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

Solution :

Step 1: Identifying the Oxidation State Change

To determine the amount of electricity required, we must first determine the change in the oxidation state of Iron ($Fe$) during the transformation of $FeO$ to $Fe_{2}O_{3}$.

  • In $FeO$: Let the oxidation state of $Fe$ be $x$. Since oxygen is $-2$, $x + (-2) = 0$, therefore $x = +2$. [Per Oxidation Number Rules].
  • In $Fe_{2}O_{3}$: Let the oxidation state of $Fe$ be $y$. Since oxygen is $-2$, $2y + 3(-2) = 0 \Rightarrow 2y = 6$, therefore $y = +3$. [Per Oxidation Number Rules].

Step 2: Writing the Half-Reaction

The transformation involves the oxidation of $Fe^{2+}$ to $Fe^{3+}$. For 1 mole of $FeO$, the chemical equation is:

$FeO \rightarrow \frac{1}{2} Fe_{2}O_{3} + e^{-}$

Alternatively, looking at the oxidation of the iron atom: $Fe^{2+} \rightarrow Fe^{3+} + 1e^{-}$.

Since we are oxidizing 1 mole of $FeO$, which contains 1 mole of $Fe^{2+}$ ions, 1 mole of electrons is involved in the transfer. [By Law of Conservation of Charge].

Step 3: Calculating Charge in Coulombs

The total charge ($Q$) required is given by the product of the number of moles of electrons ($n$) and Faraday's constant ($F$).

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

Applying the Faraday's Law of Electrolysis relationship:

$Q = n \times F$

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

$Q = 96500 \text{ C}$

Visual Representation of Charge Transfer

FeO (Fe²⁺) ½Fe₂O₃ (Fe³⁺) - 1 e⁻ (Oxidation) Charge = 96500 C

Final Solution: The electricity required for the oxidation of 1 mol of FeO to Fe_{2}O_{3} is 96500 C.


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