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)

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llltkl | College Teacher | (Level 3) Valedictorian

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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` .

Then, `2x+4*(+1)=0`

`rArr x=-2`

Again, let the oxidation no. of N in `N_2O_4` be `y` .

So, `2y+4*(-2)=0`

`rArr y=+4`

Let the oxidation no. of N in `N_2` be `z` .

So,` 2z=0`

`rArr z=0`

`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.

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