Chemistry, asked by ansarisaif3215, 5 months ago

write electrode reactions for electrolysis of aqueous NaCl​

Answers

Answered by AbhiThakur07
4

Explanation:

=>Electrolysis of sodium chloride is an important process to manufacture many bulk chemicals of commercial utility, like chlorine, sodium hydroxide etc. Sodium chloride is electrolyzed either in a molten state or in aqueous solutions. Electrolysis is, also done in the presence of additional salts to aid the redox reactions.

=>Electrolysis of Molten Sodium Chloride

Electrolysis involves the movement of ions to the electrode. Solid-state does not allow the movement of ions and unsuitable for electrolysis. When melted at high temperature, sodium chloride separates into sodium and chloride ions, so that, electrolysis can take place to form sodium atom and chlorine gas.

Read More In Detail: Electrolysis

Electrolysis of Molten Sodium Chloride

Electrolysis of Sodium Chloride

NaCl → Na +(l) + Cl–(l)

At cathode: reduction of 2Na+(l) + e– → Na(l)

At anode: oxidation of 2Cl–(l) → Cl2(g) + 2e–

Net Reaction is written as: 2Na +(l) + 2Cl–(l) → 2Na(l) + Cl2(g)

Down’s Process: Sodium chloride melts at a very high temperature of 801°C. Addition of anhydride calcium chloride in the ratio of CaCl2: NaCl = 3:2 reduces the melting point to 580°C. Electrolysis is done with an iron cathode and graphite anode and iron gauze preventing the mixing of products chlorine and sodium. The products of molten sodium chloride are sodium metal and chlorine gas.

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Answered by rahul123437
5

Net cell reaction for electrolysis of aqueous NaCl is

2H₂O +2Cl⁻ →  H₂ +Cl₂

Explanation:

Electrolysis of aqueous NaCl​

Half reaction at cathode -At cathode Sodium ions and hydrogen ions are present. Hydrogen ions are more reducible than Sodium ions.

Reduction reaction takes place liberating hydrogen ions and sodium ions

2H₂O +2e⁻→ H₂ +2OH⁻

Half reaction at anode:-

There are Chlorine ion and hydroxide ion. Chlorine ion are oxidised due to decomposition

2Cl⁻ →Cl₂+2e⁻

Net Cell reaction:-Two electrons are gained at cathode and two electrons are released at anode.

Na⁺ and OH⁻ are in the solution to form NaOH.

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