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Chapter 19: Problem 41

Identify the species oxidized and the species reduced in each of these redox equations. $$ \begin{array}{l}{\text { a. } 3 \mathrm{Br}_{2}+2 \mathrm{Ga} \rightarrow 2 \mathrm{GaBr}_{3}} \\ {\text { b. } \mathrm{HCl}+\mathrm{Zn} \rightarrow \mathrm{ZnCl}_{2}+\mathrm{H}_{2}} \\ {\text { c. } \mathrm{Mg}+\mathrm{N}_{2} \rightarrow \mathrm{Mg}_{3} \mathrm{N}_{2}}\end{array} $$

Short Answer

Expert verified
a) Br is reduced, Ga is oxidized. b) H is reduced, Zn is oxidized. c) N is reduced, Mg is oxidized.

Step by step solution

01

Write down the oxidation states

Identify the oxidation states of each element in the reactants and products.a. In \(3 \text{Br}_2 + 2\text{Ga} \rightarrow 2 \text{GaBr}_3\), Br starts as 0 (elemental state) and becomes -1; Ga starts as 0 and becomes +3.b. In \(\text{HCl} + \text{Zn} \rightarrow \text{ZnCl}_2 + \text{H}_2\), H in HCl is +1 and becomes 0; Zn starts 0 and becomes +2.c. In \(\text{Mg} + \text{N}_2 \rightarrow \text{Mg}_3\text{N}_2\), Mg starts as 0 and becomes +2; N starts as 0 and becomes -3.
02

Identify the changes in oxidation states

Determine which elements have an increase or decrease in their oxidation states, indicating oxidation or reduction. a. Bromine decreases its oxidation state from 0 to -1 (reduced), and Gallium increases from 0 to +3 (oxidized). b. Hydrogen decreases from +1 to 0 (reduced), and Zinc increases from 0 to +2 (oxidized). c. Nitrogen decreases from 0 to -3 (reduced), and Magnesium increases from 0 to +2 (oxidized).
03

Identify the oxidized and reduced species

Based on changes in oxidation states, confirm which species are oxidized and which are reduced. a. Bromine is reduced and Gallium is oxidized. b. Hydrogen is reduced and Zinc is oxidized. c. Nitrogen is reduced and Magnesium is oxidized.

