# Question #be447

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!

${H}_{2} O + 2 F {e}_{3} {O}_{4} \to 3 F {e}_{2} {O}_{3} + 2 {H}^{+} + 2 {e}^{-}$
$4 {H}^{+} + M n {O}_{4}^{-} + 3 {e}^{-} \to M n {O}_{2} + 2 {H}_{2} O$

$\left[2 O {H}^{-} + 2 F {e}_{3} {O}_{4} \to 3 F {e}_{2} {O}_{3} + {H}_{2} O + 2 {e}^{-}\right] \cdot 3$
$\left[2 {H}_{2} O + M n {O}_{4}^{-} + 3 {e}^{-} \to M n {O}_{2} + 4 O {H}^{-}\right] \cdot 2$

$6 O {H}^{-} + 6 F {e}_{3} {O}_{4} \to 9 F {e}_{2} {O}_{3} + 3 {H}_{2} O + 6 {e}^{-}$
$4 {H}_{2} O + 2 M n {O}_{4}^{-} + 6 {e}^{-} \to 2 M n {O}_{2} + 8 O {H}^{-}$
$- - - - - - - - - - - - - - - -$

$6 F {e}_{3} {O}_{4} + {H}_{2} O + 2 M n {O}_{4}^{-} \to 2 M n {O}_{2} + 9 F {e}_{2} {O}_{3} + 2 O {H}^{-}$