During the electrolysis of brine, by using inert electrodes,
- O2 liberates at anode
- H2 liberates at anode
- Na deposits on cathode
- Cl2 liberates at anode
The Correct Option is D
Approach Solution - 1
During the electrolysis of brine (aqueous NaCl solution) using inert electrodes, the ions present are:
- At Cathode (Reduction): Na+, H2O molecules
- At Anode (Oxidation): Cl−, H2O molecules
At Cathode (negative electrode):
Reduction potentials:
- Na+(aq) + e− → Na(s) (Highly negative potential)
- 2H2O(l) + 2e− → H2(g) + 2OH−(aq) (Lower negative potential)
H₂ gas is liberated at cathode since reduction of water is more favorable than Na⁺ reduction.
At Anode (positive electrode):
Oxidation potentials:
- 2Cl−(aq) → Cl2(g) + 2e− (Favorable oxidation)
- 2H2O(l) → O2(g) + 4H+(aq) + 4e− (Less favorable than chloride oxidation in concentrated solutions)
Cl₂ gas is liberated at anode, as oxidation of chloride ions is preferred due to the higher concentration and easier oxidation potential.
Overall reaction for brine electrolysis:
2NaCl(aq) + 2H₂O(l) → Cl₂(g) + H₂(g) + 2NaOH(aq)
Conclusion:
The correct option is: (D) Cl₂ liberates at anode
Correct Answer: Option (D)
Approach Solution -2
During the electrolysis of brine (aqueous sodium chloride, \( \text{NaCl} \)), the reactions that occur at the anode and cathode depend on the ions present in the solution:
At the anode (positive electrode), oxidation occurs. The chloride ions \( \text{Cl}^- \) from sodium chloride undergo oxidation to form chlorine gas \( \text{Cl}_2 \). \[ 2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^- \] So, chlorine gas (\( \text{Cl}_2 \)) is liberated at the anode.
At the cathode (negative electrode), reduction occurs. The hydrogen ions \( \text{H}^+ \) (from water) are reduced to form hydrogen gas \( \text{H}_2 \): \[ 2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \]
Sodium ions (\( \text{Na}^+ \)) do not get deposited on the cathode because the reduction potential for sodium is very high compared to that of hydrogen, so sodium remains in the solution.
Thus, the correct answer is \( \text{Cl}_2 \) liberates at the anode.
Top Questions on Electrolysis
A solution of aluminium chloride is electrolyzed for 30 minutes using a current of 2A. The amount of the aluminium deposited at the cathode is _________
- JEE Main - 2025
- Chemistry
- Electrolysis
O\(_2\) gas will be evolved as a product of electrolysis of:
(A) an aqueous solution of AgNO3 using silver electrodes.
(B) an aqueous solution of AgNO3 using platinum electrodes.
(C) a dilute solution of H2SO4 using platinum electrodes.
(D) a high concentration solution of H2SO4 using platinum electrodes.
Choose the correct answer from the options given below :- JEE Main - 2025
- Chemistry
- Electrolysis
Electrolysis of 600 mL aqueous solution of NaCl for 5 min changes the pH of the solution to 12. The current in Amperes used for the given electrolysis is ….. (Nearest integer).
- JEE Main - 2025
- Chemistry
- Electrolysis
- In electrolysis of dilute H\(_2\)SO\(_4\), what is liberated at anode?
- BCECE - 2025
- Chemistry
- Electrolysis
If \( E^\circ_{Fe^{2+}/Fe} = -0.441 \, \text{V} \) and \( E^\circ_{Fe^{3+}/Fe^{2+}} = 0.771 \, \text{V} \),
the standard emf of the cell reaction \( Fe(s) + 2Fe^{3+}(aq) \rightarrow 3Fe^{2+}(aq) \) is:\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] For the reaction, \( Fe^{3+} \) is reduced to \( Fe^{2+} \) (reduction at the cathode), and \( Fe \) is oxidized to \( Fe^{2+} \) (oxidation at the anode). So: \[ E^\circ_{\text{cell}} = E^\circ_{Fe^{3+}/Fe^{2+}} - E^\circ_{Fe^{2+}/Fe} \] \[ E^\circ_{\text{cell}} = 0.771 \, \text{V} - (-0.441 \, \text{V}) = 0.771 + 0.441 = 1.212 \, \text{V} \] Hence, the standard emf of the cell reaction is \( 1.212 \, \text{V} \).
- AP EAPCET - 2025
- Chemistry
- Electrolysis
Questions Asked in KCET exam
- If the number of terms in the binomial expansion of \((2x + 3)^n\) is 22, then the value of \(n\) is:
- KCET - 2025
- Binomial theorem
- A random experiment has five outcomes \(w_1, w_2, w_3, w_4, w_5\). The probabilities of the occurrence of the outcomes \(w_1, w_2, w_4, w_5\) are respectively \( \frac{1}{6}, a, b, \frac{1}{12} \) such that \(12a + 12b - 1 = 0\). Then the probabilities of occurrence of the outcome \(w_3\) is:
- KCET - 2025
- Probability
- If 'a' and 'b' are the order and degree respectively of the differentiable equation \[ \frac{d^2 y}{dx^2} + \left(\frac{dy}{dx}\right)^3 + x^4 = 0, \quad \text{then} \, a - b = \, \_ \_ \]
- KCET - 2025
- Differential equations
- Meera visits only one of the two temples A and B in her locality. Probability that she visits temple A is \( \frac{2}{5} \). If she visits temple A, the probability that she meets her friend is \( \frac{1}{3} \). The probability that she meets her friend, whereas it is \( \frac{2}{7} \) if she visits temple B. Meera met her friend at one of the two temples. The probability that she met her friend at temple B is:
- KCET - 2025
- Probability
In a practical examination, the following pedigree chart was given as a spotter for identification. The students identify the given pedigree chart as
- KCET - 2025
- Genetics