Question

The electric charge required to oxidise 1 mole of ${{\text{H}}_{\text{2}}}{\text{O}}$ to ${\text{O}}$ is
A. 96500 Coulomb
B. 193000 Coulomb
C. 386000 Coulomb
D. 48250 Coulomb

Answer 1

Write the reaction for the oxidation of 1 mole of ${{\text{H}}_{\text{2}}}{\text{O}}$. Determine the number of electron transfers. Using the relation between moles of electron and charge calculate the electric charge required to oxidise 1 mole of ${{\text{H}}_{\text{2}}}{\text{O}}$ to ${\text{O}}$.

Complete Step by step answer: The oxidation reaction for 1 mole of ${{\text{H}}_{\text{2}}}{\text{O}}$ is as follows:
${{\text{H}}_{\text{2}}}{\text{O}} \to {\text{ }}{{\text{H}}_{\text{2}}}{\text{ + }}\dfrac{1}{2}{{\text{O}}_{\text{2}}}$
From the reaction, we can say that 1 mole of ${{\text{H}}_{\text{2}}}{\text{O}}$ oxidised to give 1 atom${\text{O}}$ . In this reaction ${\text{O}}$ of ${{\text{H}}_{\text{2}}}{\text{O}}$
oxidised from -2 to 0.
So we can write the reaction as follows:
${{\text{O}}^{\text{2}}}^{\text{ - }} \to \dfrac{1}{2}{{\text{O}}_{\text{2}}}{\text{ + 2}}{{\text{e}}^{\text{ - }}}$
Using this reaction we can calculate the electric charge required to oxidise 1 mole of ${{\text{H}}_{\text{2}}}{\text{O}}$ to ${\text{O}}$ as follows:
${\text{Electric charge = }}nF$
Where,
$n$ = number of electron transfer
$F$ = Faraday constant = 96500 Coulomb
From the reaction we can say that there is a transfer of 2 electrons so we can calculate the electric charge as follows:
${\text{Electric charge = 2}} \times {\text{96500 Coulomb = 193000 Coulomb}}$
The electric charge required to oxidise 1 mole of ${{\text{H}}_{\text{2}}}{\text{O}}$ to ${\text{O}}$is 193000 Coulomb.

Thus, the correct option is (B).

Note: Oxidation is the loss of electrons. To calculate the electric charge required to oxidise 1 mole of ${{\text{H}}_{\text{2}}}{\text{O}}$ it is very important to determine the loss of electrons correctly. The electric charge required to oxidise any species is directly proportional to the number of electrons transferred. The greater the number of electrons transferred greater is the electric charge.