Bi2(g) + 2KI(ag) -> 2KBr(ag) + I2(s)
• Br2 gains electrons and is oxidized. I- (in KI) loses electrons and is reduced.
• Br2 loses electrons and is oxidized. I- (in KI) gains electrons and is reduced.
• I- (in KI) gains electrons and is oxidized. Br loses electrons and is reduced.
• I- (in KI loses electrons and is oxidized. Br gains electrons and is reduced.
The correct statement is:
• Br2 gains electrons and is reduced. I- (in KI) loses electrons and is oxidized.
The correct statement is:
• Br2 gains electrons and is reduced. I- (in KI) loses electrons and is oxidized.
The correct answer is:
• Br2 gains electrons and is reduced. I- (in KI) loses electrons and is oxidized.
To identify which substance is oxidized and which substance is reduced in a chemical equation, you need to analyze the changes in oxidation states.
In this reaction, Bi2 is not involved in any oxidation or reduction because the states of Bi do not change from the reactant side to the product side.
On the other hand, I- (in KI) has an oxidation state of -1 in the reactant side and 0 in the product side. Thus, I- is oxidized (it goes from an oxidation state of -1 to 0) and loses electrons.
Additionally, Br2 has an oxidation state of 0 in the reactant side and -1 in the product side. Therefore, Br2 is reduced (it goes from an oxidation state of 0 to -1) and gains electrons.
Remember, oxidation is the loss of electrons, while reduction is the gain of electrons.