Exams › JEE Main › Chemistry

The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are:

1. ClO− and ClO3−
2. ClO2 and ClO3−
3. Cl− and ClO−
4. Cl− and ClO2−

Correct answer: Cl− and ClO−

Solution

When chlorine gas reacts with cold and dilute aqueous sodium hydroxide, it undergoes a disproportionation reaction, resulting in the formation of chloride ions (Cl−) and hypochlorite ions (ClO−). This reaction occurs under mild conditions, favoring the production of these specific products.

Related JEE Main Chemistry questions

• What volume of 20-volume hydrogen peroxide is needed to produce 5 L of oxygen gas at STP?
• Which one of the following reactions shows sulfuric acid acting as an oxidizing agent?
• Which of the following statements about redox behavior are true? (i) A metal M whose standard reduction potential for the half-reaction Mⁿ⁺ + ne⁻ ⇌ M is highly negative acts as a strong reducing agent. (ii) The oxidizing strength of the halogens falls in the order chlorine to iodine. (iii) The reducing strength of hydrogen halides rises from hydrogen chloride to hydrogen iodide.
• For the balanced redox equation X MnO4⁻ + Y C2O4²⁻ + Z H⁺ → X Mn²⁺ + 2Y CO2 + Z/2 H2O, what are the values of X, Y, and Z, respectively?
• Balance the redox equation below and identify the coefficients X, Y and Z in order: XNa2HAsO3 + YNaBrO3 + ZHCl → NaBr + H3AsO4 + NaCl
• Arrange the following sulphur-containing anions in order of increasing oxidation state of sulphur: SO3²⁻, S2O4²⁻ and S2O6²⁻.

⚔️ Practice JEE Main Chemistry free + battle 1v1 →