For example, in $\ce{H2 + Cl2 -> 2 HCl}$, how do we find the oxidation numbers of $\ce{H2}$, $\ce{Cl2}$ and of $\ce{H}$ and $\ce{Cl}$ in $\ce{HCl}$?
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1$\begingroup$ You need to look at the electronegativity $\endgroup$– Ggjj11Commented Sep 4 at 6:59
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$\begingroup$ how do i do that? $\endgroup$– Aditya SinhaCommented Sep 4 at 7:12
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3$\begingroup$ Be careful to use proper capitalization of element symbols. $\endgroup$– PoutnikCommented Sep 4 at 7:52
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2 Answers
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You have free reactants $\ce{H2 and Cl2}$ so their Oxidation state is 0.
Hydrogen has a positive oxidation state when bonded to nonmetals, and the sum of the oxidation states in a compound without charge always adds up to 0.
Now, we can consider your product $\ce{HCl}$
So H has a oxidation state of +1 whereas Cl has an oxidation state of -1. (based on the no of electrons that are involved in bond formation)
(-1) + 1 = 0. (Showing that sum of oxidation states = 0)
An increase in oxidation number is oxidation and a decrease is reduction i.e. the increase or decrease in comparison to the original oxidation states in the reactant side
so in this case, H undergoes oxidation and Cl undergoes reduction.
However, your reaction is pretty simple so you can simply say that H is oxidised due to addition of an electronegative element (Cl) and Cl is reduced due to addition of Hydrogen.
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Oxidation means addition of oxygen and reduction of hydrogen to a compound. Whereas Reduction means addition of Hydrogen and removal of oxygen. Redox reactions are basically both oxidation and reduction happening simultaneously.
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2$\begingroup$ The method you suggest is essentially useless for many kinds of reactions. $\endgroup$– PAEPCommented Sep 7 at 10:57