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Arrange the following substances in order of increasing strength as an oxidizing agent, and account for the trend: (a) \(\mathrm{Mn}^{2+}\) (b) \(\mathrm{MnO}_{2}\) (c) \(\mathrm{MnO}_{4}^{-}\)

Short Answer

Expert verified
Order: \\( \text{Mn}^{2+} < \text{MnO}_{2} < \text{MnO}_{4}^{-} \\). MnO4⁻ is the strongest oxidizing agent.

Step by step solution

01

Understanding Oxidizing Agents

Oxidizing agents gain electrons and are reduced during a chemical reaction. The strength of an oxidizing agent is related to its ability to accept electrons.
02

Redox Potential Review

Redox potential (also known as standard reduction potential) indicates the tendency of a chemical species to be reduced. The more positive the potential, the stronger the oxidizing agent.
03

Consult the Standard Potential Table

Use the standard reduction potential table to identify the potentials for the given forms of manganese: - \( \text{Mn}^{2+} \)- \( \text{MnO}_{2} \)- \( \text{MnO}_{4}^{-} \)
04

Determine Values

- \( \text{Mn}^{2+} \): Typically part of the reaction \( \text{Mn}^{3+}/\text{Mn}^{2+} \, E^\circ = +1.51 \text{ V} \)- \( \text{MnO}_{2} \): Part of the reaction \( \text{MnO}_{2}/\text{Mn}^{2+} \, E^\circ = +1.23 \text{ V} \)- \( \text{MnO}_{4}^{-} \): As part of \( \text{MnO}_{4}^{-}/\text{Mn}^{2+} \, E^\circ = +1.51 \text{ V} \)
05

Arranging in Order

Arrange the substances by increasing order of their standard reduction potentials: \( \text{Mn}^{2+} < \text{MnO}_{2} < \text{MnO}_{4}^{-} \). Thus, \( \text{Mn}^{2+} \) is the weakest and \( \text{MnO}_{4}^{-} \) is the strongest oxidizing agent.

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Key Concepts

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

Standard Reduction Potential
Standard reduction potential, often referred to as redox potential, is a crucial concept in understanding how substances behave as oxidizing agents. It measures the tendency of a chemical species to gain electrons and thus be reduced. A more positive standard reduction potential indicates a greater ability to accept electrons, which in turn signifies a stronger oxidizing agent. To determine the potential of different substances, chemists use standard conditions and refer to tables that list these values for various chemical species.
For instance, in the given exercise,
  • \( \text{Mn}^{2+} \) typically involves a potential of +1.51 V in the context of reactions involving manganese.
  • \( \text{MnO}_{2} \) features a potential of +1.23 V.
  • \( \text{MnO}_{4}^{-} \) also shows a potential of +1.51 V.
By referring to these standard potentials, we can compare the strengths of different oxidizing agents and predict the direction of redox reactions.
Manganese Oxidation States
Manganese is known for its ability to exist in multiple oxidation states, which is why it plays a significant role in redox reactions. The oxidation state of a manganese compound determines its behavior as an oxidizing or reducing agent. In this context, an oxidation state indicates the number of electrons an atom gains or loses in a reaction.
Key manganese oxidation states include:
  • \( \text{Mn}^{2+} \): This is one of the more stable oxidation states of manganese, present in various compounds and reactions.
  • \( \text{MnO}_{2} \): Here, manganese is in the +4 oxidation state, a common state for manganese oxides, known for moderate oxidizing power.
  • \( \text{MnO}_{4}^{-} \): Known as the permanganate ion, this is a very strong oxidizing agent with manganese in the +7 oxidation state.
Understanding these states helps in predicting the behavior of manganese in chemical reactions and evaluating its potential as an oxidizing agent.
Redox Reactions
Redox reactions, short for reduction-oxidation reactions, are chemical processes in which there is a transfer of electrons between two substances. These reactions are fundamental to electrochemistry and involve the change in oxidation states of the involved species.
In a redox reaction:
  • Oxidation refers to the loss of electrons.
  • Reduction involves the gain of electrons.
  • Oxidizing agents are the ones that get reduced by accepting electrons.
  • Reducing agents donate electrons and get oxidized in the process.
By analyzing the standard reduction potentials of various substances, chemists can predict which species will act as oxidizing and reducing agents in a given reaction.
Electrochemistry
Electrochemistry is the branch of chemistry that deals with the relationship between electrical energy and chemical changes. It explores how reactions involving electron transfer can generate electricity, which is the basis for many applications like batteries and electroplating.
Electrochemical cells are devices that convert chemical energy into electrical energy or vice versa. Understanding the principles of redox reactions and standard reduction potentials is crucial in designing these cells. In the context of the exercise, learning how manganese species behave as oxidizing agents can help in constructing efficient electrochemical cells.
Through electrochemistry, we can also calculate the feasibility of a redox reaction occurring. If the standard reduction potential of the oxidizing agent is higher than that of the reducing agent, the reaction is likely to be spontaneous under standard conditions.
Electron Transfer
Electron transfer is the movement of electrons from one atom or molecule to another and is a fundamental concept in both chemistry and biology. This transfer is what takes place in redox reactions, where one species donates electrons and the other accepts them.
In terms of manganese in the exercise, different species like \(\text{Mn}^{2+}\), \(\text{MnO}_{2}\), and \(\text{MnO}_{4}^{-}\) participate in electron transfer, altering their oxidation states in the process. The number of electrons exchanged and the standard reduction potential of each species dictate the direction and extent of the electron transfer.
  • When \(\text{Mn}^{2+}\) gains electrons, it might convert to a lower oxidation state.
  • \(\text{MnO}_{4}^{-}\) can receive electrons readily due to its high reduction potential, making it a potent oxidizing agent.
Analyzing electron transfer is essential to understanding how substances react and why some compounds are more effective oxidizing agents than others.

