Question

identify the elements in the following reaction:
2 k + f₂ ⟶ 2 kf

○ oxidized: kf; reduced: f₂

○ oxidized: f₂; reduced: kf

○ oxidized: k; reduced: f₂

○ oxidized: f₂; reduced: k

Explanation:

Step1: Recall Oxidation/Reduction Definitions

Oxidation is loss of electrons (increase in oxidation state), reduction is gain of electrons (decrease in oxidation state).

Step2: Analyze K in Reaction

In \( \text{K} \) (elemental form), oxidation state is \( 0 \). In \( \text{KF} \), K has oxidation state \( +1 \) (since F is \( -1 \)). K loses an electron (oxidation state increases from \( 0 \) to \( +1 \)) → K is oxidized.

Step3: Analyze \( \text{F}_2 \) in Reaction

In \( \text{F}_2 \) (elemental form), oxidation state is \( 0 \). In \( \text{KF} \), F has oxidation state \( -1 \). \( \text{F}_2 \) gains electrons (oxidation state decreases from \( 0 \) to \( -1 \)) → \( \text{F}_2 \) is reduced.

Answer:

C. oxidized: K; reduced: \( \text{F}_2 \) (Note: Using option label as per the third choice in the list: "oxidized: K; reduced: \( \text{F}_2 \)")