# Question cc55e

May 8, 2016

$4 M {n}^{2 +} + 5 B i {O}_{3}^{+} + {H}_{2} O \to 4 M n {O}_{4}^{-} + 5 B {i}^{3 +} + 2 {H}^{+}$

#### Explanation:

To balance the following redox equation:

$M {n}^{2 +} + B i {O}_{3}^{+} \to M n {O}_{4}^{-} + B {i}^{3 +}$

we can split this reaction into two half equations:
Oxidation: $M {n}^{2 +} \to M n {O}_{4}^{-}$

Reduction: $B i {O}_{3}^{+} \to B {i}^{3 +}$

I am going to use the following rules:

1. Balance the elements other than oxygen and hydrogen
2. Balance oxygen using water ${H}_{2} O$
3. Balance Hydrogen using ${H}^{+}$
4. Balance the charge using electrons ${e}^{-}$

Oxidation: Mn^(2+)+4H_2O->MnO_4^(-)+8H^(+)+5e^(-)color(red)(xx4

Reduction: BiO_3^(+)+6H^(+)+4e^(-)->Bi^(3+)+3H_2Ocolor(red)(xx5#

Redox: $4 M {n}^{2 +} + 5 B i {O}_{3}^{+} + {H}_{2} O \to 4 M n {O}_{4}^{-} + 5 B {i}^{3 +} + 2 {H}^{+}$

Here is a video that explains in details the balancing process:
Balancing Redox Reactions | Acidic Medium.