Question

What is the new oxidation number for the atom that is oxidized in the following redox reaction?

Cr2O72- + 8H+ + 3SO32-  2Cr3+ + 3SO42- + 4H2O

Answer 1

Step by step please?

Answer 2

I'm assuming you're referring to Cr? lol

Answer 3

o.o I honestly don't even know how to find the oxidation number let alone the new one... sorries I'm confused... lol

Answer 4

Cr2O7{2-} + 8 H{+} + 3 SO3{2-} → 2 Cr{3+} + 3 SO4{2-} + 4 H2O
First you need to identify "the atom that is oxidized". Cr2O7{2-} is famously an oxidizing agent and H has the same oxidation number on both sides, so it must be the SO3{2-} ion that is oxidized. O doesn't usually change oxidation numbers, but S often does, so S is the oxidized atom. The old oxidation number is -2 - (-6) = +4. The new oxidation number for S (in SO4{2-}) is -2 - (-8) = +6
(The fact that S went from +4 to +6 confirms that it is the oxidized atom.)