Question

${H_2}{O_2}$ disproportionate into ${H_2}O$ and ${O_2}$. The equivalent weight of ${H_2}{O_2}$ in this reaction is:
A) $34$
B) $17$
C) $68$
D) $8.5$

Answer 1

Hint: The given reaction is a type of decomposition reaction. The equivalent weight of a molecule can be calculated as its molecular weight divided by the number of electrons transferred by that reactant. By using this data one can find an approach to find the solution.

Complete step by step answer:
1) In the above reaction hydrogen peroxide is decomposed into water and oxygen molecules. By applying the definition of equivalent weight we get the following formula,
\[{\text{Equivalent weight = }}\dfrac{{{\text{Molecular weight}}}}{{{\text{Number of electrons transferred per mole of reactant}}}}\]
Which also can be written in short as, ${\text{Equivalent weight = }}\dfrac{{MM}}{n}$

2) To find out the number of electrons transferred per mole of reactant we need to find out based on the ion-electron method by balanced equation method as follows,
The unbalanced equation for the reaction,
${H_2}{O_2} \to {H_2}O + {O_2}$
The equation for reduction reaction, $1 \times \left[ {{H_2}{O_2} + 2{H^ + } + 2{e^ - } \to 2{H_2}O} \right]$
The equation for oxidation reaction, $1 \times \left[ {{H_2}{O_2} \to {O_2} + 2{H^ + } + 2{e^ - }} \right]$
Hence, the balanced equation, $\overline {2{H_2}{O_2} \to {O_2} + 2{H_2}O} $
As we have observed in the above reaction there is a transfer of two electrons.

3) The reaction involves the 2 moles ${H_2}{O_2}$. So, to find out the number of electrons used per mole of molecule ${H_2}{O_2}$ we can use the equation for n,
$n = \dfrac{{{\text{number of electrons transferred}}}}{{{\text{number of moles of molecule}}}}$
$n = \dfrac{{2{e^ - }}}{{2{\text{ mol }}{H_2}{O_2}}} = 1$
Therefore, $1{e^ - }$ is used per $1$ mol of ${H_2}{O_2}$
4) Next, we need to find out the molecular weight for ${H_2}{O_2}$,
${\text{Molecular weight of }}{{\text{H}}_2}{{\text{O}}_2} = 2 \times 1 + 2 \times 16 = 34g$
Hence, the calculated molecular weight of the molecule ${H_2}{O_2}$ is $34g$
Now after getting the necessary values and putting it in the equation we can calculate the equivalent weight,
${\text{Equivalent weight = }}\dfrac{{MM}}{n} = \dfrac{{34g}}{1} = 34g$

Hence, option A is marked as the correct choice.

Note:
The equivalent weight of a molecule can also be defined as the amount of that molecule which exactly reacts with another random molecule’s fixed quantity. Whenever the number of electrons transferred and the number of moles of the molecule is the same then the equivalent weight will be the same as the molecular weight of the molecule.