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Chapter 20: Problem 74

What is a disproportionation reaction? Use the following reduction potentials $$ \begin{aligned} \mathrm{ClO}_{3}^{-} &+3 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}_{2}+\mathrm{H}_{2} \mathrm{O} & & \mathscr{E}^{\circ}=1.21 \mathrm{~V} \\ \mathrm{HClO}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}+\mathrm{H}_{2} \mathrm{O} & & \mathscr{E}^{\circ}=1.65 \mathrm{~V} \end{aligned} $$ to predict whether \(\mathrm{HClO}_{2}\) will disproportionate.

Short Answer

Expert verified
A disproportionation reaction is a type of redox reaction in which a single species is both oxidized and reduced. To see if \(\mathrm{HClO}_{2}\) will disproportionate, we analyze the given reduction potentials. We find the oxidation half-reaction of \(\mathrm{HClO}_{2}\) and the reduction half-reaction of \(\mathrm{HClO}_{2}\), then we calculate the overall potential of the reaction. The reaction involving oxidation of \(\mathrm{HClO}_{2}\) to \(\mathrm{ClO}_{3}^{-}\) and reduction of \(\mathrm{HClO}_{2}\) to \(\mathrm{HClO}\) has an overall potential \(\mathscr{E}_{2}^{\circ}= 0.44\mathrm{~V}\), which is positive. This indicates that the reaction is energetically favorable, and thus, \(\mathrm{HClO}_{2}\) will disproportionate.

Step by step solution

01

Identify the possible oxidation and reduction half-reactions involving HClO2

In this exercise, we can see that \(\mathrm{HClO}_{2}\) is the product of the first reaction and the reactant in the second reaction. Therefore, the possible oxidation half-reaction is: $$ \mathrm{HClO}_{2} \longrightarrow \mathrm{ClO}_{3}^{-} + 3\mathrm{H}^{+} + 2 \mathrm{e}^{-} $$ and the possible reduction half-reaction is: $$ \mathrm{HClO}_{2} + 2\mathrm{H}^{+} + 2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO} + \mathrm{H}_{2}\mathrm{O} $$
02

Calculate reduction potential for the oxidation half-reaction

To find the reduction potential for the oxidation half-reaction, we need to reverse the first given reaction and add together the potentials of the reversed reaction and the reduction half-reaction. Reverse the first reaction to get: $$ \mathrm{HClO}_{2} - 3\mathrm{H}^{+} - 2\mathrm{e}^{-} \longrightarrow \mathrm{ClO}_{3}^{-} + \mathrm{H}_{2}\mathrm{O} $$ with the potential \(\mathscr{E}_{1}^{\circ}= -1.21\mathrm{~V}\). Now, add this reversed reaction to the reduction half-reaction: $$ \mathrm{HClO}_{2} + 3\mathrm{H}^{+} + 2\mathrm{e}^{-} - (\mathrm{HClO}_{2} - 3\mathrm{H}^{+} - 2\mathrm{e}^{-}) \longrightarrow \mathrm{ClO}_{3}^{-} + \mathrm{H}_{2}\mathrm{O} - (\mathrm{HClO} + \mathrm{H}_{2}\mathrm{O}) $$ Combined, this results in: $$ \mathrm{HClO}_{2} - \mathrm{HClO} - \mathrm{e}^{-} \longrightarrow \mathrm{ClO}_{3}^{-} $$ with a potential \(\mathscr{E}_{2}^{\circ}= \mathscr{E}_{1}^{\circ} + \mathscr{E}_{3}^{\circ}= (-1.21 + 1.65) \mathrm{~V} = 0.44\mathrm{~V}\).
03

Determine if disproportionation is favorable

Since the overall potential of the reaction, \(\mathscr{E}_{2}^{\circ}= 0.44\mathrm{~V}\), is positive, the reaction is energetically favorable and will proceed spontaneously. Therefore, \(\mathrm{HClO}_{2}\) will disproportionate in this reaction.

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Key Concepts

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

Reduction Potential
Reduction potential is the measure of the tendency of a chemical species to gain electrons, which indicates its ability to be reduced. The more positive the reduction potential, the higher the tendency to gain electrons. This concept is crucial for predicting the direction of redox reactions.

In the given exercise, the reduction potential for the reaction involving \( \mathrm{HClO}_2 \) transitioning from different states is used to understand its reaction behavior. Two separate values are provided: 1.21 V and 1.65 V, reflecting how keen each reaction is to take electrons.

Reduction potentials help determine whether a reaction can proceed spontaneously; a more positive potential suggests a favorable process. Thus, understanding reduction potentials aids in predicting reaction spontaneity.
Oxidation and Reduction Half-Reactions
Oxidation and reduction half-reactions are the two halves of a redox reaction. Each half-reaction involves either the loss or gain of electrons. Here, oxidation is the loss of electrons, while reduction is the gain of electrons. These half-reactions are balanced in terms of both charges and atoms.

