For the following question, two statements are given — one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) given below. Assertion (A) : Electrolysis of aqueous NaCl gives $H_2$ at cathode and $Cl_2$ at anode. Reason (R) : Chlorine has higher oxidation potential than $H_2O$. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.

For the following question, two statements are given — — CBSE Class 12 Chemistry 2025 PYQ

Question

H_2

CBSECBSE Class XII Board 2025MCQ· 1mImportance★★★★★

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Concept understanding — Products of Electrolysis

Products of Electrolysis – From Intuition to Precision

Imagine you have a salt solution with two positive ions (cations) and two negative ions (anions) swimming around. When you switch on the current, both cations want to go to the negative electrode (cathode) and both anions want to go to the positive electrode (anode). But only one of each can actually react — the one that "wins" the competition. That winner is the product of electrolysis at that electrode.

The competition is decided by three factors: electrode potential (how badly a species wants to gain or lose electrons), concentration (more ions = more chances to react), and overpotential (an extra barrier some reactions face, especially gases like hydrogen and oxygen).

The Precise Rule

At the cathode (reduction), the species with the higher reduction potential (more positive $E^\circ$ ) gets reduced first — unless its concentration is very low or it has a large overpotential.

At the anode (oxidation), the species with the lower reduction potential (more negative $E^\circ$ ) gets oxidised first — again, concentration and overpotential can override this.

\text{Cathode: reduce the species with the highest } E^\circ_{\text{red}}

But real life is messier. Water itself can be reduced or oxidised, and its products (H $_2$ and O $_2$ ) often appear instead of the metal or halogen you might expect.

The Three Deciding Factors

1. Electrode potential — the standard reduction potential tells you the thermodynamic preference. For example, Cu $^{2+}$ ( $E^\circ = +0.34$ V) is reduced before Zn $^{2+}$ ( $E^\circ = -0.76$ V) at the cathode.

2. Concentration — the Nernst equation shifts the effective potential. A very dilute solution of Cu $^{2+}$ may have its reduction potential drop below that of H $_2$ O, so hydrogen gas bubbles out instead of copper depositing.

3. Overpotential — some reactions need extra voltage to start. Hydrogen evolution on many metal surfaces (like Pt) has low overpotential, but on mercury it has a very high overpotential. That's why Na metal can be produced by electrolysing aqueous NaCl using a mercury cathode — H $_2$ is thermodynamically favoured but kinetically blocked.

Watch out

Overpotential is not a thermodynamic quantity — it's a kinetic barrier. It can reverse the order predicted by standard potentials alone.

A Concrete Example: Aqueous NaCl

Species	Cathode (reduction)	$E^\circ$ (V)
Na $^+$ + e $^-$ → Na		–2.71
2H $_2$ O + 2e $^-$ → H $_2$ + 2OH $^-$		–0.83

At the cathode, water reduction has a much higher potential (–0.83 V) than Na $^+$ reduction (–2.71 V). So H $_2$ gas is produced — not sodium metal.

Species	Anode (oxidation)	$E^\circ$ (V)
2Cl $^-$ → Cl $_2$ + 2e $^-$		+1.36
2H $_2$ O → O $_2$ + 4H $^+$ + 4e $^-$		+1.23

Thermodynamically, water oxidation (lower $E^\circ$ ) should win. But Cl $_2$ evolution has a much lower overpotential on common anodes (like graphite or Pt) than O $_2$ evolution. So in concentrated NaCl, chlorine gas is the main product at the anode. …

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The assertion is true: electrolysis of aqueous NaCl gives H₂ at the cathode and Cl₂ at the anode. The reason given is false — chlorine does not have a higher oxidation potential than water; in fact, water is more easily oxidised. So the correct choice is (C).

This question tests your understanding of electrolysis of aqueous solutions — specifically, how the presence of water changes what gets discharged at the electrodes compared to molten salts. The key is to compare standard electrode potentials (or oxidation potentials) of the competing species.

Let’s break it down.

What happens at the cathode (reduction)? In aqueous NaCl, we have Na⁺ ions and H₂O molecules. Two possible reduction reactions compete:

\text{Na}^+ + e^- \rightarrow \text{Na} \quad E^\circ = -2.71\ \text{V}

2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^- \quad E^\circ = -0.83\ \text{V}

The less negative (or more positive) reduction potential is favoured. Since $-0.83\ \text{V}$ is much higher than $-2.71\ \text{V}$ , water is reduced preferentially, giving H₂ gas at the cathode. So the assertion is correct on this side.

