Assertion: Electrolysis of NaCl solution gives chlorine at anode instead of O₂. Reason: Formation of oxygen at

NCERT Exemplar · Q64

NaCl

(i) Both assertion and reason are true and the reason is the correct explanation of assertion.

(ii) Both assertion and reason are true and the reason is not the correct explanation of assertion.

(iii) Assertion is true but the reason is false.

(iv) Both assertion and reason are false.

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Concept understanding — Faradays Laws Electrolysis

Faraday's Laws of Electrolysis

Imagine you're trying to plate a copper spoon with silver. You drop the spoon into a solution containing silver ions, connect it to a battery, and wait. How much silver actually deposits? Does it depend on how long you wait? On how strong the battery is? On what metal you're using?

Faraday's laws answer exactly these questions. They connect the invisible world of electrons flowing through a wire to the visible world of atoms depositing on a surface.

The Intuition First

Think of electrolysis as a counting problem. Each silver ion ( $\text{Ag}^+$ ) needs exactly one electron to become a neutral silver atom ( $\text{Ag}$ ). So if you push a certain number of electrons through the circuit, you should get exactly that many silver atoms deposited.

The first law says: more charge → more mass deposited. Double the charge, double the mass. It's a direct proportionality.

The second law says: different elements need different amounts of charge per atom. A copper ion ( $\text{Cu}^{2+}$ ) needs two electrons to become neutral copper, so for the same amount of charge, you get half as many copper atoms as silver atoms.

The Precise Statements

m \propto Q \quad \text{or} \quad m = ZQ

\frac{m_1}{E_1} = \frac{m_2}{E_2}

Here $E$ is the equivalent weight: atomic mass divided by the number of electrons transferred per ion ( $n$ ). For silver ( $\text{Ag}^+$ , $n=1$ ), $E = 107.87$ g. For copper ( $\text{Cu}^{2+}$ , $n=2$ ), $E = 63.55/2 = 31.77$ g.

The Combined Law

These two laws merge into one powerful equation:

m = \frac{Q \times E}{F}

where $F$ is Faraday's constant — the charge carried by one mole of electrons: $F = 96485$ coulombs per mole.

Since $Q = I \times t$ (current × time), you can write:

m = \frac{I \times t \times E}{F}

This is the working formula for every electrolysis calculation in your exams.

Tip

To avoid confusion: equivalent weight $E$ is always atomic mass divided by $n$ (the number of electrons gained or lost per ion). For $\text{Al}^{3+}$ , $n=3$ ; for $\text{O}_2$ gas (from water), each oxygen atom loses 2 electrons, but the molecule has 2 atoms, so $n=4$ per $\text{O}_2$ molecule.

A Worked Example

Problem: How much copper deposits when a current of 2.0 A flows through a copper sulfate solution for 30 minutes? (Atomic mass of Cu = 63.5 g/mol, $n=2$ )

Step 1: Find the equivalent weight.

E = \frac{63.5}{2} = 31.75 \text{ g/mol}

Step 2: Find total charge.

Q = I \times t = 2.0 \times (30 \times 60) = 3600 \text{ C}

Step 3: Apply the combined law.

m = \frac{Q \times E}{F} = \frac{3600 \times 31.75}{96485} = 1.185 \text{ g}

So about 1.2 grams of copper deposits. …

Why this formula?

Faraday's Laws of Electrolysis: Why the Formulas Hold

Faraday's Laws of Electrolysis describe the quantitative relationship between the amount of electricity passed through an electrolyte and the mass of substance liberated at the electrodes. Let's build the reasoning step-by-step.

1. The Core Idea: Charge Carries Matter

Electrolysis works because ions (charged particles) move toward electrodes and undergo redox reactions.

At the cathode (negative electrode), cations gain electrons (reduction).
At the anode (positive electrode), anions lose electrons (oxidation).

The key insight: Each ion that reacts carries a fixed amount of charge.

For a monovalent ion (e.g., $\text{Na}^+$ ), charge = $1.602 \times 10^{-19} \, \text{C}$ (the elementary charge $e$ ).
For a divalent ion (e.g., $\text{Cu}^{2+}$ ), charge = $2e$ .

