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Using oxidation umbers to balance the following reaction? watch

1. BrO3- + Br- + H+ -> Br2 + h2O

I'm not sure how to approach this one ? Which atoms do I assign oxidation numbers to?
2. (Original post by APersonYo)
BrO3- + Br- + H+ -> Br2 + h2O

I'm not sure how to approach this one ? Which atoms do I assign oxidation numbers to?
It's worth assigning oxidation numbers to everything!

ions are easy, Br- has oxidation number -1 and H+ has oxidation number +1

water and Br2 are also pretty straight forward (remembering a pure element has oxidation state 0)

BrO3- is a tougher one, just use the fact that oxygen is almost always 2- and that the oxidation state of all the atoms must add up to the charge on the ion!

Once you've done the oxidation state see if you can work out what is changing oxidation state and use that to balance the equation!
3. (Original post by MexicanKeith)
It's worth assigning oxidation numbers to everything!

ions are easy, Br- has oxidation number -1 and H+ has oxidation number +1

water and Br2 are also pretty straight forward (remembering a pure element has oxidation state 0)

BrO3- is a tougher one, just use the fact that oxygen is almost always 2- and that the oxidation state of all the atoms must add up to the charge on the ion!

Once you've done the oxidation state see if you can work out what is changing oxidation state and use that to balance the equation!
That doesn't work
4. (Original post by APersonYo)
That doesn't work
Which part doesn't work? have you worked out all the oxidation states?
5. (Original post by MexicanKeith)
Which part doesn't work? have you worked out all the oxidation states?
Yes, I have. How would you be able to balance all the different oxidation states to formulate an overall equation?
6. (Original post by APersonYo)
Yes, I have. How would you be able to balance all the different oxidation states to formulate an overall equation?
Okay so you should find that only the oxidation state of Bromine changes in the reaction, on the left hand side it has oxidation state +5 and -1 then on the right hand side it is oxidation state 0. (This is an example of a comproportionation reaction).

so each Br- could achieve the oxidation state 0 by giving one electron to the Br(+5)

if 5 Br- gave up one electron each then the Br(+5) would be in the zero oxidation state too.

so we know we want soemthing along the lines of

BrO3- + 5Br- ----> 3Br2

obviously that isn't balanced yet, but we have balanced the bromines using oxidation states.

now just balance up the number of oxygens

BrO3- + 5Br- ----> 3Br2 + 3H2O

And finally, work out how many H+ you need to balance the hydrogen and the charge on each side.

BrO3- + 5Br- + 6H+ ----> 3Br2 + 3H2O

hope that makes sense
7. I find it much easier to create the two half equations and balance them, but since the Q required the use of o.n., I'd just add them above the balanced equation.

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