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Mathematics 8 Online
NvidiaIntely (nvidiaintely):

In the reaction, C (s) + O2 (g) arrow CO2 (g), the reducing agent is: the C the O2 the CO2 the C and the O2

NvidiaIntely (nvidiaintely):

@agent0smith @Awolflover1 @Awsomesauce @sshayer @sweetburger @Jaynator495

NvidiaIntely (nvidiaintely):

@Vuriffy

NvidiaIntely (nvidiaintely):

what? I did not say anything.

OpenStudy (photon336):

the reducing agent causes reduction but is itself oxidized

OpenStudy (photon336):

In ENGLISH: the reducing agent gives the other reactant electrons, causing it to be reduced. as a result, it loses electrons and is oxidized.

OpenStudy (photon336):

so first step: write out the oxidation state of each of those reactants and products.

OpenStudy (photon336):

@NvidiaIntely

NvidiaIntely (nvidiaintely):

The answer is C, correct?

OpenStudy (photon336):

let's work towards the answer first

NvidiaIntely (nvidiaintely):

ok

OpenStudy (photon336):

please write out oxidation states of C and O2

NvidiaIntely (nvidiaintely):

brb

NvidiaIntely (nvidiaintely):

The oxidation state of carbon can be very different, from -4 (in CH4) to +4.

OpenStudy (photon336):

alright so watch how I do it

NvidiaIntely (nvidiaintely):

okay.

OpenStudy (photon336):

Free elements have oxidation states of zero \[C^{0}(s)~+~ O^{0}_{2} \rightarrow CO_{2}\]

NvidiaIntely (nvidiaintely):

ok....

OpenStudy (photon336):

You should keep that in mind that will save you time

NvidiaIntely (nvidiaintely):

okay

OpenStudy (photon336):

In all compounds the sum of all oxidation states is zero unless there's an ion, if there is an ion the sum of the oxidation states equals the ion.

NvidiaIntely (nvidiaintely):

ok (I keep talking to let you know I am here)

OpenStudy (photon336):

Yeah so oxygen in a compound has an oxidation state of -2 because there are two oxygen atoms in CO2 we multiply -2*2 = -4. so carbon must be +4 the sum of oxidation states = 0 so -4+4 = 0 \[CO_{2} \rightarrow Oxygen~=-2 ; \] now Carbon was 0 and is now +4 Oxygen was 0 and is now -2

NvidiaIntely (nvidiaintely):

Okay, I get it.

OpenStudy (photon336):

So based on what I said in the beginning go back and tell me what is the reducing agent. I wont just give you the answer.

NvidiaIntely (nvidiaintely):

Okay, let me see.

NvidiaIntely (nvidiaintely):

Its C?

OpenStudy (photon336):

C as in Carbon or CO2?

NvidiaIntely (nvidiaintely):

CO2.

OpenStudy (photon336):

Why?

NvidiaIntely (nvidiaintely):

Okay, maybe I don't get it, let me read it a few times more.

OpenStudy (photon336):

I'll be on for an other hour, so run through it step by step then get back to me.

NvidiaIntely (nvidiaintely):

Okay.

NvidiaIntely (nvidiaintely):

C

OpenStudy (photon336):

that was quite fast. the answer is Carbon C

NvidiaIntely (nvidiaintely):

Yeah that's what I mean't Carbon C.

NvidiaIntely (nvidiaintely):

Thanks for you help!

OpenStudy (photon336):

So carbon goes from an oxidation state of 0 to +4 \[C^{0} \rightarrow C^{+4}~+4e^{-}\]

OpenStudy (photon336):

so carbon is giving its electrons to oxygen

OpenStudy (photon336):

so carbon is reducing oxygen. that's what the whole reducing agent thing means. it's a confusing term at first.

NvidiaIntely (nvidiaintely):

yeah, I got that now.

OpenStudy (photon336):

No problem anytime

NvidiaIntely (nvidiaintely):

Okay , thanks!

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