Chemistry people please helppp

which element is reduced in this equation:
2Cr(OH)3+3OCl^-+4OH- --> 2CrO4^-2 + 3Cl^- +5H2O
How do you tell?
Thanks so much for any help!

you have to go through each atom and determine the oxidation state.

then..
reduction = reduction in charge
oxidation = increase in charge..

ok?

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Cr in Cr(OH)3 is +3.. why? OH is -1.. Cr(OH)3 is neutral
O in Cr(OH)3 is -2.. why? OH is -1.. H is +1.. -2+1 = -1.. ok?
H in Cr(OH)3 is +1.. usually H is +1 and O is -2.. fyi

O in OCl- is -2.. why? O has the higher electronegativity and gets the electrons.
Cl in OCl- is +1.. why? -2 + 1xCl = -1.. Cl must = +1

O in OH- is -2
H in OH- is +1

on the product side...
Cr in CrO4-2 is +6.. why? 1xCr + 4x(-2) = -2.. Cr -8 = -2.. Cr = +6
O in CrO4-2 is -2.. why? O is usually -2.. O has the higher electronegativity

Cl in Cl- is -1.. obvious

H in H2O is +1
O in H2O is -2

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so..
Cr went from +3 to +6 and was "oxidized"
Cl went from +1 to -1 and was "reduced".. reduction means reduction in charge!!!!!

nothing else changed

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questions?

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there is no
O^0 ---> O-2 John W

When an element is reduced its valence decreases, because it gained electrons (Like oxygen.)
Cr3+ --> Cr6+ has lost 3 electrons, is oxidized.
2Cl 1- --> Cl2^0 has lost 2 electrons
O^0 --> O2- has gained two electrons.