There is no point in me redoing those parts you understand (and I assume you know how to do parts of it). What is it you're having trouble with?
c, d, e and f are where I am struggling.
12OH^- + Cl2 ==> 2ClO3^- + 10e + 6H2O
(Note: This is twice what you will need when you balance the entire equation so you just divide all of the coefficients by 2. That will leave just Cl (and not Cl2 but when you add it to the other half cell--likewise divided by 2--the Cl from the oxidized half cell and the Cl from the reduced half cell you get the Cl2 you're familiar with.)
Here is a link that shows how to do this as well as most any other item regarding redox equations.
e. Reduction is the gain of electrons.Which half cell gained electrons?
f. The oxidizing agent is the substance being reduced.
d. What type reaction? It's a redox equation; however, when the same substance is both oxidized and reduced it is still a redox reaction but it is given a special name of disproportionation.
By the way, when I have an equation like this, I just separate them into to the two half cells, balanced them with Cl2 ==> 2Cl^- and
Cl2 ==> ClO3^-, then add the two half cells and divide all of the coefficients by 2.
Thank you, thank you thank you
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