Animated Solution for Chemistry - General Principles and Processes of Isolation of Metals: Consider the following reactions at $1100^{\circ }\text{C}$ (I) $2\text{C}+{\text{O}}_2\to 2\text{CO}$, $\Delta G^{\circ }=-460{\text{ kJ mol}}^{-1}$ (II) $2\text{Zn}+{\text{O}}_2\to 2\text{ZnO}$, $\Delta G^{\circ }=-360{\text{ kJ mol}}^{-1}$ Based on these, select correct alternative.

Can Carbon Reduce Zinc Oxide?

Imagine you are a metallurgist trying to extract pure zinc from its ore, zinc oxide. You have carbon (coke) at your disposal, and your furnace is roaring at $1100^{\circ }\text{C}$. The burning question is: will the carbon successfully steal the oxygen from the zinc? To answer this, we don't need to guess; we just need to consult the ultimate arbiter of chemical destiny—Thermodynamics.

In the realm of chemistry, a reaction only happens on its own if it leads to a more stable, lower-energy state. We measure this using the Gibbs Free Energy change, denoted as $\Delta G^{\circ }$.

We are given two fundamental oxidation reactions at $1100^{\circ }\text{C}$:

1. The oxidation of carbon:

2. The oxidation of zinc:

Our goal is to see if carbon can reduce zinc oxide. Therefore, our target reaction must have zinc oxide and carbon as reactants, and pure zinc and carbon monoxide as products:

To build this target equation from our given data, we need to manipulate them. Notice that in our target equation, $\text{ZnO}$ is a reactant. However, in equation (2), $\text{ZnO}$ is a product.

We must reverse equation (2). When we reverse a chemical reaction, the energy flow also reverses. If forming $\text{ZnO}$ releases $360\text{ kJ/mol}$ (a negative $\Delta G^{\circ }$), then breaking it apart will require exactly $360\text{ kJ/mol}$ (a positive $\Delta G^{\circ }$).

Reversed Equation (2):

Now, we simply add our reversed equation to equation (1).

Notice how the ${\text{O}}_2$ molecules on the left and right sides perfectly cancel each other out, leaving us with our exact target equation!

To find the net Gibbs free energy change, we add the individual $\Delta G^{\circ }$ values:

Because the net $\Delta G^{\circ }$ is $-100{\text{ kJ mol}}^{-1}$, which is strictly less than zero, the reaction is thermodynamically spontaneous.

Conclusion: Zinc oxide can indeed be reduced by carbon at $1100^{\circ }\text{C}$.

This mathematical proof is the exact underlying logic of the famous Ellingham Diagram. On the diagram, at $1100^{\circ }\text{C}$, the line representing the formation of $\text{CO}$ dips below the line representing the formation of $\text{ZnO}$. The element whose oxidation line is lower always has the power to reduce the oxide of the element whose line is higher!