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Topic: redox  (Read 3974 times)

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Offline Moss

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redox
« on: October 23, 2009, 10:51:47 PM »
hey
is this oxidation or reduction?

Cr2O72- -> Cr3+
Cr=+6 O=-14 Cr=+3
oxidation number for Cr decreases, so reduction?
H2S -> S
H2S=0,  S=-2
is this reduction? since the oxidation number goes down?


help, thanks

Offline BetaAmyloid

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Re: redox
« Reply #1 on: October 23, 2009, 11:10:12 PM »
Remember this, "OIL RIG - Oxidation Involves Losing, Reduction Involves Gaining."

In both of these equations, the product loses in its oxidation state.

So, OIL RIG, Oxidation involves losing.

They are both oxidation - you just had it backwards for the oxidation states.

Hope this helps,
The Cancer Curer
Discovery consists of seeing what everybody has seen and thinking what nobody has thought - Albert Szent-Györgyi

Offline Moss

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Re: redox
« Reply #2 on: October 23, 2009, 11:34:24 PM »
the question im trying to do is
balance the fllowing useing the half equation method
Cr2O72- + H2S-> Cr3+ + S

firs step im trying to do is determining whcih one is undergoin reduction and oxidation

can both the half equation be undergoing oxidation?

Offline BetaAmyloid

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Re: redox
« Reply #3 on: October 23, 2009, 11:43:30 PM »
Well, I was judging the answer based upon your oxidation numbers - but I believe your oxidation numbers are incorrect. I'm pretty definite that one has to oxidize and one has to reduce, therefore I would conclude that your oxidation numbers are wrong.

Try looking up your oxidation numbers again maybe?

Sorry!
The Cancer Curer
Discovery consists of seeing what everybody has seen and thinking what nobody has thought - Albert Szent-Györgyi

Offline UG

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Re: redox
« Reply #4 on: October 23, 2009, 11:55:42 PM »
H2S -> S
H2S=0,  S=-2
is this reduction? since the oxidation number goes down?
No, the first one is reduction, this is oxidation. Your oxidation numbers are wrong, S in H2S is not 0, it can't be. The oxidation number of any atom(s) which exists as an element is 0.

Online Borek

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Re: redox
« Reply #5 on: October 24, 2009, 05:30:47 AM »
They are both oxidation - you just had it backwards for the oxidation states.

Please stop spreading nonsense. You are not helping.
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