Q.How much charge is required for the following reductions:

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 $F = 96485\ \text{C mol}^{-1}$.

Reasoning:

1. For each reduction, write the balanced half-reaction and find $n$.
2. Charge $Q = n \times F$.

1. $\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$ → $n = 3$
2. $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ → $n = 2$ …

Faraday’s laws of electrolysis tell us that the amount of chemical change is directly proportional to the quantity of electricity passed. The key idea is simple: each mole of electrons carries a fixed charge, called the Faraday constant ($F = 96485\ \text{C mol}^{-1}$). So to find the charge needed for a reduction, we just need to know how many moles of electrons are consumed per mole of the substance being reduced.

The number of moles of electrons comes from the change in oxidation state. For a reduction, the metal ion or oxyanion gains electrons — the difference in oxidation number tells you exactly how many electrons each formula unit takes up. Multiply that by the number of moles of the substance, and you have the total moles of electrons. Then multiply by $F$ to get the charge in coulombs.

Let’s work through each case.

1. Reduction of $Al^{3+}$ to $Al$ The half-reaction is:

Each $Al^{3+}$ ion gains 3 electrons. So for 1 mole of $Al^{3+}$, we need 3 moles of electrons.

Charge required = $3 \times F = 3 \times 96485\ \text{C} = 289455\ \text{C}$.

2. Reduction of $Cu^{2+}$ to $Cu$ Half-reaction:

Each $Cu^{2+}$ gains 2 electrons. For 1 mole of $Cu^{2+}$, we need 2 moles of electrons.

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

3. Reduction of $MnO_4^-$ to $Mn^{2+}$ This is a bit trickier because the oxidation state of Mn changes. In $MnO_4^-$, Mn is in +7 state (since O is -2, total charge -1: $x + 4(-2) = -1 \Rightarrow x = +7$). In $Mn^{2+}$, it’s +2. So each Mn atom gains 5 electrons. The balanced half-reaction in acidic medium is: …

Method: Faraday’s First Law of Electrolysis (Charge–Mole Relation)

Concept:

The amount of charge ($Q$) required to reduce a given amount of substance is directly proportional to the number of moles of electrons ($n$) involved in the half-reaction.

The fundamental relation is:

where

• $n$ = number of moles of electrons required per mole of substance reduced
• $F$ = Faraday constant = $96485\ \text{C mol}^{-1}$ (often taken as $96500\ \text{C mol}^{-1}$ in exams)

Steps

1. Write the balanced half-reaction for reduction.
2. Find the number of electrons ($n$) transferred per mole of the species.
3. Apply the formula:

(i) 1 mol of $Al^{3+}$ to $Al$

Half-reaction:

Electrons required: $n = 3$ mol e⁻ per mol Al

Charge:

Answer: $\boxed{2.89 \times 10^5\ \text{C}}$ (or $289455\ \text{C}$)

(ii) 1 mol of $Cu^{2+}$ to $Cu$

Half-reaction:

Electrons required: $n = 2$ mol e⁻ per mol Cu

Charge:

$Q = 2 \times 96485 = 192970\ \text{C}$ …

Common Mistakes & How to Avoid Them

Mistake 1: Forgetting the Number of Electrons (n-factor)

Students often use the charge on the ion directly without balancing the half-reaction.

Example of error:

• For $Al^{3+} \to Al$, some write $n = 3$ correctly, but for $MnO_4^- \to Mn^{2+}$, they incorrectly take $n = 7$ (from +7 to +2) without considering oxygen and hydrogen balance.

How to avoid:

Always write the balanced half-reaction first:

• (i) $Al^{3+} + 3e^- \to Al$ → $n = 3$
• (ii) $Cu^{2+} + 2e^- \to Cu$ → $n = 2$
• (iii) $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$ → $n = 5$

Key rule: $n$ = number of electrons gained per mole of the species.

Mistake 2: Using $n = 1$ for $MnO_4^-$ (thinking "one mole needs one Faraday")

Some students assume 1 mole of any ion requires 1 F of charge.

How to avoid:

Remember: 1 Faraday = charge of 1 mole of electrons. If the half-reaction consumes 5 electrons per $MnO_4^-$ ion, then 1 mole needs 5 F, not 1 F.

Mistake 3: Confusing $Q = nF$ with $Q = It$

Students mix up the formula for charge required in electrolysis.

Correct formula:

Where:

• $Q$ = charge in coulombs
• $n$ = number of moles of electrons (from balanced half-reaction)
• $F = 96485 \, \text{C/mol}$ (Faraday constant)

How to avoid:

• For "charge required" questions, always use $Q = nF$
• Save $Q = It$ for problems involving time and current

Mistake 4: Forgetting to Multiply by Faraday Constant

Some students stop at $n$ and write "3" as the answer for (i).

How to avoid:

Always express final answer as:

• (i) $Q = 3 \times 96485 = 2.89 \times 10^5 \, \text{C}$
• (ii) $Q = 2 \times 96485 = 1.93 \times 10^5 \, \text{C}$
• (iii) $Q = 5 \times 96485 = 4.82 \times 10^5 \, \text{C}$

Mistake 5: Incorrect Oxidation State Calculation for $MnO_4^-$ …