Production of halogens

Laboratory methods for chlorine synthesis

In laboratory conditions, chlorine ($Cl_2$) is traditionally produced by oxidizing hydrochloric acid ($HCl$) with various strong reagents.

1. Oxidation with potassium permanganate ($KMnO_4$) This is the simplest and most common method. Upon contact of crystalline potassium permanganate granules with concentrated hydrochloric acid, active foaming of the mixture is observed. As a result, a greenish gas with a pungent, suffocating odor is released. Equation: $2KMnO_4 + 16HCl \rightarrow 2MnCl_2 + 5Cl_2\uparrow + 8H_2O + 2KCl$. Modification: Instead of pure $HCl$, a mixture of sodium chloride ($NaCl$) solution and sulfuric acid ($H_2SO_4$) can be used. It is important to remember that without the addition of sulfuric acid, the reaction between $KMnO_4$ and $NaCl$ does not proceed. Safety: Chlorine is extremely toxic. The experiment is carried out using micro-quantities of reagents (no more than 5 drops) or in specialized sealed glassware.

2. Oxidation with manganese(IV) oxide ($MnO_2$) Black manganese dioxide powder reacts with hydrochloric acid only upon heating. Equation: $MnO_2 + 4HCl \xrightarrow{t^\circ} MnCl_2 + Cl_2\uparrow + 2H_2O$. A huge advantage of this method is the ability to interrupt the process at any moment: as soon as the heat source is removed, gas evolution stops.

3. Oxidation with potassium dichromate ($K_2Cr_2O_7$) Mixing orange dichromate crystals with acid leads to the gradual appearance of a chlorine odor. The solution changes color from orange to green (an indicator of chromium(III) chloride formation — $CrCl_3$). Equation: $K_2Cr_2O_7 + 14HCl \rightarrow 2CrCl_3 + 3Cl_2\uparrow + 7H_2O + 2KCl$. Since the oxidizing capacities of the products and reagents are close, the reaction is reversible and proceeds slowly. To accelerate chlorine evolution, the flask must be heated.

4. Reaction with potassium chlorate (Berthelot's salt) An example of chemical preparation based on the disproportionation principle (interaction of chlorine compounds in negative and positive oxidation states). Equation: $KClO_3 + HCl \rightarrow Cl_2\uparrow + KCl + H_2O$.

Industrial production of chlorine and fluorine

Large-scale production requires economically viable methods, which is why chemical oxidizing agents are replaced by electric current.

Chlorine is produced by the electrolysis of an aqueous sodium chloride ($NaCl$) solution. During the process, the products are distributed as follows:

• At the anode, the target gas — chlorine ($Cl_2$) — is released.
• At the cathode, hydrogen gas ($H_2$) is formed.
• In the electrolyte, sodium hydroxide ($NaOH$) accumulates.

Equation: $2NaCl + 2H_2O \xrightarrow{electrolysis} H_2\uparrow + Cl_2\uparrow + 2NaOH$. Hydrogen and alkali are not discarded, but act as valuable by-products of chemical production.

Fluorine ($F_2$) is the most labor-intensive group element to obtain. Being the strongest oxidizing agent in nature, it does not undergo chemical oxidation — there is no reagent capable of taking an electron away from a fluoride ion. Therefore, the only available method is the electrolysis of molten fluorides. Potassium hydrogen fluoride ($KF\cdot HF$) is used as a raw material, and the process is conducted at a melting point of about 240 °C.

Synthesis of bromine and iodine by displacement

Bromides and iodides are oxidized much easier than chlorides. Consequently, bromine and iodine are obtained by displacement reactions from aqueous solutions of their salts. A more active halogen — gaseous chlorine — acts as the oxidizing agent. General equation using sodium salts as an example: $2NaBr + Cl_2 \rightarrow Br_2 + 2NaCl$.

The reactions are accompanied by striking visual changes:

Upon obtaining bromine ($Br_2$):

1. The initially colorless solution turns yellow, and then a saturated orange color (accumulation of dissolved bromine).
2. If the concentration of reagents is high, a heavy red-brown liquid (pure liquid bromine) collects at the bottom of the vessel.
3. A characteristic heavy, unpleasant odor appears in the air.

Upon obtaining iodine ($I_2$):

1. The color of the solution intensifies from pale yellow to a saturated "strong tea" shade.
2. Soon, a black precipitate begins to drop out (iodine crystals).
3. After precipitation, the supernatant liquid lightens again to pale yellow.

Mechanism of iodine solution color change: The "strong tea" coloration is caused not by free iodine itself, but by the formation of a complex between iodide ions and iodine molecules: $I^- + I_2 \rightleftharpoons I_3^-$. As chlorine displaces iodine, the concentration of initial $I^-$ ions in the solution drops. The reaction equilibrium shifts to the left (towards the decomposition of the $I_3^-$ complex), and free $I_2$, which has low solubility, precipitates out.

Oxidation of halides by atmospheric oxygen

Certain halide ions are capable of oxidizing even under the action of ordinary atmospheric oxygen ($O_2$), without the addition of specialized reagents.

• Iodides: Undergo slow oxidation in air. Visually, this is manifested by solutions turning yellow over time. This process is greatly promoted by an acidic medium. If a solution of hydrogen iodide ($HI$) is taken, it darkens very rapidly due to the swift formation of a large amount of free $I_2$.
• Chlorides and bromides: Possess greater chemical stability and do not interact with atmospheric oxygen.

Class and exam questions

What else to study on this topic

• Mechanism of formation and decomposition of the triiodide ion complex
• Kinetics of hydrochloric acid oxidation by potassium dichromate
• Halogen disproportionation reactions
• Effect of medium pH on the rate of iodide oxidation by atmospheric oxygen

Other topics in the section "Chapter 9. Group VIIA Elements — Halogens"

Prepared by WhiteCoat Pro faculty in accordance with the medical school curriculum. For educational purposes only.