With the assumption that you're meaning:
I imagine the balanced version would be:
The first step is to recognise that one of the products do not contain Oxygen, but it is present in the reactants.
Therefore, we add a number of moles of water (
In this example,
Now the problem becomes that there is no balance in hydrogen. This is corrected by adding Hydrogen ions (
We add a sufficient number of them so that it balances the number of moles of hydrogen in the water molecule.
This gives us the following:
Notice theat one mole of
The final problem we need to deal with is the charge on each side. Right now it is unbalanced, but this is a simple fix!
We simply total up the charge on each side, and the side that has a more positive charge, gets a sufficient number of electrons added to it, in order to bring it down to the same charge as the opposite side.
When totalling the charge, we do not include
we total the
Finally, as it has the more positive charge, we add the difference in charge (
Start by writing the oxidation numbers of all the atoms that take part in the reaction
Now identify the oxidation and reduction half-reactions
Now take them one at a time and try to balance the oxygens and hydrogens (if needed).
SInce you're in acidic solution, you can balance the oxygen by adding one molecule of water for every oxygen atom you lack, and a proton, or
You have three oxygens on the reactants' side, but none on the products' side
Now you have six hydrogen atoms on the products' side, so add 6 protons on the reactants' side to balance them out
The oxidation half-reaction has no oxygen and hydrogen atoms, so you're good to go.
Now it's time to balance the electrons. Oxidation and reduction reactions take place at the same time, so the number of electrons lost by one species must be equal to the number of electrons gained by another.
To balance the two half-reactions, multiply the oxidation half-reaction by 3 to get a total number of
Now add the two half-reactions together
The balanced chemical equation will be
We can use oxidation numbers to help balance this equation.
Consider first the bromate(V) ion
Bromine has changed from +5 to -1. This will require the addition of 6 electrons:
To take account of 3 oxygens we will need 6H+ ions to form water:
This is our first half - equation.
Now for the second:
Antimony has gone from +3 to +5. This means it needs to give out 2 electrons:
To get the full balanced equation we can add
The problem here is that the electrons don't balance.
If we x equation
So we x
Now we can add both sides of each equation together
You can see here that the 6 electrons have now cancelled from both sides to give us:
This is a good general method that can be applied to balancing redox equations.