Thus, the total charge passed ( $Q$ ) is directly proportional to the number of ions that have reacted ( $N$ ):

Q = N \cdot z e

where:

$z$ = valency (number of electrons transferred per ion)
$e$ = elementary charge ( $1.602 \times 10^{-19} \, \text{C}$ )

2. From Number of Ions to Mass

The number of ions $N$ is related to the mass ( $m$ ) of substance liberated via Avogadro's number ( $N_A$ ) and molar mass ( $M$ ):

N = \frac{m}{M} \cdot N_A

Substitute into $Q = N z e$ :

Q = \left( \frac{m}{M} \cdot N_A \right) \cdot z e

3. Introducing Faraday's Constant

The product $N_A e$ appears repeatedly — it's called Faraday's constant ( $F$ ):

F = N_A e \approx 96485 \, \text{C mol}^{-1}

So:

Q = \frac{m}{M} \cdot z F

Rearrange for mass:

\boxed{m = \frac{Q M}{z F}}

This is the unified formula for both of Faraday's laws.

4. Why Two "Laws"? — They Are the Same Idea

Faraday originally stated two laws, but they are logical consequences of the same charge–mass relationship:

First Law (Direct Proportionality)

Mass liberated is directly proportional to the charge passed.

From $m = \frac{Q M}{z F}$ , if $M$ , $z$ , and $F$ are constant, then:

m \propto Q

Why? Because each ion needs a fixed charge to react — more charge means more ions, hence more mass.

Second Law (Electrochemical Equivalent)

For the same charge, masses liberated are proportional to equivalent weights.

Equivalent weight $E = \frac{M}{z}$ (mass per mole of electrons transferred).

From $m = \frac{Q}{F} \cdot \frac{M}{z} = \frac{Q}{F} \cdot E$ , if $Q$ is fixed:

m \propto E

Why? For the same charge, the number of electrons transferred is fixed. A substance with a smaller $z$ (fewer electrons per ion) will liberate more moles of substance, hence more mass per mole.

5. Practical Formula for Exams …

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In the electrolysis of aqueous NaCl, chlorine is produced at the anode instead of oxygen because the overvoltage for oxygen evolution makes it kinetically harder, even though thermodynamically oxygen should form first. The reason given is correct and explains the assertion.

The core idea: thermodynamics vs. kinetics in electrolysis.

When you electrolyse an aqueous solution of NaCl, you have two possible oxidation reactions at the anode (the positive electrode where oxidation occurs):
- Oxidation of chloride ions: $2Cl^- \rightarrow Cl_2 + 2e^-$
- Oxidation of water: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$
To decide which happens, you first look at the standard electrode potentials (thermodynamics). The reaction with the lower (less positive) reduction potential is easier to oxidise — or equivalently, the one with the more negative oxidation potential occurs preferentially.
Compare the standard potentials.

Standard reduction potentials (at 25°C, 1 M, 1 atm):
- $Cl_2 + 2e^- \rightarrow 2Cl^-$ : $E^\circ = +1.36\ \text{V}$
- $O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$ : $E^\circ = +1.23\ \text{V}$
For oxidation, we reverse the sign. So:
- Oxidation of $Cl^-$ : $E^\circ_{\text{ox}} = -1.36\ \text{V}$
- Oxidation of $H_2O$ : $E^\circ_{\text{ox}} = -1.23\ \text{V}$
Since $-1.23\ \text{V} > -1.36\ \text{V}$ , water oxidation is thermodynamically more favourable — it requires a smaller applied voltage. So, based on standard potentials alone, oxygen should be produced at the anode, not chlorine.
Enter overvoltage — the kinetic barrier.

The reason oxygen does not appear is overvoltage (also called overpotential). This is the extra voltage needed beyond the thermodynamic value to make a reaction proceed at a noticeable rate, due to kinetic hurdles (slow electron transfer, intermediate formation, etc.).
- Oxygen evolution on common anode materials (like platinum or graphite) has a high overvoltage — often around 0.4–0.6 V.
- Chlorine evolution has a much lower overvoltage — typically negligible on these electrodes.
So the actual potential required to evolve oxygen is roughly:

E_{\text{actual}}(O_2) = 1.23\ \text{V} + \text{overvoltage} \approx 1.23 + 0.5 = 1.73\ \text{V}

While for chlorine:

E_{\text{actual}}(Cl_2) = 1.36\ \text{V} + \text{(small overvoltage)} \approx 1.36\ \text{V}

Now, chlorine requires a lower applied voltage than oxygen in practice. Hence, chlorine is produced preferentially at the anode.

Why the reason is correct and explains the assertion. …

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Here’s a breakdown of the common mistakes students make on this concept and how to avoid each.

