Question

Predict the products of electrolysis in each of the following: (i) An aqueous solution of CuCl$_2$ with platinum electrodes. (ii) A concentrated solution of H$_2$SO$_4$ with platinum electrodes.

Hint:

For electrolysis: \[ \boxed{\text{Cathode: Reduction}} \] \[ \boxed{\text{Anode: Oxidation}} \] In aqueous CuCl$_2$: \[ Cu^{2+} \text{ is deposited and } Cl^- \text{ gives } Cl_2 \] In water or dilute acid solution: \[ \boxed{H_2 \text{ is produced at cathode and } O_2 \text{ at anode}} \]

Answer 1

Concept: During electrolysis, the products formed at the electrodes depend on:

• Nature and concentration of ions present in the electrolyte.

• Standard electrode potentials of possible oxidation and reduction reactions.

• Nature of electrodes used.
Platinum electrodes are inert electrodes, meaning they do not participate in the reaction. They only provide a surface for oxidation and reduction reactions.

Part (i): Electrolysis of aqueous CuCl$_2$ The ions present in aqueous copper chloride solution are: \[ Cu^{2+},\ Cl^-,\ H^+,\ OH^- \] At the cathode, reduction of ions occurs. Possible reduction reactions are: \[ Cu^{2+}+2e^- \rightarrow Cu \] \[ 2H^+ + 2e^- \rightarrow H_2 \] Since the reduction potential of copper ions is higher than hydrogen ions, copper ions are preferentially reduced. Therefore: \[ \boxed{Cu^{2+}+2e^- \rightarrow Cu} \] Copper is deposited at the cathode. At the anode, oxidation occurs. Possible oxidation reactions are: \[ 2Cl^- \rightarrow Cl_2+2e^- \] \[ 2H_2O \rightarrow O_2+4H^++4e^- \] Due to the high concentration of chloride ions, chloride ions are oxidised preferentially. Therefore: \[ \boxed{2Cl^- \rightarrow Cl_2+2e^-} \] Hence, copper and chlorine are obtained as products.

Part (ii): Electrolysis of concentrated H$_2$SO$_4$ Concentrated sulphuric acid contains: \[ H^+,\ SO_4^{2-},\ HSO_4^- \] At the cathode, hydrogen ions are reduced: \[ \boxed{2H^+ +2e^- \rightarrow H_2} \] Hydrogen gas is released. At the anode, sulphate ions are difficult to oxidise. Therefore, water molecules undergo oxidation: \[ \boxed{2H_2O \rightarrow O_2+4H^++4e^-} \] Oxygen gas is released. Thus, electrolysis of concentrated sulphuric acid gives hydrogen and oxygen gases.

Final Answer:

(i) \[ \boxed{ CuCl_2(aq) \rightarrow Cu(s)+Cl_2(g) } \] Cathode product: \[ \boxed{Cu} \] Anode product: \[ \boxed{Cl_2} \]

(ii) \[ \boxed{ 2H_2O \rightarrow 2H_2(g)+O_2(g) } \] Cathode product: \[ \boxed{H_2} \] Anode product: \[ \boxed{O_2} \]