Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 20: Problem 84

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

Short Answer

Expert verified
A disproportionation reaction is a redox reaction in which a single reactant is both oxidized and reduced, resulting in two different products. To predict whether \(\mathrm{HClO}_{2}\) will disproportionate, we combine the given half-reactions to form the potential disproportionation reaction: \(\mathrm{ClO}_{3}^{-}+\mathrm{HClO} \longrightarrow 2\mathrm{HClO}_{2}\) and calculate the standard cell potential: \(\mathscr{E}_{\text{disproportionation}}^{\circ} = 1.21\mathrm{~V} - 1.65\mathrm{~V} = -0.44\mathrm{~V}\) Since the standard cell potential is negative, the disproportionation of \(\mathrm{HClO}_{2}\) is not spontaneous, meaning it will not disproportionate.

Step by step solution

01

Write down the given half-reactions

The given half-reactions are: 1. \(\mathrm{ClO}_{3}^{-}+3 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}_{2}+\mathrm{H}_{2} \mathrm{O}\quad \mathscr{E}^{\circ}=1.21 \mathrm{~V}\) 2. \(\mathrm{HClO}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}+\mathrm{H}_{2} \mathrm{O}\quad \mathscr{E}^{\circ}=1.65 \mathrm{~V}\)
02

Reverse the second half-reaction to make it an oxidation reaction

To make the second half-reaction an oxidation reaction, reverse it and change the sign of the standard cell potential: \(\mathrm{HClO}+\mathrm{H}_{2} \mathrm{O}\longrightarrow \mathrm{HClO}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \quad \mathscr{E}^{\circ}=-1.65 \mathrm{~V}\)
03

Combine the half-reactions to form the potential disproportionation reaction

Now, we can combine the two half-reactions (oxidation and reduction) to form the potential disproportionation reaction: \(\mathrm{ClO}_{3}^{-}+3 \mathrm{H}^{+}+2 \mathrm{e}^{-} + \mathrm{HClO}+\mathrm{H}_{2}\mathrm{O} \longrightarrow\) \(\mathrm{HClO}_{2}+\mathrm{H}_{2}\mathrm{O} +\mathrm{HClO}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-}\) Simplifying the reaction, we get: \(\mathrm{ClO}_{3}^{-}+\mathrm{HClO} \longrightarrow 2\mathrm{HClO}_{2}\)
04

Calculate the standard cell potential for the disproportionation reaction

To calculate the standard cell potential, add the standard cell potentials of the two half-reactions: \(\mathscr{E}_{\text{disproportionation}}^{\circ} = \mathscr{E}_{1}^{\circ} + \mathscr{E}_{2}^{\circ} = 1.21\mathrm{~V} - 1.65\mathrm{~V} = -0.44\mathrm{~V}\) Since the standard cell potential is negative, the disproportionation of \(\mathrm{HClO}_{2}\) is not spontaneous, meaning it will not disproportionate.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Captain Kirk has set a trap for the Klingons who are threatening an innocent planet. He has sent small groups of fighter rockets to sites that are invisible to Klingon radar and put a decoy in the open. He calls this the "fishhook" strategy. Mr. Spock has sent a coded message to the chemists on the fighters to tell the ships what to do next. The outline of the message is Fill in the blanks of the message using the following clues. (1) Symbol of the halogen whose hydride has the second highest boiling point in the series of HX compounds that are hydrogen halides. (2) Symbol of the halogen that is the only hydrogen halide, \(\mathrm{HX}\), that is a weak acid in aqueous solution. (3) Symbol of the element whose existence on the sun was known before its existence on earth was discovered. (4) The Group \(5 \mathrm{~A}\) element in Table \(20.13\) that should have the most metallic character. (5) Symbol of the Group 6 A element that, like selenium, is a semiconductor. (6) Symbol for the element known in rhombic and monoclinic forms. (7) Symbol for the element that exists as diatomic molecules in a yellow-green gas when not combined with another element. (8) Symbol for the most abundant element in and near the earth's crust. (9) Symbol for the element that seems to give some protection against cancer when a diet rich in this element is consumed. (10) Symbol for the smallest noble gas that forms compounds with fluorine having the general formula \(\mathrm{AF}_{2}\) and \(\mathrm{AF}_{4}\) (reverse the symbol and split the letters as shown). (11) Symbol for the toxic element that, like phosphorus and antimony, forms tetrameric molecules when uncombined with other elements (split the letters of the symbol as shown). (12) Symbol for the element that occurs as an inert component of air but is a very prominent part of fertilizers and explosives.

The compound \(\mathrm{Pb}_{3} \mathrm{O}_{4}\) (red lead) contains a mixture of lead(II) and lead(IV) oxidation states. What is the mole ratio of lead(II) to lead(IV) in \(\mathrm{Pb}_{3} \mathrm{O}_{4}\) ?

Phosphoric acid \(\left(\mathrm{H}_{3} \mathrm{PO}_{4}\right)\) is a triprotic acid, phosphorous acid \(\left(\mathrm{H}_{3} \mathrm{PO}_{3}\right)\) is a diprotic acid, and hypophosphorous acid \(\left(\mathrm{H}_{3} \mathrm{PO}_{2}\right)\) is a monoprotic acid. Explain this phenomenon.

You travel to a distant, cold planet where the ammonia flows like water. In fact, the inhabitants of this planet use ammonia (an abundant liquid on their planet) much as earthlings use water. Ammonia is also similar to water in that it is amphoteric and undergoes autoionization. The \(K\) value for the autoionization of ammonia is \(1.8 \times 10^{-12}\) at the standard temperature of the planet. What is the \(\mathrm{pH}\) of ammonia at this temperature?

Lead forms compounds in the \(+2\) and \(+4\) oxidation states. All lead(II) halides are known (and are known to be ionic). Only \(\mathrm{PbF}_{4}\) and \(\mathrm{PbCl}_{4}\) are known among the possible lead(IV) halides. Presumably lead(IV) oxidizes bromide and iodide ions, producing the lead(II) halide and the free halogen: $$ \mathrm{Pb} \mathrm{X}_{4} \longrightarrow \mathrm{PbX}_{2}+\mathrm{X}_{2} $$ Suppose \(25.00 \mathrm{~g}\) of a lead(IV) halide reacts to form \(16.12 \mathrm{~g}\) of a lead(II) halide and the free halogen. Identify the halogen.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation
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