Common Mistake #1: Confusing the reason for chlorine being produced instead of oxygen

What students do wrong:

Many students think the reason is simply that “chloride ions are easier to oxidise than water” — and then they accept the given reason (“overvoltage”) as correct without checking the actual standard electrode potentials.

Why it’s wrong:

The standard reduction potentials (at 25°C, 1 M) are:

$E^\circ_{\text{Cl}_2/\text{Cl}^-} = +1.36 \, \text{V}$
$E^\circ_{\text{O}_2/\text{H}_2\text{O}} = +1.23 \, \text{V}$

Based on these values, oxygen should be produced first (lower potential means easier oxidation). But in practice, chlorine is produced because the oxidation of water to oxygen requires a high overvoltage at the anode (especially on inert electrodes like platinum or graphite). This overvoltage makes the effective potential for oxygen evolution higher than that for chlorine, so chlorine is favoured.

How to avoid:

Always check the standard potentials first. If the expected product (oxygen) is not formed, look for kinetic factors — overvoltage is the key here. Memorise that overvoltage is the extra voltage needed to overcome the activation energy barrier for gas evolution (especially O₂ and H₂).

Common Mistake #2: Thinking the reason is false because “overvoltage” sounds unfamiliar

What students do wrong:

Students who haven’t studied overvoltage in detail often mark the reason as false, assuming it’s a made-up term.

Why it’s wrong:

Overvoltage is a real electrochemical phenomenon. For oxygen evolution, the overvoltage is significant (~0.6 V on platinum), making the effective potential for O₂ evolution around $1.23 + 0.6 = 1.83 \, \text{V}$ , which is higher than the $1.36 \, \text{V}$ for chlorine. Hence, chlorine is produced.

How to avoid:

Learn the concept of overvoltage — it’s the extra voltage required to overcome the slow kinetics of gas evolution at an electrode. It is especially high for O₂ and H₂. This is a standard topic in electrochemistry for JEE/NEET.

Common Mistake #3: Confusing the correct option (A vs B)

What students do wrong:

Some students correctly identify that both assertion and reason are true, but then choose option (ii) — “reason is not the correct explanation” — because they think overvoltage is unrelated.

Why it’s wrong:

The reason is the correct explanation. The overvoltage for oxygen is the direct reason why chlorine is produced instead of oxygen. Without overvoltage, oxygen would be the product.

How to avoid: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2026Set 2026-A1 markMCQ
$\mathrm{CaCl}_2$
›Reveal solutionSolution
The key idea is that equal quantities of electricity produce equal moles of electrons, so the mass of metal produced is proportional to its equivalent weight. Since both Mg and Ca are divalent, the mass of Ca produced is (40/24) × 4.8 g = 8.0 g.

Concept and Intuition

This problem is about Faraday’s laws of electrolysis. The same amount of charge (coulombs) will liberate the same number of moles of electrons. For a metal ion $M^{n+}$ , the number of moles of metal produced is (moles of electrons) / n. So the mass of metal = (charge / (nF)) × atomic mass. Since the charge is the same for both, the mass ratio of two metals is simply (atomic mass₁ / n₁) : (atomic mass₂ / n₂). Here both Mg and Ca form +2 ions, so n is the same (2), and the mass ratio equals the atomic mass ratio.

Step-by-step reasoning

Identify the relevant half-reactions In molten $\mathrm{MgCl}_2$ , magnesium ions are reduced:

$\mathrm{Mg^{2+} + 2e^- \rightarrow Mg}$
In molten $\mathrm{CaCl}_2$ , calcium ions are reduced:
$\mathrm{Ca^{2+} + 2e^- \rightarrow Ca}$
Both require 2 moles of electrons per mole of metal.

Find moles of Mg produced Given mass of Mg = 4.8 g, atomic mass = 24 g/mol.

$\text{Moles of Mg} = \frac{4.8}{24} = 0.20 \text{ mol}$

Find moles of electrons used Each mole of Mg requires 2 moles of electrons.

$\text{Moles of electrons} = 0.20 \times 2 = 0.40 \text{ mol}$

Same charge means same moles of electrons for Ca The same quantity of electricity provides 0.40 mol of electrons. Each mole of Ca also requires 2 moles of electrons.
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COMEDK 2026Set 2026-M1 markMCQ
$3.482 \times 10^{-1} \mathrm{~g}$
›Reveal solutionSolution
This is a Faraday’s law problem where iron is deposited from Fe³⁺ ions. The key is that each Fe³⁺ requires 3 moles of electrons to become Fe(s). Using the mass deposited, time, and atomic mass, we solve for current. The answer is 1.5 A.

