Question

Out of C and CO, which is a better reducing agent for ZnO?

Answer 1

Reduction of ZnO to Zn is usually carried out at 1673 K. From the above figure, it can be observed that above 1073 K, the Gibbs free energy of formation of CO from C and above 1273 K, the Gibbs free energy of formation of CO₂ from C is less than the Gibbs free energy of formation of ZnO. Therefore, C can easily reduce ZnO to Zn.

On the other hand, the Gibbs free energy of formation of CO₂ from CO is always higher than the Gibbs free energy of formation of ZnO. Therefore, CO cannot reduce ZnO. Hence, C is a better reducing agent than CO for reducing ZnO.

Answer 2

The two reduction reactions are:

\[\ce{ZnO_{(s)} + C_{(s)} -> Zn_{(s)} + CO_{(g)}}\]    ...(i)

\[\ce{ZnO_{(s)} + CO_{(g)} -> Zn_{(s)} + CO2_{(g)}}\]    ...(ii)

In the first case, there is an increase in the magnitude of ΔS° while in the second case, it almost remains the same. In other words, ΔG° will have a more negative value in the first case when C_(s) is used as the reducing agent than in the second case when CO_(g) acts as the reducing agent. Therefore, C_(s) is a better reducing agent.

Answer 3

The C, CO curve in the Ellingham diagram is lower than the Zn, ZnO curve above 1120 K, while the C, CO₂ curve is lower above 1323 K. As a result, at 1120 K and 1323 K, respectively, the Δ_fG° of ZnO from C is higher than that of CO from C and CO₂ from C. However, even beyond 2273 K, the CO, CO₂ curve is higher than the Zn, ZnO curve. As a result, C can decrease ZnO, whereas CO cannot. Therefore, C is a better reducing agent for ZnO than CO.