Question

What is the equivalent weight of ${O_2}$ in the following reaction?
${H_2}O + \dfrac{1}{2}{O_2} + 2{e^ - } \to 2O{H^ - }$

Answer 1

The equivalent weight of the compound or atom is calculated by dividing the molecular weight or atomic weight of the compound or atom by the valency factor or number of charge transferred.

Complete step by step answer:
Equivalent weight is also known as gram equivalent. It is defined as the mass of one equivalent which is the mass of a given substance which combines with the fixed quantity or displaces the fixed quantity of the other substance.
The equivalent weight of the compound or atom is calculated by dividing the molecular weight or atomic weight of the compound or atom by the valency factor.
The formula to calculate the equivalent weight is shown below.
$E = \dfrac{M}{V}$
Where
E is the equivalent weight
M is the molecular weight
V is the valency
The given reaction is shown below.
${H_2}O + \dfrac{1}{2}{O_2} + 2{e^ - } \to 2O{H^ - }$
In this reaction, one mole of water reacts with half a mole of oxygen by gaining two electrons to form two hydroxide ions.
Multiply the reaction with 2.
${H_2}O + \dfrac{1}{2}{O_2} + 2{e^ - } \to 2O{H^ - } \times 2$
$2{H_2}O + {O_2} + 4{e^ - } \to 4O{H^ - }$
Here, two mole of water react with one mole of oxygen by gaining four electrons from four hydroxide ions.
The molecular weight of oxygen dioxide ${O_2}$ is 32 g/mol.
The valency factor is 4.
To calculate the equivalent weight of oxygen dioxide ${O_2}$, substitute the value of molecular weight and valency in the given formula.
$\Rightarrow E = \dfrac{{32}}{4}$
$\Rightarrow E = 8$
Therefore, the equivalent of ${O_2}$ is 8.

Note:
In the given reaction, the mole of oxygen is present in a ratio form we need to change it in whole numbers. Make sure to change the moles in the whole number by multiplying the whole reaction with 2.