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

Step 1: General rule with inert Pt electrodes.
At the cathode, the ion with the highest reduction potential is preferentially reduced. At the anode, the most easily oxidised species is discharged (concentration also plays a role).
Step 2: Aqueous $CuCl_2$ (i).
Cathode: $Cu^{2+}$ has a higher reduction potential than $H^+$, so copper deposits. Anode: high $Cl^-$ concentration drives chloride oxidation over water, releasing $Cl_2$. \[ \text{Cathode: }Cu^{2+}+2e^-\rightarrow Cu \quad \text{Anode: }2Cl^-\rightarrow Cl_2+2e^- \]
Step 3: Concentrated $H_2SO_4$ (ii).
Cathode: $2H^++2e^-\rightarrow H_2$. Anode: $SO_4^{2-}$ is extremely difficult to oxidise, so water is oxidised to $O_2$: $2H_2O\rightarrow O_2+4H^++4e^-$. \[ \boxed{(i)\;Cu\text{ and }Cl_2;\quad(ii)\;H_2\text{ and }O_2} \]