Activity 1K (Textbook) – Chemistry Form 5 Chapter 1 (Redox Equilibrium)

Activity 1K:
Carry out this activity collaboratively in groups.
1. Based on Figure 1.32, write the half equations for the reactions at the electrodes and explain the differences in observations of both cells.

2. Share your group’s findings with other groups using multimedia presentations.

Answer:
1.
– Electrolytic cell:
$$\text { Anode(+): } \mathrm{Cu} \rightarrow \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \quad \text { Cathode(-) }: \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}$$

– Chemistry cell:
$$\text { Anode(-) : } \mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \quad \text { Cathode }(+): \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}$$


– The blue colour of copper(II) sulphate solution remains unchanged in the electrolytic cell but the blue colour of copper(II) sulphate solution becomes paler in the voltaic cell.

– The concentration of Cu²⁺ ions remains unchanged in the electrolytic cell but the concentration of Cu²⁺ ions decreases in the voltaic cell.

– In the electrolytic cell, the discharge rate of Cu²⁺ ions at the cathode is the same as the ionisation rate of copper atoms at the anode, but in the voltaic cell, Cu²⁺ ions are discharged to form copper atoms at the cathode.