Question

identify the species oxidized, the species reduced, the oxidizing agent and the reducing agent in the following electron transfer reaction.
f₂ + cu → 2f⁻ + cu²+
as the reaction proceeds, electrons are transferred from
oxidizing agent
reducing agent
species oxidized
species reduced

Explanation:

Step1: Analyze oxidation states

In the reaction \(Cu + F_{2}
ightarrow 2F^{-}+Cu^{2 +}\), for \(Cu\), its oxidation state changes from \(0\) (in elemental \(Cu\)) to \(+ 2\) (in \(Cu^{2+}\)). For \(F\) in \(F_{2}\), its oxidation state changes from \(0\) (in elemental \(F_{2}\)) to \(-1\) (in \(F^{-}\)).

Step2: Determine oxidized and reduced species

The species that loses electrons (increases in oxidation state) is oxidized. Since \(Cu\) goes from \(0\) to \(+2\), \(Cu\) is oxidized. The species that gains electrons (decreases in oxidation state) is reduced. Since \(F\) in \(F_{2}\) goes from \(0\) to \(-1\), \(F_{2}\) is reduced.

Step3: Identify oxidizing and reducing agents

The oxidizing agent is the species that is reduced (because it causes another species to be oxidized). So \(F_{2}\) is the oxidizing agent. The reducing agent is the species that is oxidized (because it causes another species to be reduced). So \(Cu\) is the reducing agent.

Step4: Determine electron transfer

Electrons are transferred from the species that is oxidized (\(Cu\)) to the species that is reduced (\(F_{2}\)).

Answer:

Species oxidized: \(Cu\); Species reduced: \(F_{2}\); Oxidizing agent: \(F_{2}\); Reducing agent: \(Cu\); Electrons transferred from: \(Cu\); Electrons transferred to: \(F_{2}\)