Concept & Intuition

Electrolysis is about forcing a non-spontaneous redox reaction with electricity. The amount of substance deposited at an electrode is directly proportional to the charge passed. Faraday’s law says:
$m = \frac{Q \cdot M}{n \cdot F}$
where $m$ = mass deposited, $Q$ = total charge (current × time), $M$ = molar mass, $n$ = number of electrons per ion, and $F$ = Faraday constant (96485 C/mol).

Here, ferric sulphate gives Fe³⁺ ions. Each Fe³⁺ gains 3 electrons to become Fe metal:
$\text{Fe}^{3+} + 3e^- \rightarrow \text{Fe(s)}$
So $n = 3$ . The pitfall? Many students mistakenly use $n=2$ (thinking of Fe²⁺) or forget to convert time to seconds.

Step-by-step solution

Identify the half-reaction and $n$ Ferric ion is Fe³⁺. Reduction to metal:

$\text{Fe}^{3+} + 3e^- \rightarrow \text{Fe}$
Hence, $n = 3$ moles of electrons per mole of Fe.

Write Faraday’s law

$m = \frac{I \cdot t \cdot M}{n \cdot F}$
where $I$ = current (A), $t$ = time (s), $M$ = atomic mass (g/mol), $F = 96485 \, \text{C/mol}$ .

Convert given data to consistent units

Mass: $m = 3.482 \times 10^{-1} \, \text{g} = 0.3482 \, \text{g}$

Time: $t = 20 \, \text{min} = 20 \times 60 = 1200 \, \text{s}$

Atomic mass: $M = 56 \, \text{g/mol}$

$n = 3$

$F = 96485 \, \text{C/mol}$

Solve for current $I$

Rearranging:

$I = \frac{m \cdot n \cdot F}{t \cdot M}$
Substitute values:
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COMEDK 2025Set 2025-A1 markMCQ
$\mathrm{PdX} \mathrm{n}_{\mathrm{n}}$
›Reveal solutionSolution
The charge on the Pd ion is found by equating the total charge passed (from current and time) to the charge required to deposit the given mass of Pd, using Faraday’s laws. The result shows the ion carries a +4 charge, so the correct option is (A).

Concept & Intuition

Electrolysis is governed by Faraday’s laws: the mass of a substance deposited at an electrode is proportional to the total charge passed. For a metal ion $\mathrm{Pd}^{n+}$ , the reduction reaction is
$\mathrm{Pd}^{n+} + n e^- \rightarrow \mathrm{Pd}$
So each mole of Pd deposited consumes $n$ moles of electrons. By calculating the total charge passed and the moles of Pd deposited, we can solve for $n$ , the charge number (oxidation state) of the palladium ion.

Calculate total charge passed

Current $I = 1.5 \, \text{A}$ , time $t = 2 \, \text{hours} = 2 \times 3600 = 7200 \, \text{s}$ .

Charge $Q = I \times t = 1.5 \times 7200 = 10800 \, \text{C}$ .

Calculate moles of Pd deposited

Mass deposited $m = 2.977 \, \text{g}$ , atomic mass $M = 106.4 \, \text{g/mol}$ .

Moles of Pd:

$\text{moles} = \frac{2.977}{106.4} \approx 0.02798 \, \text{mol}$

Relate charge to moles of electrons Faraday’s constant $F = 96485 \, \text{C/mol e}^-$ . Total moles of electrons used:
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COMEDK 2025Set 2025-M1 markMCQ
$\mathrm{CuSO}_4$
›Reveal solutionSolution

Effective charge $=7200\ \text{C}$ deposits $0.0373\ \text{mol Cu}^{2+}$ ; remaining $\text{Cu}^{2+}$ in the unchanged 0.8 L gives molarity $\approx 0.433\ \text{M}$ — option (C).

Effective charge passed (80% current efficiency)
$Q = I\,t\,\eta = 2.5\ \text{A} \times 3600\ \text{s} \times 0.80 = 7200\ \text{C}.$
Moles of $\text{Cu}^{2+}$ deposited — $\text{Cu}^{2+} + 2e^{-} \rightarrow \text{Cu}$ :
$n_{e^-} = \frac{7200}{96500} = 0.0746\ \text{mol}, \qquad n_{\text{Cu}} = \frac{0.0746}{2} = 0.0373\ \text{mol}.$
Remaining $\text{Cu}^{2+}$ …
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COMEDK 2024Set 2024-A1 markMCQ
$\mathrm{MnO}_4{ }^{2-}$
›Reveal solutionSolution

$\mathrm{MnO_4^{2-}\to MnO_4^{-}}$ is a one-electron oxidation (Mn goes +6 → +7), so 0.1 mol requires $0.1\times F = 0.1\times96500 = 9650$ C.

