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Chapter 17: Problem 53

Use this unbalanced redox equation $$ \mathrm{KMnO}_{4}+\mathrm{HCl} \longrightarrow \mathrm{KCl}+\mathrm{MnCl}_{2}+\mathrm{H}_{2}+\mathrm{Cl}_{2} $$ to indicate (a) the oxidizing agent (b) the reducing agent (c) the number of electrons that are transferred per mole of oxid izing agcnt

Short Answer

Expert verified
Oxidizing agent: KMnO\(_4\); Reducing agent: HCl; Electrons transferred: 10 per mole of KMnO\(_4\).

Step by step solution

01

Identify the oxidation states of elements

Assign oxidation states to each element in the equation \( \text{KMnO}_4 + \text{HCl} \rightarrow \text{KCl} + \text{MnCl}_2 + \text{H}_2 + \text{Cl}_2 \). Manganese (Mn) in permanganate (KMnO\(_4\)) has an oxidation state of +7, while in MnCl\(_2\), it is +2. Chlorine (Cl) in HCl has an oxidation state of -1, while in Cl\(_2\), it is 0.
02

Determine the changes in oxidation states

Note the changes in oxidation states for manganese and chlorine. Manganese changes from +7 in KMnO\(_4\) to +2 in MnCl\(_2\), indicating it is reduced. Chlorine changes from -1 in HCl to 0 in Cl\(_2\), indicating it is oxidized.
03

Identify the oxidizing and reducing agents

The substance that is reduced (KMnO\(_4\)) is the oxidizing agent, and the one that is oxidized (HCl) is the reducing agent.
04

Calculate the number of electrons transferred

Determine the number of electrons transferred for the changes in oxidation states. Manganese changes from +7 to +2, a gain of 5 electrons per Mn atom. Chlorine changes from -1 to 0, a loss of 1 electron per Cl atom. Since there are 2 chlorine atoms in one Cl\(_2\) molecule, the total number of electrons lost by chlorine is 2. The reduction and oxidation must balance, so the number of electrons transferred is the least common multiple of 5 (Mn) and 2 (Cl), which is 10.

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Key Concepts

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

oxidizing agent
An oxidizing agent is a substance that gains electrons during a chemical reaction. It helps another substance to lose electrons. Therefore, the oxidizing agent itself gets reduced in the process.

In the provided redox equation, \textbf{KMnO}\(_4\) is the oxidizing agent. Let's break this down:

- \textbf{KMnO}\(_4\) contains manganese with a +7 oxidation state.
- Manganese in \textbf{KMnO}\(_4\) gets reduced to a +2 oxidation state in \textbf{MnCl}\(_2\).
- Hence, \textbf{KMnO}\(_4\) accepts 5 electrons per manganese atom, making it reduced and acting as the oxidizing agent.

Oxidizing agents are crucial in various chemical processes, including combustion, respiration, and industrial applications. When using oxidizing agents, ensure to follow safety guidelines as they can be reactive and sometimes hazardous.
reducing agent
A reducing agent is a substance that loses electrons in a chemical reaction. It donates electrons to another substance, thus causing that substance to be reduced. Consequently, the reducing agent itself becomes oxidized.

In our redox reaction, \textbf{HCl} serves as the reducing agent. Here's why:

- \textbf{HCl} contains chlorine with a -1 oxidation state.
- Chlorine in \textbf{HCl} is oxidized to a 0 oxidation state in \textbf{Cl}\(_2\).
- This indicates that \textbf{HCl} loses electrons (1 electron per chlorine atom, but since there are 2 chlorine atoms involved in forming \textbf{Cl}\(_2\), the total is 2 electrons lost).

Reducing agents are essential in processes such as metal extraction and organic synthesis. Understanding the role of reducing agents can help in better grasping chemical reactions and reactivity trends.
electron transfer
Electron transfer is a fundamental concept in redox reactions. It involves the movement of electrons from the reducing agent to the oxidizing agent.

Let's recap the electron transfer in the given redox reaction:

- Manganese in \textbf{KMnO}\(_4\) gains 5 electrons as it reduces from an oxidation state of +7 to +2.
- Chlorine in \textbf{HCl} loses 1 electron per atom (2 electrons total since \textbf{Cl}\(_2\) forms) as it oxidizes from an oxidation state of -1 to 0.
- To balance the electron transfer, we calculate the least common multiple of 5 (electrons gained by Mn) and 2 (electrons lost by Cl), which is 10.