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Key Concepts

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

Oxidation States
Oxidation states are like labels that help us keep track of electrons during chemical reactions. They tell us what happens to each atom: whether it gains or loses electrons. Think of oxidation states as the "charge" an atom would have if the compound it is part of was composed of ions. To assign oxidation states, we follow simple rules:
  • For a free element (one not combined with any other element), the oxidation state is zero. For example, in \( \text{Br}_2 \), each bromine has an oxidation state of 0.
  • For a simple ion, the oxidation state is the charge of the ion.
  • In compounds, hydrogen is usually +1 and oxygen is usually -2, except in peroxides or when bonded to fluorine.
  • The sum of the oxidation states in a neutral compound is zero, while for a polyatomic ion it equals the ion's charge.
By using these guidelines, we can determine the oxidation states for elements in a reaction like \( 3 \text{Br}_2 + 2\text{Ga} \rightarrow 2 \text{GaBr}_3 \), where bromine shifts from oxidation state 0 to -1, and gallium from 0 to +3.
Species Oxidized
A species is oxidized when it loses electrons, resulting in an increase in its oxidation state. In a redox reaction, the species that undergoes this change is referred to as the "reducing agent" because it donates electrons to another species.Consider the equation \( \text{HCl} + \text{Zn} \rightarrow \text{ZnCl}_2 + \text{H}_2 \). Here, zinc (Zn) starts with an oxidation state of 0 in its elemental form. As it forms \( \text{ZnCl}_2 \), its oxidation state increases to +2, signifying the loss of electrons and hence, oxidation.When identifying the "species oxidized," focus on the reactant side.
  • Note which atoms have an increased oxidation state in the products.
  • For example: In the \( \text{Mg} + \text{N}_2 \rightarrow \text{Mg}_3 \text{N}_2 \) reaction, magnesium (Mg) is oxidized because it goes from 0 to +2.
Species Reduced
In a redox process, reduction refers to the gain of electrons, which causes a decrease in oxidation state. The species that undergoes reduction is known as the "oxidizing agent" because it accepts electrons from another species.Using \( 3 \text{Br}_2 + 2 \text{Ga} \rightarrow 2 \text{GaBr}_3 \) as an example:
  • Bromine (Br) starts in molecule \( \text{Br}_2 \) with an oxidation state of 0.
  • Its oxidation state becomes -1 in \( \text{GaBr}_3 \), showing that it has gained electrons and been reduced.
For any reaction, identify which element's oxidation state decreases to determine the species that is reduced. Such analyses help reveal which reactants serve as oxidizing agents in the reaction.
Oxidation
Oxidation involves the loss of electrons, which results in an increase in the oxidation state of an element. It can often be memorized with the acronym LEO ("Loss of Electrons is Oxidation").Consider the reaction: \( \text{HCl} + \text{Zn} \rightarrow \text{ZnCl}_2 + \text{H}_2 \). Here, zinc (Zn) undergoes oxidation:
  • Starting oxidation state: 0 (in elemental state)
  • Ending oxidation state: +2 (when forming \( \text{ZnCl}_2 \))
This increase denotes that zinc has lost electrons and been oxidized. Identifying the oxidized species in redox reactions helps understand which elements act as reducing agents, crucial for balancing equations and predicting reaction outcomes.
Reduction
Reduction is the other half of redox reactions, which involves the gain of electrons and a decrease in oxidation state. The saying "GER" ("Gain of Electrons is Reduction") can help remember this concept.Take for example the reaction \( \text{Mg} + \text{N}_2 \rightarrow \text{Mg}_3 \text{N}_2 \):
  • Nitrogen (N) starts with an oxidation state of 0 when in \( \text{N}_2 \).
  • In \( \text{Mg}_3 \text{N}_2 \), the oxidation state decreases to -3, indicating electron gain.
Recognizing reduction within a redox reaction highlights the elements acting as oxidizing agents, crucial for chemists to decipher reactivity patterns and control chemical processes effectively.

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Most popular questions from this chapter

Write a balanced ionic redox equation using the following pairs of redox half- reactions. \begin{equation} \begin{array}{l}{\text { a. } \mathrm{Fe} \rightarrow \mathrm{Fe}^{2+}+2 \mathrm{e}^{-}} \\ {\text { } \quad \mathrm{Te}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Te}} \\ {\text { b. } \mathrm{IO}_{4}^{-}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}} \\ {\quad \text { Al } \rightarrow \mathrm{Al}^{3+}+3 \mathrm{e}^{-}(\text { in acid solution })} \\\ {\text { c. } \mathrm{I}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{I}^{-}} \\\\\ \quad{\mathrm{N}_{2} \mathrm{O} \rightarrow \mathrm{NO}_{3}^{-}+4 \mathrm{e}^{-}(\text { in acid solution })}\end{array} \end{equation}

Determine the oxidation number of the boldface element in the following formulas for compounds. a. Na\(Cl\)O\(_{4}\) b. Al\(P\)O\(_4\) c. H\(N\)O\(_{2}\)

A gaseous sample occupies 32.4 \(\mathrm{mL}\) at \(-23^{\circ} \mathrm{C}\) and 0.75 atm. What volume will it occupy at \(\mathrm{STP}\)? (Chapter 13\()\)

Use the oxidation-number method to balance the following ionic redox equations. \begin{equation} \begin{array}{l}{\text { a. } \mathrm{Al}+\mathrm{I}_{2} \rightarrow \mathrm{Al}^{3+}+\mathrm{I}^{-}} \\ {\text { b. } \mathrm{MnO}_{2}+\mathrm{Br}^{-} \rightarrow \mathrm{Mn}^{2+}+\mathrm{Br}_{2}(\text { in acid solution })}\end{array} \end{equation}

Define oxidation number.
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