Question

Question #be447

Answer 1

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^(-)`