What happens at the anode (oxidation)? Here, Cl⁻ ions and H₂O molecules compete to be oxidised. We compare oxidation potentials (the reverse of reduction potentials). The relevant half-reactions written as oxidations:

2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^- \quad E^\circ_{\text{ox}} = -1.36\ \text{V}

2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^- \quad E^\circ_{\text{ox}} = -1.23\ \text{V}

A higher (less negative) oxidation potential means the species is more easily oxidised. Here, water has an oxidation potential of $-1.23\ \text{V}$ , which is higher than that of Cl⁻ ( $-1.36\ \text{V}$ ). So water should be oxidised more easily, giving O₂ at the anode — not Cl₂.

Watch out

A common mistake is to think that because Cl⁻ is a halide, it always gets discharged first. But in dilute aqueous solutions, water’s oxidation potential is actually higher. However, in concentrated NaCl (brine), the situation changes due to overpotential — oxygen evolution has a high overpotential on inert electrodes like platinum or graphite, so Cl₂ is produced instead. The assertion in the question refers to the typical industrial electrolysis of brine, where Cl₂ is indeed obtained at the anode.

So why is the assertion true? …

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

CBSE 2026Set ANNUAL1 markMCQ
$Q. In the electrolysis of molten NaCl, the substance liberated at the cathode is :$
(a) Cl2
(b) Na
(c) H2
(d) O2
›Reveal solutionSolution
The cathode is the site of reduction; Na+ + e- -> Na, so sodium metal is liberated there. Answer: (b) Na.

In molten NaCl only Na+ and Cl- ions are present (no water).

At the cathode (reduction): Na+ + e- -> Na (sodium metal).

At the anode (oxidation): 2Cl- -> Cl2 + 2e- (chlorine gas). …
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CBSE 2025Set 56/4/11 markMCQ
$H_2$
›Reveal solutionSolution
The assertion is true: electrolysis of aqueous NaCl gives H₂ at the cathode and Cl₂ at the anode. The reason given is false — chlorine does not have a higher oxidation potential than water; in fact, water is more easily oxidised. So the correct choice is (C).

This question tests your understanding of electrolysis of aqueous solutions — specifically, how the presence of water changes what gets discharged at the electrodes compared to molten salts. The key is to compare standard electrode potentials (or oxidation potentials) of the competing species.

Let’s break it down.

What happens at the cathode (reduction)? In aqueous NaCl, we have Na⁺ ions and H₂O molecules. Two possible reduction reactions compete:

$\text{Na}^+ + e^- \rightarrow \text{Na} \quad E^\circ = -2.71\ \text{V}$ $2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^- \quad E^\circ = -0.83\ \text{V}$
The less negative (or more positive) reduction potential is favoured. Since $-0.83\ \text{V}$ is much higher than $-2.71\ \text{V}$ , water is reduced preferentially, giving H₂ gas at the cathode. So the assertion is correct on this side.

What happens at the anode (oxidation)? Here, Cl⁻ ions and H₂O molecules compete to be oxidised. We compare oxidation potentials (the reverse of reduction potentials). The relevant half-reactions written as oxidations:

$2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^- \quad E^\circ_{\text{ox}} = -1.36\ \text{V}$ $2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^- \quad E^\circ_{\text{ox}} = -1.23\ \text{V}$
A higher (less negative) oxidation potential means the species is more easily oxidised. Here, water has an oxidation potential of $-1.23\ \text{V}$ , which is higher than that of Cl⁻ ( $-1.36\ \text{V}$ ). So water should be oxidised more easily, giving O₂ at the anode — not Cl₂.

Watch out

A common mistake is to think that because Cl⁻ is a halide, it always gets discharged first. But in dilute aqueous solutions, water’s oxidation potential is actually higher. However, in concentrated NaCl (brine), the situation changes due to overpotential — oxygen evolution has a high overpotential on inert electrodes like platinum or graphite, so Cl₂ is produced instead. The assertion in the question refers to the typical industrial electrolysis of brine, where Cl₂ is indeed obtained at the anode.

So why is the assertion true? …
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CBSE 2025Set D1 markMCQ
$Q. In electrolysis, oxidation takes place at$
(a) Anode
(b) Cathode
(c) Both anode and cathode
(d) Depends upon electrolyte used
›Reveal solutionSolution
Oxidation occurs at the anode and reduction at the cathode, in both electrolytic and galvanic cells.

Definition of electrodes is by the process occurring, not by sign:

Anode: the electrode where oxidation (loss of electrons) takes place.

