3 years ago

Identify the oxidizing agent and the reducing agent for 4Li(s) + O_2 (g) to 2Li_2O(s).

1 answer:

7 0

The reaction is:

<span>4Li(s) + O2 (g) = 2Li+ + 2O-2(s).

The oxidizing agent is the one that is being reduced which is oxygen where the charge changed from neutral to -2 while the reducing agent is the on being oxidized which is lithium where the charge change from neutral to +1.</span>

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Hope this answers the question. Have a nice day.

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