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.

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 Vaia!

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.

One App. One Place for Learning.

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

Get started for free

Most popular questions from this chapter

Hydrazine is somewhat toxic. Use the half-reactions shown below to explain why household bleach (a highly alkaline solution of sodium hypochlorite) should not be mixed with household ammonia or glass cleansers that contain ammonia. \(\mathrm{ClO}^{-}+\mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{OH}^{-}+\mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=0.90 \mathrm{~V}\) \(\mathrm{N}_{2} \mathrm{H}_{4}+2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{NH}_{3}+2 \mathrm{OH}^{-} \quad \mathscr{E}^{\circ}=-0.10 \mathrm{~V}\)

In making a specific galvanic cell, explain how one decides on the electrodes and the solutions to use in the cell.

The black silver sulfide discoloration of silverware can be removed by heating the silver article in a sodium carbonate solution in an aluminum pan. The reaction is $$3 \mathrm{Ag}_{2} \mathrm{~S}(s)+2 \mathrm{Al}(s) \rightleftharpoons 6 \mathrm{Ag}(s)+3 \mathrm{~S}^{2-}(a q)+2 \mathrm{Al}^{3+}(a q)$$ a. Using data in Appendix 4 , calculate \(\Delta G^{\circ}, K\), and \(\mathscr{C}^{\circ}\) for the above reaction at \(25^{\circ} \mathrm{C}\). (For \(\mathrm{Al}^{3+}(a q), \Delta G_{\mathrm{f}}^{\circ}=-480 . \mathrm{kJ} / \mathrm{mol}\).) b. Calculate the value of the standard reduction potential for the following half-reaction: $$2 \mathrm{e}^{-}+\mathrm{Ag}_{2} \mathrm{~S}(s) \longrightarrow 2 \mathrm{Ag}(s)+\mathrm{S}^{2-}(a q)$$

Assign oxidation numbers to all the atoms in each of the following. a. \(\mathrm{HNO}_{3}\) e. \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) i. \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) b. \(\mathrm{CuCl}_{2}\) f. Ag j. \(\mathrm{CO}_{2}\) c. \(\mathrm{O}_{2}\) g. \(\mathrm{PbSO}_{4}\) k. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{3}\) d. \(\mathrm{H}_{2} \mathrm{O}_{2}\) h. \(\mathrm{PbO}_{2}\) 1\. \(\mathrm{Cr}_{2} \mathrm{O}_{3}\)

You have a concentration cell in which the cathode has a silver electrode with \(0.10 \mathrm{MAg}^{+}\). The anode also has a silver electrode with \(\mathrm{Ag}^{+}(a q), 0.050 \mathrm{M} \mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}\), and \(1.0 \times 10^{-3} \mathrm{M} \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\). You read the voltage to be \(0.76 \mathrm{~V}\). a. Calculate the concentration of \(\mathrm{Ag}^{+}\) at the anode. b. Determine the value of the equilibrium constant for the formation of \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\) \(\mathrm{Ag}^{+}(a q)+2 \mathrm{~S}_{2} \mathrm{O}_{3}{ }^{2-}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}(a q) \quad K=?\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free