$H_2O_{(g)} + C_{(s)} → H_{2(g)} + CO_{(g)}$ The oxidizing agent in the reaction above is

Analysis of the Question

The question asks us to identify the oxidizing agent in the chemical reaction: $H_2O_{(g)} + C_{(s)} \rightarrow H_{2(g)} + CO_{(g)}$

Explaining the Concept

To identify the oxidizing agent, we must track the changes in the oxidation states of the elements involved:

1. Oxidation: This is the gain of oxygen, loss of hydrogen, or an increase in oxidation number. The substance that undergoes oxidation is called the reducing agent.
2. Reduction: This is the loss of oxygen, gain of hydrogen, or a decrease in oxidation number. The substance that undergoes reduction is called the oxidizing agent.

Assigning Oxidation Numbers:

• In $H_2O$: Hydrogen is $+1$, Oxygen is $-2$.
• In $C$ (elemental state): Oxidation number is $0$.
• In $H_2$ (elemental state): Oxidation number is $0$.
• In $CO$: Carbon is $+2$, Oxygen is $-2$.

Step-by-Step Solution

1. Track Carbon ($C$): Carbon goes from $0$ (in $C_{(s)}$) to $+2$ (in $CO_{(g)}$). Since its oxidation number increased, Carbon is oxidized. Therefore, Carbon acts as the reducing agent.
2. Track Hydrogen ($H$): Hydrogen goes from $+1$ (in $H_2O$) to $0$ (in $H_2$). Since its oxidation number decreased, Hydrogen is reduced.
3. Identify the Oxidizing Agent: Because $H_2O$ provides the oxygen that oxidizes the carbon and contains the hydrogen that gets reduced, $H_2O$ is the substance that causes oxidation. Therefore, $H_2O$ is the oxidizing agent.

Conclusion

The substance that is reduced is the oxidizing agent. Since $H_2O$ is reduced to $H_2$, it is the oxidizing agent.

Exam Tip

A simple way to remember this is the acronym OIL RIG:

• Oxidation Is Loss (of electrons/increase in oxidation state).
• Reduction Is Gain (of electrons/decrease in oxidation state). The substance that gets reduced is the oxidizing agent, and the substance that gets oxidized is the reducing agent.