Determine oxidation numbers for all elements in reactans and products and which are oxidized and which are reduced: 2N2H4(l) + N2O4(g) yields 3N2(g) + 4H2O (g)
Barring peroxides and superoxides Oxygen has a oxidation no. of -2.
Except metallic hydrides, Hydrogen has an oxidation no. of +1.
`2N_2H_4(l) + N_2O_4(g) rArr 3N_2(g) + 4H_2O (g)`
Here, Hydrogen has an oxidation no. of +1 and Oxygen -2 in all the molecules.
Let the oxidation no. of N in `N_2H_4` be `x` .
Again, let the oxidation no. of N in `N_2O_4` be `y` .
Let the oxidation no. of N in `N_2` be `z` .
`N_2H_4 rarr N_2`
-2 0 (increase in oxidation no.)
Hence, `N_2H_4` is being oxidized.
`N_2O_4 rarr N_2`
+4 0 (decrease in oxidation no.)
Hence, `N_2O_4` is being reduced.
This chemical reaction where two reactants, each containing the same element but with a different oxidation no., forms a product in which the elements involved reaches the same oxidation number in the product is known as comproportionation reaction.