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QUESTION

BrO3-(aq) + Sb3+(aq) = Br-(aq) + Sb5+(aq) (acidic solution) ??? Balance the above equation. Thank you

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.