Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 10, Problem 10.121AP
Interpretation Introduction

Interpretation: The balance form of given chemical reaction occurring in basic solution is to be stated.

Concept introduction: In balanced form of chemical reaction number of atoms and total charge of reactants and products are same.

To determine: The balance form of given chemical reaction occurring in basic solution.

Expert Solution & Answer
Check Mark

Answer to Problem 10.121AP

The balanced reaction is,

IO3(aq)+3HSO3(aq)+3OH(aq)I(aq)+3SO42(aq)+3H2O(l)

Explanation of Solution

The given unbalanced chemical reaction is,

IO3(aq)+HSO3(aq)I(aq)+SO42(aq) (1)

In balanced form of chemical reaction number of atoms and total charge of reactants and products are same.

To balance the above equation (1) following steps are followed.

Step 1:

Calculate the oxidation state of each atom of given charge compounds.

IO3

The oxidation state of oxygen is 2.

The oxidation state of iodine is assumed to be x. Thus, the oxidation state of iodine in IO3 is calculated as,

1×x+3×(2)=1x6=1x=+5

HSO3

The oxidation state of H is +1. The oxidation state of oxygen is 2. The oxidation state of sulfur is assumed to be x. Thus, the oxidation state of sulfur in HSO3 is calculated as,

1×(+1)+1×x+3×(2)=1x5=1x=+4

SO42

The oxidation state of oxygen is 2.

The oxidation state of sulfur is assumed to be x. Thus, the oxidation state of sulfur in SO42 is calculated as,

1×x+4×(2)=2x8=2x=+6

I

The oxidation state of I is 1.

Step 2:

Identify the oxidation and reduction reaction and separate them as half reactions.

In oxidation process, oxidation state of atoms increased by some number.

In reduction process, oxidation state of atoms decreased by some number.

Oxidation:HSO3(aq)SO42(aq)+4+6Reduction:IO3(aq)I(aq)+51

Step 3:

Balance the atoms of reactant and product.

Balance O by the addition of one H2O in oxidation half reaction on the right side.

Balance H by the addition of one H+ in oxidation half reaction on the left side. To neutralize H+ add same number of OH on both sides.

Balance O by the addition of three H2O in reduction half reaction on the right side.

Balance H by the addition of six H+ in reduction half reaction on the left side. To neutralize H+ add same number of OH on both sides.

Oxidation:HSO3(aq)+H2O(l)SO42(aq)HSO3(aq)+H2O(l)+3OH(aq)SO42(aq)+3H+(aq)+3OH(aq)Reduction:IO3(aq)I(aq)+3H2O(l)IO3(aq)+6H+(aq)+6OH(aq)I(aq)+3H2O(l)+6OH(aq)

Step 4:

Balance each half reaction with respect to charge.

Balance the charge of oxidation half reaction by the addition of 2e on right side.

Balance the charge of reduction half reaction by the addition of 6e on left side.

Oxidation:HSO3(aq)+H2O(l)+3OH(aq)SO42(aq)+3H+(aq)+3OH(aq)+2eReduction:IO3(aq)+6H+(aq)+6OH(aq)+6eI(aq)+3H2O(l)+6OH(aq)IO3(aq)+6H2O(l)+6eI(aq)+3H2O(l)+6OH(aq)

Step 5:

Make the number of electrons equal on both sides of oxidation and reduction half reaction.

Multiply oxidation half reaction with three on both sides.

Oxidation:3×[HSO3(aq)+H2O(l)+3OH(aq)SO42(aq)+3H+(aq)+3OH(aq)+2e]3HSO3(aq)+3H2O(l)+9OH(aq)3SO42(aq)+9H+(aq)+9OH(aq)+6e3HSO3(aq)+3H2O(l)+9OH(aq)3SO42(aq)+9H2O(l)+6eReduction:IO3(aq)+6H2O(l)+6eI(aq)+3H2O(l)+6OH(aq)

Step 6:

To get overall reaction add the half reactions together and remove common terms.

Oxidation:3HSO3(aq)+3H2O(l)+9OH(aq)3SO42(aq)+9H2O(l)+6eReduction:IO3(aq)+6H2O(l)+6eI(aq)+3H2O(l)+6OH(aq)Overall:IO3(aq)+3HSO3(aq)+3OH(aq)I(aq)+3SO42(aq)+3H2O(l)_¯

Thus, the balanced reaction is,

IO3(aq)+3HSO3(aq)+3OH(aq)I(aq)+3SO42(aq)+3H2O(l)

Conclusion

The balanced reaction is,

IO3(aq)+3HSO3(aq)+3OH(aq)I(aq)+3SO42(aq)+3H2O(l)

