What is the process by which sodium hydroxide is prepared? | General Science MCQ | Examoogle

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What is the process by which sodium hydroxide is prepared?

Solvay’s process

Chlor-alkali process

Haber’s process

Contact process

Correct Answer

🔬 SC • Science Concept & PrincipleGeneral SciencePrevious Year Paper 2024

Option B

Chlor-alkali process

Quick Summary:

Sodium hydroxide ( $NaOH$ ), commonly known as caustic soda, is manufactured industrially by the electrolysis of an aqueous solution of sodium chloride ( $\text{Na Cl}$ ), which is called brine. This method is called the Chlor-alkali process because the main products formed are chlorine ( $Cl_2$ , 'chlor') and sodium hydroxide ( $NaOH$ , 'alkali').

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💡 Explanation

💡 Everyday Analogy (Real-World Intuition)

Imagine a solution filled with salt ( $\text{Na Cl}$ ) and water ( $H_2O$ ). Passing electric current acts like a sorting magnet—pulling chlorine out as a gas at one post, venting off hydrogen gas at the other, and leaving behind a concentrated liquid soap base ( $NaOH$ ) in the tank.

🧠 Memory Mnemonic / Shortcut Aid

Remember 'CAN-HAC': Chlorine at Anode (+), Hydrogen at Cathode (-).

🔢 Key Formulas

$2NaCl(aq) + 2H_2O(l) \xrightarrow{\text{Electrolysis}} 2NaOH(aq) + Cl_2(g) + H_2(g)$ — Chlor-alkali process overall reaction

$\text{At Anode (+): } 2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-$ — Oxidation of chloride ions

$\text{At Cathode (-): } 2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$ — Reduction of water molecules

⚙️ Working Principle

When electricity is passed through an aqueous solution of sodium chloride (brine), it decomposes to form sodium hydroxide, chlorine gas, and hydrogen gas. At the anode (+), chloride ions lose electrons to form chlorine gas ( $Cl_2$ ). At the cathode (-), hydrogen ions gain electrons to form hydrogen gas ( $H_2$ ), leaving hydroxide ions ( $OH^-$ ) in solution which combine with $Na^+$ to form $NaOH$ . $\text{Overall Reaction: } 2NaCl(aq) + 2H_2O(l) \xrightarrow{\text{Electrolysis}} 2NaOH(aq) + Cl_2(g) \uparrow + H_2(g) \uparrow$

📌 Key Points

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Products obtained are Chlorine ( $Cl_2$ ) gas, Hydrogen ( $H_2$ ) gas, and Sodium Hydroxide ( $NaOH$ ) solution.
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Chlorine gas is released at the anode (positively charged electrode).
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Hydrogen gas is evolved at the cathode (negatively charged electrode).
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Sodium Hydroxide ( $NaOH$ ) solution is formed near the cathode.
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The process uses brine, which is a concentrated aqueous solution of Sodium Chloride ( $\text{Na Cl}$ ).

✅ Advantages

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Produces three commercially useful products ( $NaOH$ , $Cl_2$ , $H_2$ ) simultaneously.
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High purity of $NaOH$ is achieved, especially when using modern membrane cell technology.

❌ Disadvantages / Limitations

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Highly energy-intensive process requiring large amounts of electricity.
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Older technologies like the mercury cell method pose serious environmental contamination hazards.

🛠️ Applications / Uses

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Sodium Hydroxide ( $NaOH$ ): Used in soap and detergent manufacturing, paper making, and petroleum refining.
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Chlorine ( $Cl_2$ ): Used in water purification, PVC manufacture, disinfectants, and bleaching powder.
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Hydrogen ( $H_2$ ): Used in fuel cells, margarine production (hydrogenation), and ammonia synthesis for fertilizers.

📄 Additional Information

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Option A (Solvay’s Process): Used for industrial preparation of Sodium Carbonate ( $Na_2CO_3$ , washing soda) and Sodium Bicarbonate ( $NaHCO_3$ ).
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Option C (Haber’s Process): Used for industrial synthesis of Ammonia ( $NH_3$ ) from nitrogen and hydrogen gases.
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Option D (Contact Process): Used for industrial manufacture of Sulfuric Acid ( $H_2SO_4$ ).

📊 Diagram / Illustration

✅

B is correct — Sodium hydroxide is industrially prepared by the Chlor-alkali process via the electrolysis of brine solution.

Core Concepts Used

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Chlor-alkali Process Electrolysis of Brine Sodium Hydroxide Production Industrial Chemical Processes

💡 EXAM TIP

In competitive exams, electrode product locations are heavily tested: Remember Chlorine evolves at Anode ( $Cl_2$ at +) and Hydrogen evolves at Cathode ( $H_2$ at -), while $NaOH$ remains near the Cathode.

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