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.

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.

thank you! that helps a lot.

Here is the site.

http://www.chemteam.info/Redox/Redox-Rules.html

To identify the change in oxidation number for the involved elements, we need to compare the oxidation states of the elements in the reactants and products. The oxidation state, also known as the oxidation number, is a hypothetical charge assigned to an atom in a compound or ion.

Let's start by looking at each element and their respective oxidation states in the reactants and products:

1. K2Cr2O7:
- In K2Cr2O7, the oxidation state of K is +1, and the oxidation state of O is -2 (since oxygen is almost always -2 in compounds).
- The total charge of K2Cr2O7 is zero, so we can calculate the oxidation state of Cr.
- Let's assume the oxidation state of Cr is x. Therefore, 2(+1) + 2(x) + 7(-2) = 0. Solving this equation gives us x = +6.
- In the product Cr3+, the oxidation state of Cr is +3. So, the change in oxidation number for Cr is -6.

2. H2O2:
- In H2O2, the oxidation state of H is +1.
- The total charge of H2O2 is zero, so we can calculate the oxidation state of O.
- Let's assume the oxidation state of O is y. Therefore, 2(+1) + y + 2(-2) = 0. Solving this equation gives us y = -1.
- In the product H2O, the oxidation state of O is -2. So, the change in oxidation number for O is +1.

3. KMnO4:
- In KMnO4, the oxidation state of K is +1, and the oxidation state of O is -2 (as mentioned earlier).
- The total charge of KMnO4 is zero, so we can calculate the oxidation state of Mn.
- Let's assume the oxidation state of Mn is z. Therefore, (+1) + z + 4(-2) = 0. Solving this equation gives us z = +7.
- In the product Mn2+, the oxidation state of Mn is +2. So, the change in oxidation number for Mn is -5.

To summarize the changes in oxidation numbers:
- The oxidation number of Cr changes from +6 to +3 (a decrease of 3).
- The oxidation number of O changes from -1 to -2 (an increase of 1).
- The oxidation number of Mn changes from +7 to +2 (a decrease of 5).