Question

How much electricity is required in coulomb for the oxidation of:
(i) 1 mol of ${{H}_{2}}O$ to ${{O}_{2}}$?
(ii) 1 mol of $FeO$to $F{{e}_{2}}{{O}_{3}}$?

Answer 1

Hint: When electricity is passed through an aqueous solution, the ions present in the solution get oxidized or reduced depending on their sign of charge. The quantitative relationship between the reactants and products can be done by applying Faraday's first law of electrolysis.

Complete step by step answer:
Faraday’s First Law of Electrolysis states that the chemical deposition of a substance due to the flow of current through an electrolyte is directly proportional to the quantity of electricity passed through it.
$\begin{align}
  & {{\text{m}}_{{}}}{{\alpha }_{{}}}Q \\
 & {{m}_{{}}}{{=}_{{}}}Z.Q \\
\end{align}$
Where,
m is the mass of electrolyte deposited,
Q is the quantity of electricity
Z is the constant of proportionality and is known as electrochemical equivalent.
Let us now write the chemical reaction for (i) 1 mol of ${{H}_{2}}O$ to ${{O}_{2}}$:
${{\text{H}}_{\text{2}}}\text{O }\to \text{ 2}{{\text{H}}^{+}}\text{ + }\dfrac{1}{2}{{\text{O}}_{2}}\text{ + 2}{{\text{e}}^{-}}$
In the above reaction we see that 2 moles of electrons are released when 1 mole of ${{H}_{2}}O$ decomposes to give ${{O}_{2}}$. The amount of charge required is 2 F. The value of F is 96500 C.

So, the amount of charge required is 193000 C.

The chemical reaction for (ii) 1 mol of $FeO$to $F{{e}_{2}}{{O}_{3}}$ is given below:
$\text{F}{{\text{e}}^{2+}}\text{ }\to \text{ F}{{\text{e}}^{3+}}\text{ + }{{\text{e}}^{-}}$
In the above reaction we observe that 1 mole of electron is released for one mole of reactant. The charge required is 1 F.

So, the amount of charge required is 96500 C.

Note: Faraday had given two laws on electrolysis. The first law is discussed above and the second law is given below:
Faraday’s second law states that the amount of substances deposited due to passage of the same amount of electric current will be proportional to their respective equivalent weights.
Equivalent weight is the molar mass of the substance divided by the n-factor of the substance.