Question

what is oxidation and reduction? identify the reagent. undergo oxidation and reduction in given equation.CH4 + O2 → CO2 + H2O.​

Answer 1

Hlo mate here is ur ans....

It is UNUSUAL to assign oxidation numbers to carbons in organic compounds, but nevertheless it can be done… Carbon is MORE electronegative than hydrogen, and so for methane, CH4 , we got C(−IV) , and 4×H(+I) ….for ethane we got C(−III) … and for propane we got … 2×C(−III) , C(−II) , and 8×H(+I) . As always, the WEIGHTED sum of the oxidation numbers EQUALS the ELECTRIC CHARGE of the original species. And what do I mean by this?

And so methane is OXIDIZED to CO2 , i.e. C(−IV)→C(+IV) …

CH4(g)+2H2O(l)→CO2(g)+8H++8e−8 electron oxidation

And what is reduced…? Well clearly, dioxygen, a POTENT oxidant, is…

12O2(g)+2H++2e−→H2O(l)2 electron oxidation of dioxygen gas

…and so we add the former to FOUR of the latter to retire the electrons…

CH4(g)+2H2O(l)+2O2(g)+8H++8e−→4H2O(l)+CO2(g)+8H++8e−almost balanced redox equation

…and cancel away…

CH4(g)+2O2(g)→2H2O(l)+CO2(g)balanced redox equation

And of course this is the same result if had balanced the hydrocarbon combustion in the usual way … balance the carbons as carbon dioxide, then balance the hydrogens as water, and then balance the oxygens as dioxygen gas…

Capisce?

❤️❤️Hope this helps u dear❤️❤️

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