Question

Question #cc55e

Answer 1

`4Mn^(2+)+5BiO_3^(+)+H_2O->4MnO_4^(-)+5Bi^(3+)+2H^(+)`

Explanation:

To balance the following redox equation:

`Mn^(2+)+BiO_3^(+)->MnO_4^(-)+Bi^(3+)`

we can split this reaction into two half equations:
Oxidation: `Mn^(2+)->MnO_4^(-)`

Reduction: `BiO_3^(+)->Bi^(3+)`

I am going to use the following rules:

1. Balance the elements other than oxygen and hydrogen
2. Balance oxygen using water `H_2O`
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: `4Mn^(2+)+5BiO_3^(+)+H_2O->4MnO_4^(-)+5Bi^(3+)+2H^(+)`

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