Question

K2Cr2O7, H2O2 and KMnO4 are all oxidizing agents. Common products of their reduction are Cr3+, H2O and Mn2+. Identify the change in oxidation number for the involved elements.

Answer 1

Here is a good site that tells you how to determine oxidation numbers. I'll do the first one for you.
Cr in K2Cr2O7 is +6 for each Cr.
K is +1; 2 x 1 = +2.
O is -2: 7 x -2 = -14
So total Cr must be +12 to make K2Cr2O7 zero. If both Cr atoms = +12 then each must be +6.
For Cr^3+, the oxidation state is +3 for an ion.
So the change is from +6 for each Cr to +3 for each Cr or +3 for the change (6-3 = 3). It is +6 for both Cr atoms; i.e., from +12 to +6 or 12-6=6.

Answer 2

thank you! that helps a lot.

Answer 3

Here is the site.
http://www.chemteam.info/Redox/Redox-Rules.html