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Q2.14(ii):
How much electricity is required in coulomb for the oxidation of 1 mol of FeO to \(Fe_{2}O_{3}\)?

Solution :

Step 1: Identifying the Chemical Reaction and Oxidation States

To determine the amount of electricity required, we must first determine the change in the oxidation state of iron ($Fe$) during the conversion of $FeO$ to $Fe_2O_3$.

  • In $FeO$: Let the oxidation state of $Fe$ be $x$. Since oxygen is $-2$, $x + (-2) = 0$, so $x = +2$.
  • In $Fe_2O_3$: Let the oxidation state of $Fe$ be $y$. Since oxygen is $-2$, $2y + 3(-2) = 0 \implies 2y = 6$, so $y = +3$.

The half-reaction for the oxidation of $Fe^{2+}$ to $Fe^{3+}$ is:

$Fe^{2+} \rightarrow Fe^{3+} + e^-$

Step 2: Determining Moles of Electrons Transferred

The problem asks for the oxidation of $1\text{ mol}$ of $FeO$. Based on the stoichiometry of the reaction:

$FeO \rightarrow \frac{1}{2}Fe_2O_3 + \frac{1}{2}O^{2-}$ [Charge balance: $Fe^{2+} \rightarrow Fe^{3+} + e^-$]

Since $1\text{ mol}$ of $FeO$ contains $1\text{ mol}$ of $Fe^{2+}$, the oxidation of $1\text{ mol}$ of $FeO$ to $Fe_2O_3$ involves the loss of exactly $1\text{ mole of electrons}$ ($n = 1$).

Step 3: Calculating Charge using Faraday's Law

Per Faraday's First Law of Electrolysis, the quantity of electricity ($Q$) in coulombs is given by the product of the number of moles of electrons ($n$) and the Faraday constant ($F$):

$Q = n \times F$

Where:

  • $n = 1\text{ mol of } e^-$
  • $F \approx 96500\text{ C mol}^{-1}$ [The charge of one mole of electrons]

Calculation:

$Q = 1\text{ mol} \times 96500\text{ C mol}^{-1} = 96500\text{ C}$

Visual Representation of Redox Process

Fe²⁺ (in FeO) - e⁻ Fe³⁺ (in Fe₂O₃) 1 mole of electrons = 96500 C

Final Solution: The amount of electricity required for the oxidation of 1 mol of FeO to Fe₂O₃ is 96,500 C.


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