H^(+)+2e^(-)#
#4H^(+)+MnO_4^(-) +3e^(-) to MnO_2 + 2H_2O#

#[2OH^(-)+2Fe_3O_4 to 3Fe_2O_3+H_2O+2e^(-)]*3#
#[2H_2O+MnO_4^(-) +3e^(-) to MnO_2 + 4OH^(-)]*2#

#6OH^(-)+6Fe_3O_4 to 9Fe_2O_3+3H_2O+6e^(-)#
#4H_2O+2MnO_4^(-)+6e^(-) to 2MnO_2 + 8OH^(-)#
#----------------#

#6Fe_3O_4 +H_2O+2MnO_4^(-) to 2MnO_2+9Fe_2O_3+2OH^(-)#

Related questions
See all questions in Balancing Redox Equations Using the Oxidation Number Method
Impact of this question
1492 views around the world
You can reuse this answer
Creative Commons License