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Chapter 20: Problem 97

A disproportionation reaction is an oxidation-reduction reaction in which the same substance is oxidized and reduced. Complete and balance the following disproportionation reactions: (a) \(\mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Fe}(s)+\mathrm{Fe}^{3+}(a q)\) (b) \(\mathrm{Br}_{2}(l) \longrightarrow \mathrm{Br}^{-}(a q)+\mathrm{BrO}_{3}^{-}(a q)\) (acidic solution) (c) \(\mathrm{Cr}^{3+}(a q) \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cr}(s)\) (acidic solution) (d) \(\mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{3}^{-}(a q)\) (acidic solution)

Short Answer

Expert verified
The balanced disproportionation reactions are: (a) \( 2Fe^{2+}(aq) \longrightarrow Fe(s) + 2Fe^{3+}(aq) \). (b) \(3Br_{2}(l) + 12H^{+}(aq) \longrightarrow 6Br^{-}(aq) + 2BrO_3^{-}(aq) + 6H_{2}O(l)\). (c) \( 3Cr^{3+}(aq) + 14H^{+}(aq) \longrightarrow Cr_{2}O_{7}^{2-}(aq) + 2Cr(s) + 7H_{2}O(l)\). (d) \(6NO(g) + 6H_{2}O(l) \longrightarrow 2N_{2}(g) + 6NO_{3}^{-}(aq) + 12H^{+}(aq)\).

Step by step solution

01

Work on reaction (a)

\(Fe^{2+}(aq) \longrightarrow Fe(s) + Fe^{3+}(aq)\) First, identify the changes in oxidation numbers for the iron atoms: \( Fe^{2+}(aq)\) is reduced to \( Fe(s)\): Oxidation number decreases from +2 to 0. \( Fe^{2+}(aq)\) is oxidized to \( Fe^{3+}(aq)\): Oxidation number increases from +2 to +3. To balance the reaction, we have to make sure that the number of electrons transferred is the same during oxidation and reduction. The difference in electrons here is 1: Reduction: \( Fe^{2+}(aq) \longrightarrow Fe(s) + 1e^{-}\). Oxidation: \( Fe^{2+}(aq) \longrightarrow Fe^{3+}(aq) + 1e^{-}\). Now we can write the balanced disproportionation reaction: \( 2Fe^{2+}(aq) \longrightarrow Fe(s) + 2Fe^{3+}(aq) \).
02

Work on reaction (b)

\(Br_{2}(l) \longrightarrow Br^{-}(aq) + BrO_3^{-}(aq)\) (acidic solution) First, identify the changes in oxidation numbers for the bromine atoms: \( Br_{2}(l)\) is reduced to \( Br^{-}(aq)\): Oxidation number decreases from 0 to -1. \( Br_{2}(l)\) is oxidized to \( BrO_3^{-}(aq)\): Oxidation number increases from 0 to +5. To balance the reaction: Reduction: \( Br_{2}(l) \rightarrow 2Br^{-}(aq) + 2e^{-} \). Oxidation: \( Br_{2}(l) + 10e^{-} \rightarrow 2BrO{_3}^{-}(aq) + 12H^{+}(aq) \) (Considering the acidic solution). To balance the electrons, we must multiply the reduction half-reaction by 5: \(5 (Br_{2}(l) \rightarrow 2Br^{-}(aq) + 2e^{-})\). Now, we can write the balanced disproportionation reaction in acidic solution: \(3Br_{2}(l) + 12H^{+}(aq) \longrightarrow 6Br^{-}(aq) + 2BrO_3^{-}(aq) + 6H_{2}O(l)\).
03

Work on reaction (c)

