2Li(s) + F2(g) -> 2LiF (s)
• Li loses electrons and is oxidized. F2 gains electrons and is reduced.
Li gains electrons and is oxidized. F2 loses electrons and is reduced.
• F2 loses electrons and is oxidized. Li gains electrons and is reduced.
• F2 gains electrons and is oxidized. Li loses electrons and is reduced.
The correct answer is:
Li loses electrons and is oxidized. F2 gains electrons and is reduced.
The correct answer is:
• Li loses electrons and is oxidized. F2 gains electrons and is reduced.
In order to determine which species is oxidized and which is reduced in the reaction 2Li(s) + F2(g) -> 2LiF(s), we need to identify the changes in oxidation states of each element.
The oxidation state of an element represents the number of electrons gained or lost by that element during a chemical reaction. Oxidation is the process where an element loses electrons and its oxidation state increases, while reduction is the process where an element gains electrons and its oxidation state decreases.
In this reaction, we have Li and F2 as the reactants, and LiF as the product. The oxidation state of Li in LiF is +1 because Group 1A elements, like Li, typically have an oxidation state of +1. The oxidation state of F in LiF is -1 because Group 7A elements, like F, typically have an oxidation state of -1.
Since the oxidation state of Li increases from 0 to +1, we can conclude that Li loses electrons and is oxidized. On the other hand, the oxidation state of F decreases from 0 to -1, so F2 gains electrons and is reduced.
Therefore, the correct answer is: Li loses electrons and is oxidized. F2 gains electrons and is reduced.