During the electrolysis of copper II sulphate between platinum electrodes, if litmus solution is added to the anode compartment

Analysis

To solve this, we need to understand the electrolysis of aqueous copper(II) tetraoxosulphate(VI) ($CuSO_4$) using inert electrodes (platinum).

Concept

In the electrolysis of $CuSO_4(aq)$:

1. Ions present: $Cu^{2+}$, $SO_4^{2-}$, $H^+$, and $OH^-$ (from water).
2. At the Cathode (Negative Electrode): $Cu^{2+}$ ions are preferentially discharged over $H^+$ ions because copper is lower in the electrochemical series. • Equation: $Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}$
3. At the Anode (Positive Electrode): $OH^-$ ions are preferentially discharged over $SO_4^{2-}$ ions. • Equation: $4OH^-_{(aq)} \rightarrow 2H_2O_{(l)} + O_{2(g)} + 4e^-$
4. The Resulting Solution: As $OH^-$ ions are removed from the solution to form oxygen gas, the concentration of $H^+$ ions increases relative to $OH^-$ ions. This makes the solution around the anode acidic.

Step-by-Step Solution

• Step 1: Identify the product at the anode. As established, $OH^-$ ions are discharged to produce oxygen gas ($O_2$).
• Step 2: Determine the effect on the solution. The removal of $OH^-$ ions leaves behind an excess of $H^+$ ions, which increases the acidity of the electrolyte near the anode.
• Step 3: Apply the litmus test. Litmus solution is an indicator that turns red in an acidic environment and blue in an alkaline environment. Since the anode compartment becomes acidic, the litmus will turn red.
• Step 4: Combine the findings. The litmus turns red, and oxygen gas is evolved at the anode.

Conclusion

Comparing our findings to the options provided:

• A, B, and C are incorrect because the solution becomes acidic (turning litmus red), not alkaline.
• Option D correctly identifies that the litmus turns red due to the acidity and oxygen gas is evolved.

Exam Tip

Remember the "Anode-Acidic" rule for inert electrodes: When you electrolyze a salt of an oxyacid (like $SO_4^{2-}$ or $NO_3^-$) using inert electrodes, the anode compartment almost always becomes acidic because $OH^-$ is discharged, leaving $H^+$ ions behind.