Q.Consider the reaction:
$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
What is the quantity of electricity in coulombs needed to reduce 1 mol of $Cr_2O_7^{2-}$?

The key idea is Faraday’s laws of electrolysis: the charge required to reduce one mole of an ion equals the number of moles of electrons transferred multiplied by Faraday’s constant ($F = 96485\ \text{C mol}^{-1}$).

From the balanced half-reaction:

each mole of $Cr_2O_7^{2-}$ consumes 6 moles of electrons. …

This is a straightforward application of Faraday’s laws of electrolysis. The core idea: the quantity of electricity (charge) needed to reduce a substance is directly proportional to the number of moles of electrons transferred in the balanced half-reaction. One mole of electrons carries a fixed charge — Faraday’s constant, $F = 96485$ C/mol.

The given half-reaction is already balanced:

Notice the stoichiometric coefficient of electrons: 6. That means for every 1 mole of dichromate ion reduced, 6 moles of electrons are consumed.

Now, the charge carried by 1 mole of electrons is $F = 96485$ C. So for 6 moles:

$\text{Charge} = 6 \times 96485 \text{ C} = 578910 \text{ C}$ …

Method: Faraday’s First Law of Electrolysis

Concept: The amount of chemical change is directly proportional to the quantity of electricity passed.

For a given reaction, the charge needed is:

where

• $n$ = number of moles of electrons required per mole of substance
• $F$ = Faraday constant = 96500 C/mol (exact value for exams)

Steps

1. Identify the electron requirement From the balanced half-reaction:

Each mole of $\text{Cr}_2\text{O}_7^{2-}$ consumes 6 moles of electrons. …

Here are the most common mistakes students make with this exact type of Faraday’s Law problem, and how to avoid each one.

Mistake 1: Confusing “moles of electrons” with “moles of substance”

The error:

Students see “1 mol of $Cr_2O_7^{2-}$” and immediately plug $n = 1$ into $Q = nF$, forgetting that $n$ in Faraday’s law is moles of electrons, not moles of the ion.

How to avoid:

Always read the half-reaction carefully. The balanced equation shows:

For 1 mol of $Cr_2O_7^{2-}$, the equation consumes 6 mol of electrons.

So $n_{\text{electrons}} = 6$.

Mistake 2: Using the wrong value of Faraday’s constant

The error:

Using $F = 96500$ C when the problem expects $96500$ C/mol, or mixing up $F$ with the gas constant $R$.

How to avoid:

Memorise:

This is the charge of 1 mole of electrons. Always write the unit to check:

$Q = n \times F$ gives coulombs when $n$ is in mol and $F$ in C/mol.

Mistake 3: Forgetting to multiply by the stoichiometric coefficient

The error:

If the reaction were written as $2Cr_2O_7^{2-} + \dots$, some students still use $n = 6$ for 2 moles of dichromate.

How to avoid:

Always scale the electron count to 1 mole of the substance asked.

Here the question says “1 mol of $Cr_2O_7^{2-}$” — the given equation is already for 1 mol, so $n = 6$ directly.

Mistake 4: Confusing “quantity of electricity” with “current” or “time”

The error:

Students try to use $Q = I \times t$ when no current or time is given.

How to avoid: …