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Most popular questions from this chapter

The reaction of the octahedral complex \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{NO}_{2}\right)_{3}\) with \(\mathrm{HCl}\) yields a complex \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}_{2}\right]^{+}\) in which the two chloride ligands are trans to one another. (a) Draw the two possible stereoisomers of the starting material \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{NO}_{2}\right)_{3}\right] .\) (All three \(\mathrm{NO}_{2}^{-}\) ligands are bonded to Co through the \(\mathrm{N}\) atom. ) (b) Assuming that the \(\mathrm{NH}_{3}\) groups remain in place, which of the two starting isomers could give rise to the observed product?

The \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) ion is violet, and \(\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{3-}\) is yellow. Ex- plain this difference using crystal field theory. Use the colors to order \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CN}^{-}\) in the spectrochemical series.

Formation constants for the ammonia and ethylenediamine complexes of nickel(II) indicate that \(\mathrm{Ni}(\mathrm{en})_{3}{ }^{2+}\) is much more stable than \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) : (1) \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \rightleftharpoons\) \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) \quad K_{\mathrm{f}}=2.0 \times 10^{8}\) (2) \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}(a q)+3 \mathrm{en}(a q) \rightleftharpoons\) \(\mathrm{Ni}(\mathrm{en})_{3}^{2+}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) \quad K_{\mathrm{f}}=4 \times 10^{17}\) The enthalpy changes for the two reactions, \(\Delta H_{1}^{\circ}\) and \(\Delta H_{2}^{\circ}\), should be about the same because both complexes have six Ni-N bonds. (a) Which of the two reactions should have the larger entropy change, \(\Delta S^{\circ} ?\) Explain. (b) Account for the greater stability of \(\mathrm{Ni}(\mathrm{en})_{3}^{2+}\) in terms of the relative values of \(\Delta S^{\circ}\) for the two reactions. (c) Assuming that \(\Delta H_{2}^{\circ}-\Delta H^{\circ}\) is zero, calculate the value of \(\Delta S_{2}^{\circ}-\Delta S_{1}^{\circ} .\)

Write the formula for each of the following compounds: (a) Tetraammineplatinum(II) chloride (b) Sodium hexacyanoferrate(III) (c) Tris(ethylenediamine)platinum(IV) sulfate (d) Triamminetrithiocyanatorhodium(III)

Draw a crystal field energy-level diagram, and predict the number of unpaired electrons for each of the following: (a) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (b) \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\) (c) \(\left[\mathrm{FeO}_{4}\right]^{2}\) (d) \(\left[\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) (low-spin)

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