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Balance the following oxidation-reduction reactions that occur in basic solution. a. \(\mathrm{Al}(s)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnO}_{2}(s)+\mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)\) b. \(\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{Cl}^{-}(a q)+\mathrm{OCl}^{-}(a q)\) c. \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Al}(s) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{AlO}_{2}^{-}(a q)\)

Short Answer

Expert verified
The balanced equations are: a. \(3\mathrm{Al}(s)+4\mathrm{MnO}_{4}^{-}(a q)\rightarrow 4\mathrm{MnO}_{2}(s)+2\mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)\) b. \(2\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2\mathrm{Cl}^{-}(a q)+2\mathrm{OCl}^{-}(a q)\) c. \(3\mathrm{NO}_{2}^{-}(a q)+2\mathrm{Al}(s) \rightarrow 3\mathrm{NH}_{3}(g)+2\mathrm{AlO}_{2}^{-}(a q)\)

Step by step solution

01

Assign oxidation numbers

Assign oxidation numbers to each element in the reactants and products. Al(s): 0 Mn in MnO4^-: +7 O in MnO4^-: -2 Mn in MnO2: +4 O in MnO2: -2 Al in Al(OH)4^-: +3 O in Al(OH)4^-: -2 H in Al(OH)4^-: +1
02

Identify species being oxidized and reduced

Check which elements' oxidation numbers have changed: Al: 0 → +3 (oxidized, loses 3 e-) Mn: +7 → +4 (reduced, gains 3 e-)
03

Balance half-reactions

First balance the atoms and then the charges using H2O or OH- ions. Oxidation half-reaction: Al → Al^3+ + 3e^- Reduction half-reaction: MnO4^- + 3e^- → MnO2 + 2H2O (adding 4OH^- to both sides to balance O atoms) MnO4^- + 3e^- + 4OH^- → MnO2 + 2H2O + 4OH^-
04

Combine half-reactions

Now, combine the two half-reactions, making sure that the same number of electrons are transferred in both sides of the equation. 1Al + 1MnO4^- + 4OH^- → 1MnO2 + 1Al^3+ + 4OH^- + 4e^- + 2H2O 1Al + 1MnO4^- → 1MnO2 + 1Al^3+ + 4e^- + 2H2O 1Al + 1MnO4^- → 1MnO2 + 1Al^3+ + 2H2O (since Al gains 3e^-, only one Al needed here) Finally, add the Al(OH)4^- ion to the last product: Al + MnO4^- → MnO2 + Al(OH)4^- #Reaction b: Balancing Cl2(g) → Cl^-(aq) + OCl^-(aq)# Repeat steps 1-4 for reaction b. #Reaction c: Balancing NO2^-(aq) + Al(s) → NH3(g) + AlO2^-(aq)# Repeat steps 1-4 for reaction c.

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Key Concepts

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

Oxidation Numbers
In chemistry, oxidation numbers offer us a way to keep track of electrons in atoms, molecules, or ions, particularly in redox (reduction-oxidation) reactions. This is essential because it helps us determine which species is oxidized and which is reduced during a reaction.

For example, in a substance such as water (H2O), hydrogen has an oxidation number of +1, and oxygen has an oxidation number of -2. These values adhere to certain rules: The oxidation number of an element in its standard state is always zero, like Aluminum (Al) in its solid form. In compounds, Alkali metals always have an oxidation number of +1, and alkaline earth metals +2. Oxygen typically has an oxidation number of -2, except in peroxides. By assigning these oxidation numbers, you can start to see how electrons are transferred during a redox reaction, providing a guide to balance the reaction.
Half-Reaction Method
The half-reaction method is a systematic approach for balancing redox equations, particularly useful in basic solutions. It involves breaking down complex redox reactions into two simpler parts: oxidation and reduction half-reactions. Each half-reaction is balanced separately for mass and charge.

