Q.How much electricity is required in coulomb for the oxidation of

Concept: Faraday’s laws of electrolysis — the charge required is $nF$, where $n$ is the number of moles of electrons transferred per mole of substance and $F = 96485\ \text{C mol}^{-1}$.

(i) Oxidation of $\text{H}_2\text{O}$ to $\text{O}_2$:

$2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$

So for 1 mol $\text{H}_2\text{O}$, $n = 2$ (since 4 electrons come from 2 water molecules).

Charge = $2 \times 96485 = 192970\ \text{C}$.

(ii) Oxidation of $\text{FeO}$ to $\text{Fe}_2\text{O}_3$: …

Principle

By Faraday's law, charge $= (\text{moles of } e^-)\times F$, where $F = 96500\ \text{C mol}^{-1}$. The number of electrons follows from the change in oxidation state.

(i) $H_2O \to O_2$

Oxygen goes from $-2$ (in $H_2O$) to $0$ (in $O_2$):

So 1 mol of $H_2O$ loses 2 mol of electrons:

(ii) $FeO \to Fe_2O_3$ …

Faraday’s Laws of Electrolysis — Step-by-Step Method

Method: Faraday’s First Law & Mole-Electron Stoichiometry

Core idea: The quantity of electricity (charge) required is directly proportional to the number of moles of electrons transferred in the balanced half-reaction.

Formula:

$Q = n \times F$

where

• $Q$ = charge in coulombs (C)
• $n$ = number of moles of electrons transferred per mole of substance
• $F$ = Faraday constant = 96485 C mol⁻¹ (often taken as 96500 C mol⁻¹ in exams)

(i) Oxidation of 1 mol H₂O to O₂

Step 1 — Write the balanced half-reaction (oxidation)

Water is oxidised to oxygen gas:

Step 2 — Find moles of electrons per mole of H₂O

From the equation:

• 2 moles of H₂O release 4 moles of electrons
• So, 1 mole of H₂O releases 2 moles of electrons

Step 3 — Apply Faraday’s law

(In exams, using 96500 gives 1.93 × 10⁵ C)

(ii) Oxidation of 1 mol FeO to Fe₂O₃

Step 1 — Write the balanced half-reaction

FeO contains Fe²⁺, Fe₂O₃ contains Fe³⁺.

Oxidation: Fe²⁺ → Fe³⁺ + e⁻

But we have 1 mole of FeO → 1 mole of Fe²⁺ ions.

Step 2 — Find moles of electrons per mole of FeO

Each Fe²⁺ loses 1 electron to become Fe³⁺. …

Here are the most common mistakes students make on this exact type of Faraday’s law electrolysis problem, along with how to avoid each.

Mistake 1: Forgetting to write the balanced half-reaction first

The error:

Students jump straight to “1 mol of $H_2O$ needs 2 mol $e^-$” without writing the half-reaction. This leads to wrong electron counts.

How to avoid:

Always write the balanced half-reaction in acidic or basic medium before counting electrons.

For (i):

So 1 mol $H_2O$ is involved in a reaction that produces 4 mol $e^-$ per 2 mol $H_2O$.

Therefore, per 1 mol $H_2O$, electrons required = $2$ mol $e^-$.

Key result: $Q = 2 \times 96485 = 1.93 \times 10^5 \, \text{C}$

Mistake 2: Incorrect oxidation state change for $FeO \rightarrow Fe_2O_3$

The error:

Students think $Fe$ goes from $+2$ to $+3$ (change of 1 electron per Fe atom) but forget to account for how many Fe atoms are in 1 mol of the reactant.

How to avoid:

Write the half-reaction for 1 mol of the given compound, not per atom.

For (ii):

• $FeO$: Fe is $+2$
• $Fe_2O_3$: Fe is $+3$
• Change per Fe atom: $+1$ electron lost

But 1 mol $FeO$ contains 1 mol Fe atoms.

However, the product $Fe_2O_3$ has 2 Fe atoms — so to balance, you need 2 mol $FeO$ to make 1 mol $Fe_2O_3$.

Balanced half-reaction:

So for 2 mol $FeO$, electrons = $2$ mol $e^-$

Thus for 1 mol $FeO$, electrons = $1$ mol $e^-$

Key result: $Q = 1 \times 96485 = 9.65 \times 10^4 \, \text{C}$

Mistake 3: Using $n = 1$ for $H_2O$ oxidation (wrong stoichiometry)

The error:

Students treat “1 mol of $H_2O$” as if it directly gives 4 mol $e^-$, forgetting the coefficient in the balanced equation.

How to avoid:

From the half-reaction:

• 2 mol $H_2O$ give 4 mol $e^-$
• So 1 mol $H_2O$ gives 2 mol $e^-$

Formula to remember:

Mistake 4: Confusing coulombs with faradays

The error:

Students write the answer in faradays (F) instead of coulombs (C), or forget to multiply by $F = 96485 \, \text{C/mol}$.

How to avoid:

• Faraday’s law: $Q = n \times F$
• $n$ = number of moles of electrons
• $F = 96485 \, \text{C mol}^{-1}$
• Always state the unit as coulombs (C) in the final answer.

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