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When balancing aqueous redox reactions, charge is balanced using ____

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro ISBN: 9780321910295 34

Solution for problem 15Q Chapter 16

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

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Problem 15Q

Problem 15Q

When balancing aqueous redox reactions, charge is balanced using ____.

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to fill in the blank.

Step1:

Redox reactions are the reactions in which oxidation(loss of electrons) and reduction(gain of electrons) occurs simultaneously. While balancing the redox reaction, several steps are followed:

Suppose we are to balance the equation showing the oxidation of Fe2+ ions to Fe3+ ions by dichromate ions (Cr2O7)2– in acidic medium, wherein, (Cr2O7)2– ions are reduced to Cr3+ ions. The following steps are involved in this task.

  •  Produce unbalanced equation for the reaction in ionic form :

 Fe2+(aq) + (Cr2O7)2– (aq) → Fe3+ (aq) + Cr3+(aq)

  • Separate the equation into half reactions:

 Oxidation half : Fe2+ (aq) → Fe3+(aq)

        Reduction half : (Cr2O7)2–(aq) → Cr3+(aq)

  • Balance the atoms other than O and H in each half reaction individually. Here the oxidation half reaction is already balanced with respect to Fe atoms. For the reduction half reaction, we multiply the Cr3+ by 2 to balance Cr atoms

(Cr2O7)2–(aq) → 2Cr3+(aq)

  • For reactions occurring in acidic medium, add H2O to balance O atoms and H+ to balance H atoms. Thus, we get :

(Cr2O7)2– (aq) + 14H+ (aq) → 2Cr3+(aq) + 7H2O

  • Add electrons to one side of the half reaction to balance the charges. If need be, make the number of electrons equal in the two half reactions by multiplying one or both half reactions by appropriate coefficients. The oxidation half reaction is thus rewritten to balance the charge:

Fe2+ (aq) → Fe3+ (aq) + e 

Now in the reduction half reaction there are net twelve positive charges on the left hand

side and only six positive charges on the right hand side. Therefore, we add six electrons

on the left side.

(Cr2O7)2– (aq) + 14H+ (aq) + 6e → 2Cr3+(aq) + 7H2O

To equalise the number of electrons in both the half reactions, we multiply the oxidation

half reaction by 6 and write as :

6Fe2+ (aq) → 6Fe3+(aq) + 6e

  • We add the two half reactions to achieve the overall reaction and cancel the electrons on each side. This gives the net ionic equation as :

 6Fe2+(aq) + (Cr2O7)2–(aq) + 14H+ (aq) → 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)

  • Verify that the equation contains the same type and number of atoms and the same charges on both sides of the equation. This last check reveals that the equation is fully balanced with respect to number of atoms and the charges.

For the reaction in a basic medium, first balance the atoms as is done in acidic medium.

Then for each H+ ion, add an equal number of OH ions to both sides of the equation.

Where H+ and OH appear on the same side of the equation, combine these to give H2O.

Step2:

Thus, we have seen that:

When balancing aqueous redox reactions, charge is balanced using electrons.

                                ---------------------------

Step 2 of 3

Chapter 16, Problem 15Q is Solved
Step 3 of 3

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

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When balancing aqueous redox reactions, charge is balanced using ____