Question

If an aqueous solution of NaF is electrolyzed between insert electrodes, the product obtained at anode is

• A. O₂
• B. F₂
• C. H₂
• D. Na

Answer 1

The Correct Option is A

In the electrolysis of an aqueous solution of NaF, the reactions at the anode and cathode are important. The anode is where oxidation occurs, and the cathode is where reduction takes place. At the anode, water molecules can also be oxidized because \( \text{OH}^- \) ions are present. The possible oxidation reactions at the anode are:
1. Oxidation of \( \text{OH}^- \) to \( \text{O}_2 \): \[ 4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4e^- \] 2. Oxidation of \( \text{F}^- \) to \( \text{F}_2 \): \[ 2\text{F}^- \rightarrow \text{F}_2 + 2e^- \] However, oxygen gas (\( \text{O}_2 \)) is generally produced at the anode in aqueous electrolysis, as the oxidation of water is easier than the oxidation of fluoride ions under normal conditions. Thus, the product obtained at the anode is \( \text{O}_2 \).

The correct option is (A) : \(O_2\)

Answer 2

When an aqueous solution of sodium fluoride (NaF) is electrolyzed between inert electrodes, the reactions at the electrodes can be explained as follows: At the cathode (reduction reaction), sodium ions (Na\(^+\)) migrate towards the cathode and gain electrons to form sodium metal (Na). However, the reduction of water molecules can also take place at the cathode to form hydrogen gas (H\(_2\)): \[ \text{At Cathode:} \quad 2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq) \] At the anode (oxidation reaction), fluoride ions (F\(^-\)) would migrate towards the anode to release electrons and form fluorine gas (F\(_2\)). However, water molecules are more easily oxidized than fluoride ions. Therefore, oxidation of water takes place at the anode, leading to the formation of oxygen gas (O\(_2\)) instead of fluorine gas: \[ \text{At Anode:} \quad 2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^- \] Thus, the product obtained at the anode is oxygen gas (O\(_2\)), not fluorine gas. Hence, the correct answer is O\(_2\).