\(Cr^{3+}(aq) \longrightarrow Cr_{2}O_{7}^{2-}(aq) + Cr(s)\) (acidic solution) First, identify the changes in oxidation numbers for the chromium atoms: \( Cr^{3+}(aq)\) is reduced to \( Cr(s)\): Oxidation number decreases from +3 to 0. \( Cr^{3+}(aq)\) is oxidized to \( Cr_{2}O_{7}^{2-}(aq)\): Oxidation number increases from +3 to +6. To balance the reaction: Reduction: \(2Cr^{3+}(aq) + 6e^{-} \rightarrow 2Cr(s)\). Oxidation: \(2Cr^{3+}(aq) \rightarrow Cr_{2}O_{7}^{2-}(aq) + 6e^{-} + 14H^{+}(aq)\) (Considering the acidic solution). Now we can write the balanced disproportionation reaction in acidic solution: \( 3Cr^{3+}(aq) + 14H^{+}(aq) \longrightarrow Cr_{2}O_{7}^{2-}(aq) + 2Cr(s) + 7H_{2}O(l)\).
04

Work on reaction (d)

\(NO(g) \longrightarrow N_{2}(g) + NO_{3}^{-}(aq)\) (acidic solution) First, identify the changes in oxidation numbers for the nitrogen atoms: \( NO(g)\) is reduced to \( N_{2}(g)\): Oxidation number decreases from +2 to 0. \( NO(g)\) is oxidized to \( NO_{3}^{-}(aq)\): Oxidation number increases from +2 to +5. To balance the reaction: Reduction: \(2NO(g) + 2e^{-} \rightarrow N_{2}(g)\). Oxidation: \(4NO(g) + 6H_2O(l) \rightarrow 4NO_{3}^{-}(aq) + 12H^{+}(aq) + 6e^{-}\) (Considering the acidic solution). To balance the electrons, we must multiply the reduction half-reaction by 3: \(3 (2NO(g) + 2e^{-} \rightarrow N_{2}(g))\). Now, we can write the balanced disproportionation reaction in acidic solution: \(6NO(g) + 6H_{2}O(l) \longrightarrow 2N_{2}(g) + 6NO_{3}^{-}(aq) + 12H^{+}(aq)\).

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Key Concepts

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

Oxidation-Reduction Reactions
Oxidation-reduction reactions, often referred to as redox reactions, are a fundamental type of chemical reaction where there is a transfer of electrons between two species. These reactions are characterized by changes in oxidation states. The substance that loses electrons is oxidized, while the substance that gains electrons is reduced.
In a disproportionation reaction, which is a special type of redox reaction, a single substance undergoes both oxidation and reduction simultaneously. For example, in reaction (a):
  • The iron ion, Fe^{2+}$, is both oxidized to \( ext{Fe}^{3+} \) and reduced to \( ext{Fe}(s)\).
Understanding these changes requires recognizing how to track oxidation numbers and balance the resulting equation to reflect the conservation of mass and charge.
Balancing Chemical Equations
Balancing chemical equations ensures that the same number of each type of atom appears on both sides of the reaction. The process respects the Law of Conservation of Mass, where mass is neither created nor destroyed during a chemical reaction.
When balancing a disproportionation reaction, one must balance both mass and charge. In reaction (b), \( ext{Br}_2 \) disproportionates into \( ext{Br}^- \) and \( ext{BrO}_3^- \), requiring careful adjustment of coefficients to match electron transfer between reactants and products:
  • Balance the electrons transferred: Ensure the electrons lost during oxidation match those gained during reduction.
  • Add necessary ions (such as \( ext{H}^+ \)) and water molecules to balance hydrogen and oxygen in the context of an acidic medium.
This comprehensive balancing makes sure no excess charge is left unaccounted for.
Oxidation Numbers
Oxidation numbers act as guidelines to understanding electron transfer in redox reactions. Each atom in a chemical compound is assigned an oxidation number that reflects its electron density relative to a pure element state. The key rules for determining oxidation numbers include:
  • The oxidation number of a free element (not combined with other elements) is always 0.
  • For ions, it is equal to the charge of the ion.
  • In compounds, the sum of oxidation numbers for all atoms equals the compound鈥檚 overall charge.
For instance, in reaction (c), the chromium ion \( ext{Cr}^{3+}\) has an oxidation number of +3, which decreases to 0 when reduced to \( ext{Cr}(s)\) and increases to +6 when it forms \( ext{Cr}_2 ext{O}_7^{2-}\). These variations signify electron gain or loss and are essential for reaction balancing.
Half-Reactions in Acidic Solutions
Redox reactions in acidic solutions often involve breaking the overall reaction into half-reactions to simplify the balancing process. A half-reaction reveals either the oxidation or reduction part of the full reaction. Typically, one starts by identifying the species undergoing oxidation and reduction and then explicitly writing out the half-reactions:
  • For the reduction half: Electrons are gained by the molecule or ion.
  • For the oxidation half: Electrons are lost by the molecule or ion.
In reaction (d), the nitric oxide \( ext{NO}(g)\) reduces to \( ext{N}_2(g)\) and oxidizes to \( ext{NO}_3^-\), essentially performing both functions.
After writing each half-reaction, balance the atoms and charges by adding \( ext{H}^+ \) ions (to balance hydrogen) and \( ext{H}_2 ext{O} \) (to balance oxygen). Finally, adjust for equal electron flow between oxidation and reduction to maintain electroneutrality.