Manganese changes oxidation state from +6 in $\mathrm{MnO_4^{2-}}$ to +7 in $\mathrm{MnO_4^{-}}$ , i.e. loss of 1 electron per ion: …
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COMEDK 2024Set 2024-E1 markMCQ
$\mathrm{Cr}_2 \mathrm{O} 7^{2-}$
›Reveal solutionSolution
The key is to find the total change in oxidation number per mole of dichromate, then multiply by moles and convert to Faradays. For 3.5 moles of $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ reduced to $\mathrm{Cr}^{3+}$ in acid, the charge required is 21.0 Faradays, so the correct option is (C).

Concept & Intuition

In electrochemistry, one Faraday (F) is the charge carried by one mole of electrons — about 96,485 coulombs. When we talk about "reduction" of an ion, we mean it gains electrons. The number of Faradays needed equals the total moles of electrons transferred. So the problem reduces to: How many electrons does one dichromate ion take up when it is reduced in acid? That number comes from the change in oxidation state of chromium.

Step-by-step solution

Determine the oxidation state of Cr in dichromate

In $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ , oxygen is always -2 (except in peroxides). Let the oxidation state of Cr be $x$ .

For the ion: $2x + 7(-2) = -2$

$\Rightarrow 2x - 14 = -2 \Rightarrow 2x = +12 \Rightarrow x = +6$ .

So each Cr is in the +6 state.

Identify the reduction product in acid medium

In acidic solution, dichromate is reduced to chromium(III) ions, $\mathrm{Cr}^{3+}$ .

So each Cr goes from +6 to +3, a gain of 3 electrons per Cr atom.

Electrons per dichromate ion

Since one $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ contains two Cr atoms, the total electrons gained per ion is $2 \times 3 = 6$ electrons.

Total electrons for 3.5 moles

For 3.5 moles of dichromate:

$\text{moles of electrons} = 3.5 \times 6 = 21.0$ moles of electrons. …
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COMEDK 2024Set 2024-E1 markMCQ
$_4$
›Reveal solutionSolution

Effective charge 19440 C deposits 0.1007 mol Cu; remaining $\mathrm{Cu^{2+}}=0.3375-0.1007=0.2368$ mol in 0.75 L → 0.316 M.

Effective charge passed (90% efficiency):
$Q=I\,t\,\eta=3.0\times(2\times3600)\times0.90=19440\ \mathrm{C}$
Moles of electrons:
$n_e=\frac{19440}{96500}=0.2015\ \mathrm{mol}$
Copper deposited ( $\mathrm{Cu^{2+}}+2e^-\to\mathrm{Cu}$ ):

$n_{Cu}=\frac{0.2015}{2}=0.1007\ \mathrm{mol}$ …
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COMEDK 2024Set 2024-M1 markMCQ
$\mathrm{O}_2$
›Reveal solutionSolution

Burning $44\ \text{g}$ ( $1\ \text{mol}$ ) of propane needs $5\ \text{mol}$ of $\text{O}_2$ ; producing that much $\text{O}_2$ by electrolysis takes $20\ \text{mol}$ of electrons, i.e. $t = \dfrac{20 \times 96500}{200} = 9650\ \text{s} = \mathbf{2.68\ hours}$ — option (A).

Step 1 — Combustion stoichiometry
$\text{C}_3\text{H}_8 + 5\,\text{O}_2 \rightarrow 3\,\text{CO}_2 + 4\,\text{H}_2\text{O}$
Moles of propane $= \dfrac{44}{44} = 1\ \text{mol}$ , so $\text{O}_2$ required $= 5\ \text{mol}$ .

Step 2 — Electrons needed to make $5\ \text{mol}\ \text{O}_2$

At the anode during water electrolysis:

$2\,\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$ …
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COMEDK 2023Set 2023-E1 markMCQ
$100 \mathrm{~ml}$
›Reveal solutionSolution

Current (t = 2 h = 7200 s): I = Q/t = 1723 / 7200 = 0.2393 A

Concept: Faraday's laws. At the anode in the electrolysis of water:

2 H2O -> O2 + 4 H+ + 4 e-

So 4 mol of electrons (4 F) liberate 1 mol O2.

