anonymous
  • anonymous
Can you find what is being oxidized and reduced (including their oxidation numbers), the oxidizing agent, and reducing agent for CH4 + 2O2 CO2 + 2H2O.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
The equation has the correct numbers, you dont need to change anyghing. CH4 is oxidized. O2 is the oxidised agent CO2 is the reducing agent I think.
anonymous
  • anonymous
So nothing has oxidation numbers?
anonymous
  • anonymous
CH4 + 2O2 ---> CO2 + 2H2O sorry if me forgetting the yield sign confused you

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anonymous
  • anonymous
Oh wait, I though you were talking about the equation numbers. I am not sure about the axidation numbers.
anonymous
  • anonymous
Hm, no the oxidation numbers :/ I'm not sure. Like, would I solve the oxidation numbers for each one.. Like CH4 and then 2O2 and so on?
anonymous
  • anonymous
I m not sure about the oxidation numbers either, sry... gtg, I have some math exams right now:-p. I think I m gonna fail so hard :-P
anonymous
  • anonymous
its fine. and thanks a lot. Good luck! You seem really smart (: You'll do fine
anonymous
  • anonymous
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anonymous
  • anonymous
hii, if u need any clarifications then let me know !!
anonymous
  • anonymous
here , C has oxidation no. = -4 in CH4 (since H has +1 oxidation state), while it's +4 in CO2 ( since O has -2 oxidation state) while O is in zero oxidation state in O2 (as oxidation states of elements in free state is zero) and -2 in H2O

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