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Most popular questions from this chapter

In the Br酶nsted-Lowry concept of acids and bases, acidbase reactions are viewed as proton-transfer reactions. The stronger the acid, the weaker is its conjugate base. If we were to think of redox reactions in a similar way, what particle would be analogous to the proton? Would strong oxidizing agents be analogous to strong acids or strong bases?

A voltaic cell utilizes the following reaction: \(4 \mathrm{Fe}^{2+}(a q)+\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q) \longrightarrow 4 \mathrm{Fe}^{3+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\) (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when \(\left[\mathrm{Fe}^{2+}\right]=1.3 \mathrm{M},\left[\mathrm{Fe}^{3+}\right]=\) \(0.010 \mathrm{M}, P_{\mathrm{O}_{2}}=50.7 \mathrm{kPa},\) and the \(\mathrm{pH}\) of the solution in the cathode half-cell is \(3.50 ?\)

The capacity of batteries such as a lithium-ion battery is expressed in units of milliamp-hours (mAh). A typical battery of this type yields a nominal capacity of \(2000 \mathrm{mAh}\). (a) What quantity of interest to the consumer is being expressed by the units of \(\mathrm{mAh}\) ? (b) The starting voltage of a fresh lithium-ion battery is \(3.60 \mathrm{~V}\). The voltage decreases during discharge and is \(3.20 \mathrm{~V}\) when the battery has delivered its rated capacity. If we assume that the voltage declines linearly as current is withdrawn, estimate the total maximum electrical work the battery could perform during discharge.

From each of the following pairs of substances, use data in Appendix \(\mathrm{E}\) to choose the one that is the stronger reducing agent: (a) \(\mathrm{Al}(s)\) or \(\mathrm{Mg}(s)\) (b) \(\mathrm{Fe}(s)\) or \(\mathrm{Ni}(s)\) (c) \(\mathrm{H}_{2}(g\), acidic solution) or \(\operatorname{Sn}(s)\) (d) \(\mathrm{I}^{-}(a q)\) or \(\mathrm{Br}^{-}(a q)\)

Magnesium is obtained by electrolysis of molten \(\mathrm{MgCl}_{2}\). (a) Why is an aqueous solution of \(\mathrm{MgCl}_{2}\) not used in the electrolysis? (b) Several cells are connected in parallel by very large copper bars that convey current to the cells. Assuming that the cells are \(96 \%\) efficient in producing the desired products in electrolysis, what mass of \(\mathrm{Mg}\) is formed by passing a current of 97,000 A for a period of 24 h?
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