In the example problem, we identify potential half-reactions for \( \mathrm{HClO}_2 \). During oxidation, \( \mathrm{HClO}_2 \) loses electrons to form \( \mathrm{ClO}_3^- \) and during reduction, it gains electrons to form \( \mathrm{HClO} \).

Balancing and understanding these half-reactions are vital because they allow us to see the complete electron transfer process. This knowledge is necessary for calculating the overall reaction potential, which tells us if the reaction is favorable.
Spontaneous Reaction
A spontaneous reaction occurs naturally under given conditions, characterized by a negative change in free energy or a positive reaction potential. These reactions proceed without the need for additional energy input once they have started.

In this exercise, the combined oxidation and reduction half-reaction results in a positive potential of 0.44 V, indicating that the disproportionation of \( \mathrm{HClO}_2 \) is spontaneous.

Recognition of a spontaneous reaction is important in chemistry as it determines the feasibility of chemical reactions occurring under specified conditions. It helps chemists predict whether a process can occur naturally or if external energy is required.
Redox Reactions
Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between substances. One substance gets oxidized by losing electrons, while another is reduced by gaining those electrons.

The disproportionation reaction of \( \mathrm{HClO}_2 \) is an excellent example of a redox process, as it involves the simultaneous oxidation and reduction of the same species. In this reaction, \( \mathrm{HClO}_2 \) both donates and receives electrons, splitting into two different products.

Understanding redox reactions is fundamental because they are prevalent in various chemical processes, including cellular respiration and corrosion. Identifying redox changes allows chemists to comprehend reaction mechanisms and predict product formation.

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Most popular questions from this chapter

The synthesis of ammonia gas from nitrogen gas and hydrogen gas represents a classic case in which a knowledge of kinetics and equilibrium was used to make a desired chemical reaction economically feasible. Explain how each of the following conditions helps to maximize the yield of ammonia. a. running the reaction at an elevated temperature b. removing the ammonia from the reaction mixture as it forms c. using a catalyst d. running the reaction at high pressure

Nitric acid is produced commercially by the Ostwald process, represented by the following equations: $$\begin{aligned}4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)\end{aligned}$$ What mass of \(\mathrm{NH}_{3}\) must be used to produce \(1.0 \times 10^{6} \mathrm{~kg} \mathrm{HNO}_{3}\) by the Ostwald process? Assume \(100 \%\) yield in each reaction and assume that the NO produced in the third step is not recycled.

The compound with the formula \(\mathrm{TII}_{3}\) is a black solid. Given the following standard reduction potentials: $$\begin{aligned}\mathrm{Tl}^{3+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Tl}^{+} & \mathscr{E}^{\circ}=+1.25 \mathrm{~V} \\ \mathrm{I}_{3}^{-}+2 \mathrm{e}^{-} \longrightarrow 3 \mathrm{I}^{-} & \mathscr{E}^{\circ}=+0.55 \mathrm{~V} \end{aligned}$$ would you formulate this compound as thallium(III) iodide or thallium(I) triiodide?

a. Many biochemical reactions that occur in cells require relatively high concentrations of potassium ion \(\left(\mathrm{K}^{+}\right) .\) The concentration of \(\mathrm{K}^{+}\) in muscle cells is about \(0.15 \mathrm{M}\). The concentration of \(\mathrm{K}^{+}\) in blood plasma is about \(0.0050 M\). The high internal concentration in cells is maintained by pumping \(\mathrm{K}^{+}\) from the plasma. How much work must be done to transport \(1.0 \mathrm{~mol} \mathrm{~K}^{+}\) from the blood to the inside of a muscle cell at \(37^{\circ} \mathrm{C}\) (normal body temperature)? b. When \(1.0 \mathrm{~mol} \mathrm{~K}^{+}\) is transferred from blood to the cells, do any other ions have to be transported? Why or why not? c. Cells use the hydrolysis of adenosine triphosphate, abbreviated ATP, as a source of energy. Symbolically, this reaction can be represented as $$\operatorname{ATP}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{ADP}(a q)+\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)$$ where ADP represents adenosine diphosphate. For this reaction at \(37^{\circ} \mathrm{C}, K=1.7 \times 10^{5}\). How many moles of ATP must be hydrolyzed to provide the energy for the transport of \(1.0 \mathrm{~mol}\) \(\mathrm{K}^{+}\) ? Assume standard conditions for the ATP hydrolysis reaction.

One reason suggested to account for the instability of long chains of silicon atoms is that the decomposition involves the transition state shown below: The activation energy for such a process is \(210 \mathrm{~kJ} / \mathrm{mol}\), which is less than either the \(\mathrm{Si}-\mathrm{Si}\) or \(\mathrm{Si}-\mathrm{H}\) energy. Why would a similar mechanism not be expected to be very important in the decomposition of long carbon chains?
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