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Chapter 20: Problem 93

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) \(\mathrm{Ni}^{+}(a q)-\rightarrow \rightarrow \mathrm{Ni}^{2+}(a q)+\mathrm{Ni}(s)\) (acidic solution) (b) \(\mathrm{MnO}_{4}^{2-}(a q) \longrightarrow \mathrm{MnO}_{4}^{-}(a q)+\mathrm{MnO}_{2}(s)\) (acidic solution) (c) \(\mathrm{H}_{2} \mathrm{SO}_{3}(a q) \longrightarrow \mathrm{S}(s)+\mathrm{HSO}_{4}^{-}(a q)\) (acidic solution) (d) \(\mathrm{Cl}_{2}(a q)-\longrightarrow \mathrm{Cl}^{-}(a q)+\mathrm{ClO}^{-}(a q)\) (basic solution)

Short Answer

Expert verified
The short answers for the disproportionation reactions are: (a) \(2Ni^{+}(aq) \rightarrow Ni^{2+}(aq) + Ni(s)\) (b) \(4MnO_4^{2-}(aq) + 10H^+(aq) \rightarrow 3MnO_4^-(aq) + MnO_2(s) + 5H_2O(l)\) For reactions (c) and (d), the same step-by-step methodology should be followed to balance the equations using the half-reaction method.

Step by step solution

01

Separate the half-reactions.

In this reaction, \(Ni^{+}\) is both oxidized and reduced. So we have: Oxidation half-reaction: \(Ni^{+}(aq) \rightarrow Ni^{2+}(aq)\) Reduction half-reaction: \(Ni^{+}(aq) \rightarrow Ni(s)\)
02

Balance the atoms in each half-reaction.

Atoms are already balanced in both half-reactions, as there is only one nickel atom on each side.
03

Balance the charges in each half-reaction.

Add electrons in each half-reaction to balance the charges. Oxidation half-reaction: \(Ni^{+}(aq) \rightarrow Ni^{2+}(aq) + e^-\) Reduction half-reaction: \(Ni^{+}(aq) + e^- \rightarrow Ni(s)\)
04

Make the number of electrons equal in both half-reactions.

The number of electrons is already equal in both half-reactions.
05

Add the half-reactions back together.

Combine the half-reactions back together to get the balanced disproportionation reaction: \(2Ni^{+}(aq) \rightarrow Ni^{2+}(aq) + Ni(s)\) The balanced equation for (a) is: \(2Ni^{+}(aq) \rightarrow Ni^{2+}(aq) + Ni(s)\) (b) \(MnO_4^{2-}(aq) \rightarrow MnO_4^-(aq) + MnO_2(s)\) (acidic solution)
06

Separate the half-reactions.

In this reaction, \(MnO_4^{2-}\) is both oxidized and reduced. So we have: Oxidation half-reaction: \(MnO_4^{2-}(aq) \rightarrow MnO_4^-(aq)\) Reduction half-reaction: \(MnO_4^{2-}(aq) \rightarrow MnO_2(s)\)
07

Balance the atoms in each half-reaction.

In the oxidation half-reaction, we have: - Balance the O atoms by adding the appropriate number of water molecules: \(MnO_4^{2-}(aq) \rightarrow MnO_4^-(aq) + H_2O(l)\) - Balance the H atoms by adding H\(^+\) ions in acidic solution: \(MnO_4^{2-}(aq) + 2H^+(aq) \rightarrow MnO_4^-(aq) + H_2O(l)\) In the reduction half-reaction, we have: - Balance the O atoms by adding water molecules: \(MnO_4^{2-}(aq) \rightarrow MnO_2(s) + 2H_2O(l)\) - Balance the H atoms by adding H\(^+\) ions: \(MnO_4^{2-}(aq) + 4H^+(aq) \rightarrow MnO_2(s) + 2H_2O(l)\)
08

Balance the charges in each half-reaction.

Add electrons to each half-reaction to balance the charges: Oxidation half-reaction: \(MnO_4^{2-}(aq) + 2H^+(aq) + e^- \rightarrow MnO_4^-(aq) + H_2O(l)\) Reduction half-reaction: \(MnO_4^{2-}(aq) + 4H^+(aq) + 3e^- \rightarrow MnO_2(s) + 2H_2O(l)\)
09

Make the number of electrons equal in both half-reactions.

Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 1 to make the number of electrons equal: 3(oxidation): \(3MnO_4^{2-}(aq) + 6H^+(aq) + 3e^- \rightarrow 3MnO_4^-(aq) + 3H_2O(l)\) 1(reduction): \(MnO_4^{2-}(aq) + 4H^+(aq) + 3e^- \rightarrow MnO_2(s) + 2H_2O(l)\)
10

Add the half-reactions back together.

