##BrO_3^(-)+3Sb^(3+)+6H^(+)rarrBr^(-)+3Sb^(5+)+3H_2O##

We can use to help balance this equation.

Consider first the bromate(V) ion ##BrO_3^-##. It is converted to ##Br^-##

##BrO_3^(-)rarrBr^-##

##+5rarr-1##

Bromine has changed from +5 to -1. This will require the addition of 6 electrons:

##BrO_3^(-)+6erarr6Br^(-)##

To take account of 3 oxygens we will need 6H+ ions to form water:

##BrO_3^(-)+6H^(+)+6erarrBr^(-)+3H_2O## ##color(red)((1))##

This is our first half – equation.

Now for the second:

##Sb^(3+)rarrSb^(5+)##

##+3rarr+5##

Antimony has gone from +3 to +5. This means it needs to give out 2 electrons:

##Sb^(3+)rarrSb^(5+)+2e## ##color(red)((2))##

To get the full balanced equation we can add ##color(red)((1))## to ##color(red)((2))##

The problem here is that the electrons don’t balance.

If we x equation ##color(red)((2))## by 3 you can see we get 6 electrons given out and 6 taken in.

So we x ##color(red)((2))## by 3 ##rArr##

##3Sb^(3+)rarr3Sb^(5+)+6e##

Now we can add both sides of each equation together ##rArr##

##BrO_3^(-)+6H^(+)+cancel(6e)+3Sb^(3+)rarrBr^(-)+3H_2O+3Sb^(5+)+cancel(6e)##

You can see here that the 6 electrons have now cancelled from both sides to give us:

##BrO_3^(-)+3Sb^(3+)+6H^(+)rarrBr^(-)+3Sb^(5+)+3H_2O##

This is a good general method that can be applied to balancing redox equations.