Question

In the electrolysis of aqueous solution of $CuS{O}_4$ using copper electrodes, the process taking place at the anode is:
(A) $S{O}_4^{2-}\to S{O}_4+2e^{-}$
(B) $Cu\to C{u}^{+}+e^{-}$
(C) $2O{H}^{-}\to H_{2}O+{}^1/{}_2{O}_2+2e^{-}$
(D) $Cu\to C{u}^{2+}+2e^{-}$

Answer 1

Hint: Think which reactions take place at anode and cathode. Write down the reactions happening during electrolysis of copper sulphate solution. Assign the reactions based on their type to anode and cathode.

Complete step by step answer:
- Electrolysis is the process in which an electrolyte, in this case copper sulphate solution, undergoes redox reactions at the electrodes due to the action of electric current.
- When copper sulphate dissociates to form cupric ions and sulphate ions. This reaction can be represented as,
$$CuS{O}_4\rightleftharpoons C{u}^{2+}+S{O}_4^{2-}$$

-At Cathode, reduction takes place. Cupric ions present in the electrolyte gain two electrons and get converted to copper metal. This copper metal is deposited on copper electrodes.
$$C{u}^{2+}+2e^{-}\to Cu$$

Since this reduction reaction occurs at cathode, copper electrode is the cathode.
-At anode, oxidation takes place. Copper sulphate solution has water as solvent. These water molecules get dissociated due to action of current to form protons and hydroxyl ions. Protons combine with sulphate ions present in the solution to form sulphuric acid. Now, the hydroxyl ions present in the solution undergo oxidation to liberate oxygen gas at anode. This reaction is represented as,
$2O{H}^{-}\to H_{2}O+{}^1/{}_2{O}_2+2e^{-}$

-Therefore, during the electrolysis of aqueous solution of $CuS{O}_4$ using copper electrodes, the process taking place at the anode is (C) $2O{H}^{-}\to H_{2}O+{}^1/{}_2{O}_2+2e^{-}$
So, the correct answer is “Option C”.

Note: Remember in electrolytic reactions, at anode only oxidation will take place and at cathode, reduction takes place. LOAN = Left oxidation anode and CAR = Cathode reduction are tips to remember this concept.