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forever_confused

  • 2 years ago

Can you find what is being oxidized and reduced (including their oxidation numbers), the oxidizing agent, and reducing agent for CH4 + 2O2 CO2 + 2H2O.

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  1. Theo177
    • 2 years ago
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    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.

  2. forever_confused
    • 2 years ago
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    So nothing has oxidation numbers?

  3. forever_confused
    • 2 years ago
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    CH4 + 2O2 ---> CO2 + 2H2O sorry if me forgetting the yield sign confused you

  4. Theo177
    • 2 years ago
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    Oh wait, I though you were talking about the equation numbers. I am not sure about the axidation numbers.

  5. forever_confused
    • 2 years ago
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    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?

  6. Theo177
    • 2 years ago
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    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

  7. forever_confused
    • 2 years ago
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    its fine. and thanks a lot. Good luck! You seem really smart (: You'll do fine

  8. meera_yadav
    • 2 years ago
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    |dw:1341216983942:dw|

  9. meera_yadav
    • 2 years ago
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    hii, if u need any clarifications then let me know !!

  10. meera_yadav
    • 2 years ago
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    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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