How do you balance this redox reaction?

Question

I have uploaded a picture so it is easier to see the question:

http://farm4.static.flickr.com/3070/2817098246_b59675827f.jpg?v=0

Can you please show me how to solve this redox reaction, giving me steps on how you decided which element stayed the same, which element undergone oxidation and which element underwent reduction. Thanks!

Answer 1

i don't think that your question is accurate, by the fact that to balance re-dox equations, you need two equations.

also, in the question, there must be a e-. the e- should be on the left for the first equation and on the right for the 2nd.

so what you just need to do (if you ever encounter a real redox equation that needs to be balanced), you make the number of e- the same (times the whole equation by that number). Once the number of electrons are the same, add all the molecules on the left side together, and all the molecules on the right side together.

yes. good luck!

Answer 2

Self explained :)

Answer 3

This is indeed a redox reaction, in which the chlorine in chlorate ion is reduced to chloride ion and the sulfur in sultur dioxide is oxidized to sulfate ion.  I would suggest the half-reaction method.  I would assume that this reaction occurs in acidic solution.  Therefore we balance oxygens with water and H+.

First, determine the oxidation numbers

+5-2.....+4-2......+6-2........-1

ClO3- + SO2 --> SO4^2- + Cl-

6H+ ClO3-  +  6e-  --> Cl- + 3H2O

3(SO2  +  2H2O  -->  SO4^2-  +  4H+ + 2e-)

--------------------------------------...

6H+ + ClO3- + 3SO2 + 6H2O --> Cl- + 3SO4^2- + 3H2O + 12H+

simplify

ClO3- + 3SO2 + 3H2O --> Cl- + 3SO4^2-  + 6H+