Given the reaction:
C (s) + O2 (g) --> CO2 (g)
Why is the oxidation state of C in CO2 4+ and why is O2 the species that is reduced, when their oxidation states as reactants and products are essentially the same?
Test 2 #1
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Re: Test 2 #1
O has a charge of -2, and since there is 2 molecules of O the total charge of the O2 species is -4. The molecule CO2 is neutral, so that would mean that C has to have a charge of +4 to balance the charge of the O.
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Re: Test 2 #1
I think O2 is reduced because the charge on the oxygens in O2 is both 0, because that is a neutral molecule. Therefore each gains two electrons (reduction) because the oxygens in CO2 are both -2.
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Re: Test 2 #1
O2 has a charge of 0. For CO2, the O has a charge of -2, and since there are 2 "O"s in CO2 the charge for O is -4. In order to cancel this -4 out, the charge on C must be +4. O2 is reduced since its charge goes from 0 to -4, and thus gained electrons.
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Re: Test 2 #1
Jeremiah Samaniego 2C wrote:Given the reaction:
C (s) + O2 (g) --> CO2 (g)
Why is the oxidation state of C in CO2 4+ and why is O2 the species that is reduced, when their oxidation states as reactants and products are essentially the same?
C has an oxidation state of 0 and is oxidized to an oxidation state of 4+ because O has an oxdidation state of 2- (2x2 = 4- and therefore needs 4+ on carbon to create neutral CO2). O2 initially has oxidation state 0 because it's a neutral diatomic gas and is reduced to oxidation state of 2- in CO2.
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Re: Test 2 #1
O2 by itself has a charge of 0 but in a compound/molecule you consider how O's charge is 2- so it goes from 0 to 4- charge.
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Re: Test 2 #1
The C in O2 is 4+ because the O is 2-, but the overall charge is 0, so 2 x -2 would mean C needs to be +4. And O2 is reduced because O in O2 has 0 charge but in CO2 it has a 2- charge, meaning it was reduced.
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