Identify the species oxidized, the species reduced, the oxidizing agent, and the reducing agent in the following electron-transfer reaction. Ni(s) + Pb2+(aq) -> Ni2+(aq) + Pb(s) Species oxidized: Ni(s) Species reduced: Pb2+(aq) Oxidizing agent: Pb2+(aq) Reducing agent: Ni(s) As the reaction proceeds, electrons are transferred from Ni(s) to Pb2+(aq). b) Identify the species oxidized, the species reduced, the oxidizing agent, and the reducing agent in the following electron-transfer reaction. Br2(l) + Pb(s) -> 2Br-(aq) + Pb2+(aq) Species oxidized: Pb(s) Species reduced: Br2(l) Oxidizing agent: Br2(l) Reducing agent: Pb(s) As the reaction proceeds, electrons are transferred from Pb(s) to Br2(l). c) Identify the species oxidized, the species reduced, the oxidizing agent, and the reducing agent in the following electron-transfer reaction. 2Al(s) + 3Sn2+(aq) -> 2Al3+(aq) + 3Sn(s) Species oxidized: Sn2+(aq) Species reduced: Al(s) Oxidizing agent: Al(s) Reducing agent: Sn2+(aq) As the reaction proceeds, electrons are transferred from Sn2+(aq) to Al(s).
The Correct Answer and Explanation is:
In each of the given reactions, we need to identify the species being oxidized and reduced, as well as the oxidizing and reducing agents. Let’s break down each reaction:
a) Ni(s) + Pb²⁺(aq) → Ni²⁺(aq) + Pb(s)
- Species oxidized: Ni(s). Oxidation is the loss of electrons. In this reaction, nickel (Ni) starts as a solid (Ni(s)) and turns into Ni²⁺ (aq), meaning it loses two electrons.
- Species reduced: Pb²⁺(aq). Reduction is the gain of electrons. Lead ions (Pb²⁺) gain electrons and are reduced to form solid lead (Pb).
- Oxidizing agent: Pb²⁺(aq). The species that gains electrons is called the oxidizing agent. Here, Pb²⁺ is reduced by gaining electrons from Ni(s), so Pb²⁺ is the oxidizing agent.
- Reducing agent: Ni(s). The species that loses electrons is called the reducing agent. Since Ni(s) loses electrons to reduce Pb²⁺, Ni(s) is the reducing agent.
In this reaction, Ni(s) is oxidized and provides the electrons for the reduction of Pb²⁺(aq).
b) Br₂(l) + Pb(s) → 2Br⁻(aq) + Pb²⁺(aq)
- Species oxidized: Pb(s). Here, solid lead (Pb) goes from an oxidation state of 0 in Pb(s) to Pb²⁺ in solution, meaning it loses two electrons and is oxidized.
- Species reduced: Br₂(l). Bromine (Br₂) starts as a liquid molecule with an oxidation state of 0 and is reduced to Br⁻ by gaining electrons.
- Oxidizing agent: Br₂(l). The species that gains electrons is the oxidizing agent. In this case, bromine (Br₂) is reduced to Br⁻, so Br₂ is the oxidizing agent.
- Reducing agent: Pb(s). The species that loses electrons is the reducing agent. Here, Pb(s) gives up electrons, so Pb(s) is the reducing agent.
In this reaction, Pb(s) is oxidized, and Br₂(l) is reduced.
c) 2Al(s) + 3Sn²⁺(aq) → 2Al³⁺(aq) + 3Sn(s)
- Species oxidized: Al(s). Aluminum (Al) starts as a solid and is oxidized to form Al³⁺, meaning it loses three electrons.
- Species reduced: Sn²⁺(aq). Tin (Sn²⁺) ions gain electrons and are reduced to solid tin (Sn).
- Oxidizing agent: Sn²⁺(aq). The species that gains electrons is the oxidizing agent. Here, Sn²⁺ is reduced to Sn(s), so Sn²⁺ is the oxidizing agent.
- Reducing agent: Al(s). The species that loses electrons is the reducing agent. Aluminum (Al) is oxidized to Al³⁺, so Al(s) is the reducing agent.
In this reaction, Al(s) is oxidized, and Sn²⁺(aq) is reduced.
Summary:
- In each reaction, the species that is oxidized loses electrons, and the species that is reduced gains electrons.
- The oxidizing agent is always the species that gets reduced (it gains electrons), and the reducing agent is the species that gets oxidized (it loses electrons).
By analyzing the electron transfer process, we can correctly identify the species involved and how they interact to facilitate the reactions.
