Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 21: Problem 18

Write a balanced net ionic equation for the disproportionation reaction of (a) hypochlorous acid to chlorine gas and chlorous acid in acidic solution. (b) chlorate ion to perchlorate and chlorite ions.

Short Answer

Expert verified
Question: Write the balanced net ionic equations for the following disproportionation reactions: (a) The reaction of hypochlorous acid to chlorine gas and chlorous acid in acidic solution. (b) The reaction of chlorate ion to perchlorate and chlorite ions. Answer: (a) 3 HOCl + 4H⁺ → Cl₂ + 2 HClO₂ + H₂O (b) 3 ClO₃⁻ → 2 ClO₄⁻ + 2 ClO₂⁻

Step by step solution

01

Write half-reactions for oxidation and reduction

The oxidation half-reaction is the conversion of hypochlorous acid to chlorine gas: \(2 \text{HOCl} \rightarrow \text{Cl}_2 + 2 \text{H}^+ +2e^-\) The reduction half-reaction is the conversion of hypochlorous acid to chlorous acid: \(\text{HOCl} + e^- \rightarrow \text{HClO}_2\)
02

Balance the half-reactions

The oxidation half-reaction is already balanced. The reduction half-reaction must be balanced by adding water. \(\text{HOCl} + e^- + 2H^+ \rightarrow \text{HClO}_2 + H_2O\)
03

Add the balanced half-reactions

To add the half-reactions, first multiply to get equal numbers of electrons, and then combine the half-reactions: \(2 (\text{HOCl} + e^- + 2H^+ \rightarrow \text{HClO}_2 + H_2O) + (2 \text{HOCl} \rightarrow \text{Cl}_2 + 2 \text{H}^+ +2e^-)\)
04

Cancel out the common species and get the net ionic equation

After cancelling out the common species, we get the net ionic equation: \(3 \text{HOCl} + 4H^+ \rightarrow \text{Cl}_2 + 2 \text{HClO}_2 + H_2O\) (b) Chlorate ion to perchlorate and chlorite ions.
05

Write half-reactions for oxidation and reduction

The oxidation half-reaction is the conversion of chlorate ion to perchlorate ion: \(2\text{ClO}_3^- \rightarrow 2\text{ClO}_4^- +2e^-\) The reduction half-reaction is the conversion of chlorate ion to chlorite ion: \(\text{ClO}_3^- + 2e^- \rightarrow \text{ClO}_2^-\)
06

Balance the half-reactions

Both oxidation and reduction half-reactions are already balanced.
07

Add the balanced half-reactions

To add both half-reactions, first multiply the reduction half-reaction by 2 to get equal numbers of electrons, and then combine the half-reactions: \((2\text{ClO}_3^- \rightarrow 2\text{ClO}_4^- +2e^-) + 2 (\text{ClO}_3^- + 2e^- \rightarrow \text{ClO}_2^-)\)
08

Cancel out the common species and get the net ionic equation

After cancelling out the common species, we get the net ionic equation: \(3 \text{ClO}_3^- \rightarrow 2 \text{ClO}_4^- + 2 \text{ClO}_2^-\)

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

Write the formula of a compound of hydrogen with (a) sulfur. (b) nitrogen, which is a liquid at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\). (c) phosphorus, which is a poisonous gas at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

Sulfur dioxide can be removed from the smokestack emissions of power plants by reacting it with hydrogen sulfide, producing sulfur and water. What volume of hydrogen sulfide at \(27^{\circ} \mathrm{C}\) and \(755 \mathrm{~mm} \mathrm{Hg}\) is required to remove the sulfur dioxide produced by a power plant that burns one metric ton of coal containing \(5.0 \%\) sulfur by mass? How many grams of sulfur are produced by the reaction of \(\mathrm{H}_{2} \mathrm{~S}\) with \(\mathrm{SO}_{2} ?\)

The amount of sodium hypochlorite in a bleach solution can be determined by using a given volume of bleach to oxidize excess iodide ion to iodine; \(\mathrm{ClO}^{-}\) is reduced to \(\mathrm{Cl}^{-}\). The amount of iodine produced by the redox reaction is determined by titration with sodium thiosulfate, \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} ; \mathrm{I}_{2}\) is reduced to \(\mathrm{I}^{-}\). The sodium thiosulfate is oxidized to sodium tetrathionate, \(\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\). In this analysis, potassium iodide was added in excess to \(5.00 \mathrm{~mL}\) of bleach \(\left(d=1.00 \mathrm{~g} / \mathrm{cm}^{3}\right)\). If \(25.00 \mathrm{~mL}\) of \(0.0700 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) was required to reduce all the iodine produced by the bleach back to iodide, what is the mass percent of \(\mathrm{NaClO}\) in the bleach?

Calculate the \(\mathrm{pH}\) and the equilibrium concentration of \(\mathrm{HClO}\) in a \(0.10 M\) solution of hypochlorous acid. \(K_{\mathrm{a}} \mathrm{HClO}=2.8 \times 10^{-8}\)

Choose the strongest acid from each group. (a) \(\mathrm{HClO}, \mathrm{HBrO}, \mathrm{HIO}\) (b) \(\mathrm{HIO}, \mathrm{HIO}_{3}, \mathrm{HIO}_{4}\) (c) \(\mathrm{HIO}, \mathrm{HBrO}_{2}, \mathrm{HBrO}_{4}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Analysis

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