When aqueous solution of cupric bromide is electrolyzed the product obtained at cathode and at anode will be respectively :

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TestHub · Accepted Answer

In the electrolysis of aqueous CuBr₂, Cu²⁺ ions are reduced at the cathode to form Cu metal, as Cu²⁺ has a higher reduction potential than H₂O. At the anode, Br⁻ ions are oxidized to Br₂, as Br⁻ has a lower oxidation potential than H₂O.

Cathode (reduction): $C u^{2 +} (a q) + 2 e^{-} \to C u (s)$

Anode (oxidation): $2 B r^{-} (a q) \to B r_{2} (l) + 2 e^{-}$

Therefore, the products are Cu at the cathode and Br₂ at the anode.

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