Moles of O2 at STP:

n(O2) = 0.100 L / 22.4 L/mol = 4.464 x 10^-3 mol

Charge required:

n(e-) = 4 x 4.464 x 10^-3 = 1.786 x 10^-2 mol …
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COMEDK 2023Set 2023-M1 markMCQ
$\mathrm{pH}=4$
›Reveal solutionSolution

Reaching $\mathrm{pH}=4$ requires generating $\mathrm{H^+}$ ( $10^{-4}\,\mathrm{M}$ in $0.1\,\mathrm{L} = 10^{-5}\,\mathrm{mol}$ ). Since each $\mathrm{H^+}$ needs one electron ( $Q = nF$ ), $t = nF/I = (10^{-5}\times96500)/0.5 \approx 1.93$ , matching option (A).

At the anode, electrolysis of the aqueous solution generates $\mathrm{H^+}$ :
$2\mathrm{H_2O} \rightarrow \mathrm{O_2} + 4\mathrm{H^+} + 4e^-,\quad \text{so moles of } \mathrm{H^+} = \text{moles of } e^-.$
Target: $\mathrm{pH}=4 \Rightarrow [\mathrm{H^+}] = 10^{-4}\,\mathrm{M}$ in $100\,\mathrm{mL} = 0.1\,\mathrm{L}$ :
$n_{\mathrm{H^+}} = 10^{-4} \times 0.1 = 10^{-5}\,\mathrm{mol}.$
Charge required (Faraday's law): …
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KCET 2022Set B-31 markMCQ
$Q. In Fuel cells ______ are used as catalysts. (A) Zinc - Mercury (B) Lead - Manganese (C) Platinum - Palladium (D) Nickel - Cadmium$
›Reveal solutionSolution

The NCERT hydrogen–oxygen fuel cell uses finely divided platinum or palladium metal, incorporated into the porous carbon electrodes, as the catalyst.

Step 1 — What a fuel cell is.

A fuel cell is a galvanic cell that converts the energy of combustion of a fuel (here $H_2$ ) directly into electrical energy, instead of burning it to run a turbine. The classic example is the $H_2$ – $O_2$ fuel cell used in the Apollo space programme.

Step 2 — The construction.

$H_2$ and $O_2$ are bubbled through porous carbon electrodes into a concentrated aqueous $NaOH$ (or $KOH$ ) electrolyte. To make the electrode reactions fast enough at ordinary temperature, catalysts such as finely divided platinum or palladium metal are incorporated into the electrodes.

Step 3 — The electrode reactions being catalysed.

Anode (oxidation of the fuel):
$2H_2(g) + 4OH^-(aq) \longrightarrow 4H_2O(l) + 4e^-$
Cathode (reduction of oxygen):
$O_2(g) + 2H_2O(l) + 4e^- \longrightarrow 4OH^-(aq)$
Overall:

$2H_2(g) + O_2(g) \longrightarrow 2H_2O(l)$ …
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Q.Assertion: Electrolysis of $NaCl$ solution gives chlorine at anode instead of $O_2$ .
Reason: Formation of oxygen at anode requires overvoltage.

Concept understanding — Faradays Laws Electrolysis

Faraday's Laws of Electrolysis

The Intuition First

The Precise Statements

The Combined Law

A Worked Example

Why this formula?

Faraday's Laws of Electrolysis: Why the Formulas Hold

1. The Core Idea: Charge Carries Matter

2. From Number of Ions to Mass

3. Introducing Faraday's Constant

4. Why Two "Laws"? — They Are the Same Idea

First Law (Direct Proportionality)

Second Law (Electrochemical Equivalent)

5. Practical Formula for Exams …

Method: Overvoltage Analysis in Electrolysis

Steps to solve:

Common Mistake #1: Confusing the reason for chlorine being produced instead of oxygen

Common Mistake #2: Thinking the reason is false because “overvoltage” sounds unfamiliar

Common Mistake #3: Confusing the correct option (A vs B)

Q.Assertion: Electrolysis of NaClNaClNaCl solution gives chlorine at anode instead of O2O_2O2​. Reason: Formation of oxygen at anode requires overvoltage.

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Q.Assertion: Electrolysis of $NaCl$ solution gives chlorine at anode instead of $O_2$ .
Reason: Formation of oxygen at anode requires overvoltage.

Unlock everything free for 14 days