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 10 Solutions

Chemistry

Ch. 10 - Prob. 10.4VPCh. 10 - Prob. 10.5VPCh. 10 - Prob. 10.6VPCh. 10 - Prob. 10.8VPCh. 10 - Prob. 10.15QPCh. 10 - Prob. 10.16QPCh. 10 - Prob. 10.17QPCh. 10 - Prob. 10.18QPCh. 10 - Prob. 10.19QPCh. 10 - Prob. 10.20QPCh. 10 - Prob. 10.23QPCh. 10 - Prob. 10.24QPCh. 10 - Prob. 10.25QPCh. 10 - Prob. 10.26QPCh. 10 - Prob. 10.27QPCh. 10 - Prob. 10.28QPCh. 10 - Prob. 10.29QPCh. 10 - Prob. 10.30QPCh. 10 - Prob. 10.31QPCh. 10 - Prob. 10.32QPCh. 10 - Prob. 10.33QPCh. 10 - Prob. 10.34QPCh. 10 - Prob. 10.35QPCh. 10 - Prob. 10.36QPCh. 10 - Prob. 10.37QPCh. 10 - Prob. 10.38QPCh. 10 - Prob. 10.39QPCh. 10 - Prob. 10.40QPCh. 10 - Prob. 10.41QPCh. 10 - Prob. 10.42QPCh. 10 - Prob. 10.43QPCh. 10 - Prob. 10.44QPCh. 10 - Prob. 10.45QPCh. 10 - Prob. 10.46QPCh. 10 - Prob. 10.47QPCh. 10 - Prob. 10.48QPCh. 10 - Prob. 10.49QPCh. 10 - Prob. 10.50QPCh. 10 - Prob. 10.51QPCh. 10 - Prob. 10.52QPCh. 10 - Prob. 10.53QPCh. 10 - Prob. 10.54QPCh. 10 - Prob. 10.55QPCh. 10 - Prob. 10.56QPCh. 10 - Prob. 10.57QPCh. 10 - Prob. 10.58QPCh. 10 - Prob. 10.59QPCh. 10 - Prob. 10.60QPCh. 10 - Prob. 10.61QPCh. 10 - Prob. 10.62QPCh. 10 - Prob. 10.63QPCh. 10 - Prob. 10.64QPCh. 10 - Prob. 10.65QPCh. 10 - Prob. 10.66QPCh. 10 - Prob. 10.67QPCh. 10 - Prob. 10.68QPCh. 10 - Prob. 10.69QPCh. 10 - Prob. 10.70QPCh. 10 - Prob. 10.71QPCh. 10 - Prob. 10.72QPCh. 10 - Prob. 10.73QPCh. 10 - Prob. 10.74QPCh. 10 - Prob. 10.75QPCh. 10 - Prob. 10.76QPCh. 10 - Prob. 10.77QPCh. 10 - Prob. 10.78QPCh. 10 - Prob. 10.79QPCh. 10 - Prob. 10.80QPCh. 10 - Prob. 10.81QPCh. 10 - Prob. 10.82QPCh. 10 - Prob. 10.83QPCh. 10 - Prob. 10.84QPCh. 10 - Prob. 10.85QPCh. 10 - Prob. 10.86QPCh. 10 - Prob. 10.87QPCh. 10 - Prob. 10.88QPCh. 10 - Prob. 10.89QPCh. 10 - Prob. 10.90QPCh. 10 - Prob. 10.91QPCh. 10 - Prob. 10.92QPCh. 10 - Prob. 10.93QPCh. 10 - Prob. 10.94QPCh. 10 - Prob. 10.95QPCh. 10 - Prob. 10.96QPCh. 10 - Prob. 10.97QPCh. 10 - Prob. 10.98QPCh. 10 - Prob. 10.99APCh. 10 - Prob. 10.100APCh. 10 - Prob. 10.101APCh. 10 - Prob. 10.102APCh. 10 - Prob. 10.103APCh. 10 - Prob. 10.104APCh. 10 - Prob. 10.105APCh. 10 - Prob. 10.106APCh. 10 - Prob. 10.107APCh. 10 - Prob. 10.108APCh. 10 - Prob. 10.109APCh. 10 - Prob. 10.110APCh. 10 - Prob. 10.111APCh. 10 - Prob. 10.112APCh. 10 - Prob. 10.113APCh. 10 - Prob. 10.114APCh. 10 - Prob. 10.115APCh. 10 - Prob. 10.116APCh. 10 - Prob. 10.117APCh. 10 - Prob. 10.118APCh. 10 - Prob. 10.119APCh. 10 - Prob. 10.120APCh. 10 - Prob. 10.121AP
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Electrolysis; Author: Tyler DeWitt;https://www.youtube.com/watch?v=dRtSjJCKkIo;License: Standard YouTube License, CC-BY