Cathode: the electrode where reduction (gain of electrons) takes place. …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. The products of the electrolysis of molten sodium chloride are -$
(a) Na(s) and H2(g)
(b) NaOH and H2SO4
(c) H2(g) and Cl2(g)
(d) Na(s) and Cl2(g)
›Reveal solutionSolution

Electrolysis of MOLTEN NaCl (no water present) gives sodium metal at the cathode and chlorine gas at the anode - this is the Down's process used industrially to manufacture sodium metal.

In molten NaCl, the only ions present are Na+ and Cl- (there is no water to supply H+/OH-, unlike electrolysis of aqueous NaCl brine).

At the cathode (reduction):

Na+ + e- -> Na(s)

At the anode (oxidation):

2Cl- -> Cl2(g) + 2e-

Overall: 2NaCl(l) --electrolysis--> 2Na(s) + Cl2(g)

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. During electrolysis of aqueous solution of NaCl, the products formed are:$
(a) Na, Cl2
(b) NaOH, Cl2, H2
(c) Na, Cl2, NaOH
(d) H2, Cl2
›Reveal solutionSolution

Electrolysis of aqueous NaCl (brine) gives H2 at the cathode, Cl2 at the anode, and leaves NaOH behind in solution — the industrial chlor-alkali process.

At the cathode, water is preferentially reduced over Na+ (Na+ would need a far more negative potential): 2H2O + 2e- → H2 + 2OH-.

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. An electrochemical cell can behave like an electrolytic cell, when :$
(a) E_cell = 0
(b) E_cell > E_ext
(c) E_ext > E_cell
(d) E_cell = E_ext
›Reveal solutionSolution
If the opposing external voltage exceeds the cell's own emf, the reaction is driven backward — this is electrolysis, i.e. electrolytic-cell behaviour.

In a galvanic (electrochemical) cell, the spontaneous redox reaction generates a potential $E_{cell}$ . If an external opposing potential $E_{ext}$ is applied:

When $E_{ext} < E_{cell}$ : the cell drives current in its normal direction (galvanic behaviour).

When $E_{ext} = E_{cell}$ : no current flows (equilibrium, used to measure emf via a potentiometer). …
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CBSE 2023Set F1 markMCQ
$Q. Which of the following is deposited at cathode on electrolysis of aqueous NaCl solution?$
(a) Chlorine
(b) Sodium
(c) Sodium amalgam
(d) Hydrogen
›Reveal solutionSolution

Water (H+) is reduced more easily than Na+, so hydrogen gas is evolved at the cathode.

In aqueous NaCl the possible cathode reductions are Na+ + e- → Na (E° = -2.71 V) and 2H2O + 2e- → H2 + 2OH- (E° ≈ -0.83 V). The species with the higher (less negative) reduction potential is discharged, so water is reduced in preference to Na+:

2H2O + 2e- → H2(g) + 2OH-

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Which of the following type of reactions occur at cathode during electrolysis?$
(a) Reduction
(b) Oxidation
(c) Association
(d) Dissociation
›Reveal solutionSolution

During electrolysis, reduction always occurs at the cathode and oxidation always occurs at the anode.

In an electrolytic cell, the cathode is connected to the negative terminal of the external battery, so it is rich in electrons. Positively charged ions (cations) in the electrolyte migrate towards the cathode and gain electrons there, i.e. they are reduced (e.g. Cu2+ + 2e- -> Cu). Meanwhile at the anode, negatively charged ions (anions) lose electrons, i.e. oxidation occurs.

…

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CBSE 2022Set ANNUAL1 mark
$MgSO_4$
›Reveal solutionSolution

Water, not Mg²⁺, gets reduced at the cathode during electrolysis of aqueous MgSO₄, so the product is H₂ gas.

During electrolysis of an aqueous solution, the species that is reduced at the cathode is the one that is easiest to reduce, i.e. has the higher (less negative) standard reduction potential, among all species present (here Mg²⁺ and H₂O).

$Mg^{2+} + 2e^- \rightarrow Mg(s), \quad E° = -2.37\ V$

$2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq), \quad E° = -0.83\ V$

…

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CBSE 2018Set ANNUAL1 markMCQ
$Q. A dilute aqueous Solution of sodium fluoride is electrolysed; the products at the anode and cathode are -$
(a) F2, Na
(b) F2, H2
(c) O2, Na
(d) O2, H2
›Reveal solutionSolution
Anode → O₂, cathode → H₂ (water is discharged in preference to F⁻ and Na⁺).

Electrolysis of dilute aqueous NaF (Na⁺, F⁻, H₂O):

Anode (oxidation): F⁻ → ½F₂ needs a huge potential (F₂/F⁻ = +2.87 V), whereas oxidation of water (2H₂O → O₂ + 4H⁺ + 4e⁻, ≈ +1.23 V) is far easier, so O₂ is released. …
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