Which atom has a change in oxidation number of - 3 in the following redox reaction K2Cr2O7 + H2O + S - - > KOH + Cr2O3 + SO2

You have to calculate the oxidation estates of the atoms in each compound.

I will start with K2Cr2O7 because I believe that Cr is the best candidate to reduce its oxidation number in 3 units.

In K2Cr2O7:

- K has oxidation state of 1+, then K2 has a charge of 2 * (1+) = 2+.

- O has oxidation state of 2*, then O7 has a charge of 7 * (2-) = 14-.

That makes that Cr2 has charge of 14 - 2 = + 12, so each Cr has + 12/2 = + 6 oxidation state.

In Cr2O3:

- O has oxidation state of 2-, then O3 has charge 3 * (2-) = - 6

- Then, Cr2 has charge 6+, and each Cr has charge 6 + / 2 = 3+.

So, we have seen that Cr reduced its oxidation state in 3 units, from 6 + to 3+.

Answer: Cr has a change in oxidation number of - 3.