Problem 46 A displacement reaction is an ox... [FREE SOLUTION]

Chapter 12: Problem 46

A displacement reaction is an oxidation-reduction reaction in which one element displaces another from solution. In each of the following displacement reactions identify the element that is oxidized and the element that is reduced: $$\begin{array}{l} \mathrm{Zn}+\mathrm{Cu}^{2+} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Cu} \\ \mathrm{Fe}+2 \mathrm{H}^{+} \longrightarrow \mathrm{Fe}^{2+}+\mathrm{H}_{2} \\\ \mathrm{Cl}_{2}+2 \mathrm{Br}^{-} \longrightarrow 2 \mathrm{Cl}^{-}+\mathrm{Br}_{2} \end{array}$$

Short Answer

Expert verified

In the given reactions: Zinc is oxidized and copper is reduced. Iron is oxidized and hydrogen is reduced. Bromine is oxidized and chlorine is reduced.

Step by step solution

- Identify the Oxidation States

For each reaction, determine the oxidation states of the elements in the reactants and products. For example, in the reaction ewline $ \text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu} $: \[Zn: 0 \ (reactant) \rightarrow Zn^{2+}: +2 \ (product)\] \[Cu^{2+}: +2 \ (reactant) \rightarrow Cu: 0 \ (product)\]

- Determine Oxidation and Reduction

Identify which elements are losing electrons (oxidation) and which are gaining electrons (reduction). In the reaction $ \text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu} $: Zinc is oxidized as it goes from $0 $ to $+2$. ewline Copper is reduced as it goes from $ +2 $ to $0$.

- Repeat for the Second Reaction

Repeat Steps 1 and 2 for the second reaction: ewline $ \text{Fe} + 2\text{H}^{+} \rightarrow \text{Fe}^{2+} + \text{H}_2 $: \[Fe: 0 \ (reactant) \rightarrow Fe^{2+}: +2 \ (product)\] ewline \[H^{+}: +1 \ (reactant) \rightarrow H_2: 0 \ (product)\] Iron is oxidized as it goes from $0 $ to $ +2 $. Hydrogen is reduced as it goes from $ +1 $ to $ 0 $.

- Repeat for the Third Reaction

Repeat Steps 1 and 2 for the third reaction: ewline $ \text{Cl}_2 + 2\text{Br}^{-} \rightarrow 2\text{Cl}^{-} + \text{Br}_2 $: \[Cl_2: 0 \ (reactant) \rightarrow Cl^{-}: -1 \ (product)\] ewline \[Br^{-}: -1 \ (reactant) \rightarrow Br_2: 0 \ (product)\] Chlorine is reduced as it goes from $0 $ to $ -1 $. Bromine is oxidized as it goes from $ -1 $ to $ 0 $.

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

oxidation

Oxidation is a core concept in chemistry, especially in displacement reactions. It refers to the loss of electrons by a molecule, atom, or ion. Whenever oxidation occurs, the oxidation state of the chemical species increases.
For example, in the reaction: \ $ \text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu} \ $, zinc atoms lose two electrons. Here, zinc is oxidized as its oxidation state goes from $ 0 $ to $ +2 $.
Remember, oxidation can be identified as:

Loss of electrons
Increase in oxidation state
Occurs in coordination with reduction (oxidation-reduction reactions)

reduction

Reduction is the opposite of oxidation. It involves the gain of electrons by a molecule, atom, or ion. Consequently, the oxidation state of the substance decreases.
Take for instance the same reaction: \ $ \text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu} \ $. Copper ions gain two electrons and are reduced, as their oxidation state goes from $ +2 $ to $ 0 $.
To identify reduction, look for:

Gain of electrons
Decrease in oxidation state
Always found alongside oxidation (paired process)

oxidation states

Oxidation states (oxidation numbers) are crucial for understanding redox reactions. These numbers indicate the degree of oxidation of an atom in a chemical compound.
To determine oxidation states:

Elements in their elemental form have an oxidation state of $ 0 $ (e.g. $ \text{Cl}_2 $).
For ions, the oxidation state equals the charge of the ion (e.g. $ \text{Cu}^{2+} $ has an oxidation state of $ +2 $).
In compounds, the general rules are followed (e.g., hydrogen is usually $ +1 $ and oxygen is $ -2 $).

Let's apply this to the given reactions:
For reaction \ $ \text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu} \ $, we see changes in oxidation states as:
鈥� $ \text{Zn} $ changes from $ 0 $ to $ +2 $
鈥� $ \text{Cu}^{2+} $ changes from $ +2 $ to $ 0 $ This insight helps us unearth which elements get oxidized and which get reduced.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

Short Answer

Step by step solution

- Identify the Oxidation States

- Determine Oxidation and Reduction

- Repeat for the Second Reaction

- Repeat for the Third Reaction

Key Concepts

oxidation

reduction

oxidation states

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Magnetism

Translational Dynamics

Astrophysics

Fluids

Space Physics

Thermodynamics

Study anywhere. Anytime. Across all devices.