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The products formed when an aqueous solution of NaBr is electrolysed in a cell having inert electrodes are

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Na and Br₂

Na and O₂

H₂, Br₂ and NaOH

H₂ and O₂

answer is C.

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Detailed Solution

NaBr⇌Na⁺+Br⁻
2H₂O+2e^-→H₂+2OH⁻
Na⁺+OH⁻→NaOH At cathode
Br⁻→Br+e⁻
Br+Br→Br₂ At anode

H₂ and NaOH (at the cathode) and Br₂ are the end products (at anode).

Hence the correct answer is C) H₂, Br₂ and NaOH

As discharge potential of H⁺ ions is lower than that of Na⁺ ions and discharge potential of Br^- ions is lower than that of OH^- ions, H₂ is liberated at cathode and Br₂ is liberated at anode.

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