Question

Identify reducing agent in the following reaction: \[ \text{CH}_4(g) + \text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \]

• A. O\(_2\)(g)
• B. H\(_2\)O(l)
• C. CH\(_4\)(g)
• D. CO\(_2\)(g)

Hint:

The reducing agent is the substance that donates electrons and is oxidized in the reaction. In this case, CH\(_4\) is oxidized, so it is the reducing agent.

Answer 1

The Correct Option is C

Step 1: Understanding reduction and oxidation.
In this reaction, methane (CH\(_4\)) is oxidized to carbon dioxide (CO\(_2\)), while oxygen (O\(_2\)) is reduced to water (H\(_2\)O). The reducing agent is the substance that donates electrons and is oxidized. Here, CH\(_4\) is oxidized, so it is the reducing agent.
Step 2: Analyzing the options.
(A) O\(_2\)(g): Incorrect. O\(_2\) is reduced in the reaction, not oxidized.
(B) H\(_2\)O(l): Incorrect. Water is formed in the reaction and does not act as the reducing agent.
(C) CH\(_4\)(g): Correct. CH\(_4\) donates electrons and gets oxidized to CO\(_2\), making it the reducing agent.
(D) CO\(_2\)(g): Incorrect. CO\(_2\) is the product and is not involved in electron donation.
Step 3: Conclusion.
The correct answer is (C) CH\(_4\)(g).