Thus, 10 electrons are transferred per mole of oxidizing agent (\textbf{KMnO}\(_4\)). This balance ensures that the number of electrons lost equals the number of electrons gained, making the process consistent with the law of conservation of mass and charge.

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Most popular questions from this chapter

Alcohols can be oxidized very easily by potassium dichromate. This fact formed the basis for many of the early breathalyzers used by law eaforcement officers to determine the approximate blood alcohol levels of suspected drunk drivers. The unbalanced equation for the reaction is $$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{Cr}_{2} \mathrm{O}^{2} \rightarrow \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{Cr}^{3} $$ In this reaction, the potassium dichromate is orange and the chromium(III) jons are blue. The bluer a solution of potassium dichromate becomes, the greater the concentration of alcohol in the breath. Write a balanced ionic cquation for this reaction in an acidic cnvironment.

The following observations were made concerning metals \(A, B, C\), and \(D\). (a) When a strip of metal \(\mathrm{A}\) is placed in a solution of \(\mathrm{B}^{2+}\) aons, no reaction is observed. (b) Similarly, \(A\) in a solution containing \(\mathrm{C}^{+}\)ions produces no reaction. (c) When a strip of metal \(D\) is placed in a solut?on of " \(\mathrm{C}^{+}\)àns, black metallic C deposits on the surface of D, and the solution tests posatively for \(\mathrm{D}^{2+}\) tons. (d) When a piece of metallic B is placed in a solution of \(D^{2+}\) ions, metallic \(D\) appears on the surface of \(\mathrm{B}\) and \(\mathrm{B}^{2+}\) jons are found in the solution. Arrange the ions \(\mathrm{A}^{4}, \mathrm{~B}^{2+}, \mathrm{C}^{4}\), and \(\mathrm{D}^{24}\) in order of their ability to attract clectrons. List them in order of increasing ability.

What mass of \(\mathrm{KMnO}_{4}\) is needed to react with \(100, \mathrm{~mL}, \mathrm{H}_{2} \mathrm{O}_{2}\) solution? ( \(a=1,031 \mathrm{~g} / \mathrm{mL}, 9.0^{3} \% \mathrm{H}_{2} \mathrm{O}_{2}\) by mass) $$ \begin{aligned} \mathrm{H}_{2} \mathrm{O}_{2}+& \mathrm{KMnO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \\ & \mathrm{O}_{2}+\mathrm{MnSO}_{4}+\mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O} \quad \text { (acídic solution) } \end{aligned} $$

In one type of alkaline cell used to power devices such as portable radios, \(\mathrm{Hg}^{20}\) ions are reduced to metallac mercury when the cell is bcing discharged. Does this reduction occur at the anode or the cathode? Explain.

Determine whether the following oxidation-reduction reactions ate balanced correctly, If they are not, provide the correct balanced reaction. (a) unbalanced: \(\mathrm{MnO}_{2}(s)+\mathrm{Al}(s) \longrightarrow \mathrm{Mn}(s)+\mathrm{Al} \mathrm{O}_{3}(s)\) balanced: \(\mathrm{MnO}_{2}(s)+2 A 1(s) \longrightarrow \mathrm{Mn}(s)+\mathrm{Al}_{2} \mathrm{O}_{3}(s)\) (b) unbalanced: \(\mathrm{Cu}(s)+\mathrm{Ag}^{4}(a q) \longrightarrow \mathrm{Cu}^{24}(a q)+A \mathrm{~g}(s)\) balanced: \(\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Ag}(s)\) (c) unbalanced: $$ \mathrm{Br}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Br}_{2}(d)+\mathrm{Mn}^{2+}(a q) $$ (icidic solution) balanced: $$ \begin{aligned} 16 \mathrm{H}^{+}(a q)+10 \mathrm{Br}^{-}(a q)+& 2 \mathrm{MnO}_{4}^{-}(a q) \longrightarrow \\ & 5 \mathrm{Br}(l)+2 \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ (d) unhalanced: $$ \begin{aligned} &\mathrm{MnO}_{4}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{MnS}(s)+\mathrm{S}(s) \quad \text { (basic solution) } \\ &\text { balanced: } \\ &8 \mathrm{H}^{-1}(a q)+\mathrm{MnO}_{4}^{-}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \\ &\qquad \mathrm{S}(s)+\mathrm{MnS}(s)+4 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$
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