Combine the half-reactions and simplify to get the balanced disproportionation reaction: \(3MnO_4^{2-}(aq) + MnO_4^{2-}(aq) + 10H^+(aq) \rightarrow 3MnO_4^-(aq) + MnO_2(s) + 5H_2O(l)\) The balanced equation for (b) is: \(4MnO_4^{2-}(aq) + 10H^+(aq) \rightarrow 3MnO_4^-(aq) + MnO_2(s) + 5H_2O(l)\) Solutions for (c) and (d) can be found using the same methodology, breaking down the reaction into half-reactions and recombining them after balancing.

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Key Concepts

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

Oxidation-Reduction Reaction
In the fascinating world of chemistry, an oxidation-reduction reaction, also known as a redox reaction, encompasses the simultaneous processes of oxidation and reduction. These key chemical reactions involve the transfer of electrons between species, leading to changes in their oxidation states.

Oxidation refers to the loss of electrons, during which the oxidation state of a substance increases. Conversely, reduction describes the gain of electrons, resulting in a decrease in the oxidation state. A helpful mnemonic to remember this concept is 'OIL RIG' - Oxidation Is Loss, Reduction Is Gain.

Disproportionation reactions, a specialized type of redox reaction, showcase a unique characteristic where a single substance undergoes both oxidation and reduction. This occurs when an intermediary oxidation state element transforms into a higher oxidation state (indicating oxidation) and a lower oxidation state (indicating reduction). An example is the transformation of nickel in different oxidation states as illustrated in the exercise provided.
Balance Chemical Equations
The art of balancing chemical equations is a fundamental skill in chemistry. It involves ensuring that the number of atoms for each element is equal on both the reactant and product sides, adhering to the Law of Conservation of Mass. This law posits that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations.

As exemplified in the given disproportionation reactions, the process begins by writing separate half-reactions for oxidation and reduction. Balancing these half-reactions starts with equalizing the number of atoms of each element and culminates in adjusting charges through the addition of electrons. The electrons must cancel out when the half-reactions are added together, creating a fully balanced chemical equation.

To empower students to excel in this process, it's essential to methodically approach balancing by first considering atoms and then charges. This step-by-step methodology prevents overlooking any component and ensures a correct final equation.
Electrochemistry
Electrochemistry is an intriguing branch of chemistry that deals with the interrelation of electrical currents and chemical reactions. This field is rooted in redox reactions, where the flow of electrons from the oxidation process to the reduction process can generate an electric current.

In relation to disproportionation reactions, electrochemistry highlights how a species can act as both an electron donor and acceptor. This dual role leads to interesting applications, such as the creation of batteries and electrochemical cells. Understanding how these reactions work requires comprehension of concepts like electrode potentials and standard reduction potentials, which predict the direction and feasibility of redox processes.

Electrochemical principles are omnipresent in various industries, from energy storage systems to corrosion prevention and beyond. They exemplify the practical significance of chemical equations, as they lay the foundation for designing systems that harness chemical reactions to produce and store energy.

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Most popular questions from this chapter

A voltaic cell utilizes the following reaction: $$ 2 \mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2}(g)-\rightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}^{+}(a q) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf for this cell when \(\left[\mathrm{Fe}^{3+}\right]=2.50 \mathrm{M}\), \(P_{\mathrm{H}_{2}}=0.85 \mathrm{~atm},\left[\mathrm{Fe}^{2+}\right]=0.0010 \mathrm{M}\), and the \(\mathrm{pH}\) in both compartments is \(5.00 ?\)

A \(1 M\) solution of \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) is placed in a beaker with a strip of Cu metal. A \(1 M\) solution of \(\mathrm{SnSO}_{4}\) is placed in a second beaker with a strip of Sn metal. A salt bridge connects the two beakers, and wires to a voltmeter link the two metal electrodes. (a) Which electrode serves as the anode, and which as the cathode? (b) Which electrode gains mass and which loses mass as the cell reaction proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions?

If the equilibrium constant for a one-electron redox reaction at \(298 \mathrm{~K}\) is \(8.7 \times 10^{4}\), calculate the corresponding \(\Delta G^{\circ}\) and \(\bar{E}_{\text {cell }}^{\circ}\).

(a) Based on standard reduction potentials, would you expect copper metal to oxidize under standard conditions in the presence of oxygen and hydrogen ions? (b) When the Statue of Liberty was refurbished, Teflon spacers were placed between the iron skeleton and the copper metal on the surface of the statue. What role do these spacers play?

A voltaic cell similar to that shown in Figure \(20.5\) is constructed. One electrode compartment consists of an aluminum strip placed in a solution of \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\), and the other has a nickel strip placed in a solution of \(\mathrm{NiSO}_{4}\). The overall cell reaction is $$ 2 \mathrm{Al}(s)+3 \mathrm{Ni}^{2+}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+3 \mathrm{Ni}(s) $$ (a) What is being oxidized, and what is being reduced? (b) Write the half-reactions that occur in the two electrode compartments. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the aluminum electrode to the nickel electrode, or from the nickel to the aluminum? (f) In which directions do the cations and anions migrate through the solution? Assume the Al is not coated with its oxide.
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