Question

On combustion of $C{H}_4$ to $$C{O}_2$$ and $H_2$,the oxidation number of carbon changes by:
a.) 8
b.) 3
c.) 0
d.) 4

Answer 1

Hint: Combustion of any substance is in which the substance reacts completely with oxygen until it is completely burnt. Combustion is oxidation. In oxidation, there is an increase in oxidation number. Oxidation number of hydrogen in methane is +1 and the oxidation state of oxygen in carbon dioxide is -2.


Complete step by step answer:
-In combustion of any substance, the substance reacts completely with oxygen until it is completely burnt.
-Oxidation is defined as the addition of oxygen. In oxidation, there is an increase in oxidation number.
-On combustion of methane, oxygen reacts with methane to produce carbon dioxide and hydrogen gas.
$C{H}_4+O_2\to C{O}_2+H_2$

Balanced reaction can be written as
$$C{H}_4+O_2\to C{O}_2+2H_2$$

Oxidation number of carbon in methane can be calculated as
Oxidation number of hydrogen in methane is +1, so
C + 4 (1) = 0
C= -4
Oxidation number of carbon in carbon dioxide can be calculated as
Oxidation number of oxygen atom in carbon dioxide is -2, so
C + 2(-2) = 0
C – 4 =0
C= +4

So, change in oxidation number of carbon will be = +4 - (-4) = 4 + 4 = 8
On combustion of $C{H}_4$ to $C{O}_2$ and $H_2$,the oxidation number of carbon changes by 8.So, the correct answer is “Option A”.

Note: Combustion is oxidation reaction. In oxidation, there is an increase in oxidation number. Oxidation number of hydrogen in methane is +1 and the oxidation state of oxygen in carbon dioxide is -2 as oxygen acts as electron acceptor. In combustion there is release of heat as combustion is an exothermic reaction.