In basic solutions, you often add OH- ions to balance oxygen atoms and H2O to balance hydrogen atoms. After the individual half-reactions are balanced, you adjust them so that the number of electrons lost in the oxidation half-reaction equals the electrons gained in the reduction half-reaction. These balanced half-reactions are then added together to form a balanced overall redox reaction. This methodical approach simplifies the complex problem of balancing redox reactions, making the process more understandable for learners.
Redox Reactions
Redox reactions are chemical processes where electrons are transferred between atoms or molecules. 'Redox' is short for reduction-oxidation. Every redox reaction is composed of two parts: reduction, where an atom or molecule gains electrons, and oxidation, where an atom or molecule loses electrons.

For instance, when Aluminum (Al) reacts with manganese oxide (MnO4-), Al loses electrons (is oxidized) while Mn gains electrons (is reduced). In balancing redox reactions, the key is ensuring that the number of electrons lost in oxidation equals the number of electrons gained in reduction, maintaining the principle of conservation of charge. Understanding redox reactions is essential for mastering concepts in electrochemistry, metabolism, and corrosion, among others.
Basic Solution Chemistry
Basic solutions are those that have a pH greater than 7 and are characterized by the presence of excess hydroxide ions (OH-). In the context of redox reactions, basic solutions necessitate specific strategies for balancing reactions, different from those used in acidic solutions.

When balancing redox reactions in a basic solution, after balancing the atoms and charges in half-reactions, additional OH- ions are used to balance any H+ ions present. This is because there are no free hydrogen ions in a basic solution, and it maintains the pH by compensating for any acidic elements introduced during the reaction. Additional water molecules may also be added to balance out the oxygen and hydrogen atoms, which should be done with care to avoid compromising the balance of the equation.

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Most popular questions from this chapter

Electrolysis of an alkaline earth metal chloride using a current of \(5.00 \mathrm{~A}\) for \(748 \mathrm{~s}\) deposits \(0.471 \mathrm{~g}\) of metal at the cathode. What is the identity of the alkaline earth metal chloride?

Sketch a galvanic cell, and explain how it works. Look at Figs. \(18.1\) and \(18.2 .\) Explain what is occurring in each container and why the cell in Fig. \(18.2\) "works" but the one in Fig. \(18.1\) does not.

Nerve impulses are electrical "signals" that pass through neurons in the body. The electrical potential is created by the differences in the concentration of \(\mathrm{Na}^{+}\) and \(\mathrm{K}^{+}\) ions across the nerve cell membrane. We can think about this potential as being caused by a concentration gradient, similar to what we see in a concentration cell (keep in mind that this is a very simple explanation of how nerves work; there is much more involved in the true biologic process). A typical nerve cell has a resting potential of about \(-70 \mathrm{mV}\). Let's assume that this resting potential is due only to the \(\mathrm{K}^{+}\) ion concentration difference. In nerve cells, the \(\mathrm{K}^{+}\) concentration inside the cell is larger than the \(\mathrm{K}^{+}\) concentration outside the cell. Calculate the \(\mathrm{K}^{+}\) ion concentration ratio necessary to produce a resting potential of \(-70 . \mathrm{mV}\). $$\frac{\left[\mathrm{K}^{+}\right]_{\text {inside }}}{\left[\mathrm{K}^{+}\right]_{\text {outside }}}=?$$

What reaction will take place at the cathode and the anode when each of the following is electrolyzed? (Assume standard conditions.) a. \(1.0 \mathrm{M}\) KF solution b. \(1.0 \mathrm{M} \mathrm{CuCl}_{2}\) solution c. \(1.0 \mathrm{M} \mathrm{MgI}_{2}\) solution

Consider only the species (at standard conditions) $$\mathrm{Na}^{+}, \mathrm{Cl}^{-}, \mathrm{Ag}^{+}, \mathrm{Ag}, \mathrm{Zn}^{2+}, \mathrm{Zn}, \mathrm{Pb}$$ in answering the following questions. Give reasons for your answers. (Use data from Table 18.1.) a. Which is the strongest oxidizing agent? b. Which is the strongest reducing agent? c. Which species can be oxidized by \(\mathrm{SO}_{4}^{2-}(a q)\) in acid? d. Which species can be reduced by \(\mathrm{Al}(s)\) ?

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