KCET · Chemistry · Redox Reactions
If an aqueous solution of \(\mathrm{NaF}\) is electrolysed between inert electrodes, the product obtained at anode is
- A \(\mathrm{F}_{2}\)
- B \(\mathrm{H}_{2}\)
- C \(\mathrm{Na}\)
- D \(\mathrm{O}_{2}\)
Answer & Solution
Correct Answer
(D) \(\mathrm{O}_{2}\)
Step-by-step Solution
Detailed explanation
Aqueous NaF, when electrolysed between inert electrodes we always get \(\mathrm{O}_{2}\) at anode, as: At cathode \(2 \mathrm{H}^{+}+2 e^{-} \longrightarrow \mathrm{H}_{2} \uparrow\)
At anode \(4 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}+4 e^{-}\)
Also, if any fluorine is liberated, that too reacts with \(\mathrm{H}_{2} \mathrm{O}\).
\(2 \mathrm{~F}_{2}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 4 \mathrm{HF}+\mathrm{O}_{2}\)
So, ultimate product is oxygen.
At anode \(4 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}+4 e^{-}\)
Also, if any fluorine is liberated, that too reacts with \(\mathrm{H}_{2} \mathrm{O}\).
\(2 \mathrm{~F}_{2}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 4 \mathrm{HF}+\mathrm{O}_{2}\)
So, ultimate product is oxygen.
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