Question

How do you balance the following redox reaction? `"MnO"_4^(-) + "I"^(-) -> "MnO"_2 + "IO"_3^(-)` ?

Answer 1

`2MnO_4^(-)+H_2O+I^(-)rarr2MnO_2+IO_3^(-)+2OH^(-)`

Set up the 1/2 equations:

`MnO_4^(-)rarrMnO_2`

Assign oxidation numbers to Mn

`+7rarr+4`

So this will need to take in 3 electrons. Add H+ to the left and water to the right to balance the oxygens`rArr`

`MnO_4^(-)+4H^(+)+3erarrMnO_2+2H_2O` `color(red)((1))`

and:

`I^(-)rarrIO_3^(-)`

Assign oxidation numbers to `I`:

`-1rarr+5`

So this will need to give up 6 electrons. Add water to the left and H+ to the right to balance the oxygens:

`I^(-)+3H_2OrarrIO_3^(-)+6H^(+)+6e` `color(red)((2))`

To balance the electrons we need to x `color(red)((1))` by 2 then add both sides of `color(red)((1))` and `color(red)((2))` together:

`2MnO_4^(-)+cancel(8H^(+))+cancel(6e)+I^(-)+cancel(3H_2O)rarrMnO_2+cancel(4H_2O)+IO_3^(-)+cancel(6H^(+))+cancel(6e)`

This cancels down to:

`2MnO_4^(-)+2H^(+)+I^(-)rarr2MnO_2+H_2O+IO_3^(-)` `color(red)((3))`

For basic conditions add 2OH- to both sides of `color(red)((3))` to remove the 2H+ ions on the left, making water:

`2MnO_4^(-)+cancel(2H_2O)+I^(-)rarrMnO_2+cancel(H_2O)+IO_3^(-)+2OH^(-)`

Cancelling out the water gives:

`2MnO_4^(-)+H_2O+I^(-)rarr2MnO_2+IO_3^(-)+2OH^(-)`