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Balance each redox reaction using the half-reaction

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

Solution for problem 62P Chapter 16

Introductory Chemistry | 5th Edition

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

Introductory Chemistry | 5th Edition

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Problem 62P

Balance each redox reaction using the half-reaction method.

(a) Zn(s) + Sn2+ (aq) → Zn2+ (aq) + Sn(s)

(b) Mg(s) + Cr3+ (aq) → Mg2+ (aq) + Cr(s)

(c) Al(s) + Ag+ (aq) → Al3+ (aq) + Ag(s)

Step-by-Step Solution:

Problem 62P :

Step 1:

Here, we have to balance each redox reaction using the half-reaction method :

Half reaction is either the oxidation or the reduction reaction of a redox reaction.

Steps to balance redox reaction using the half reaction method :

First, we assign oxidation states to all atoms and then identify the substances that are being oxidized and reduced. Then we separate the overall reaction into two half-reactions - oxidation and reduction. Next, we balance each half-reaction with respect to mass in the following order : Balance all elements other than H and O.  Balance O by adding H2O. Balance H by adding H+.Next, we balance each half-reaction with respect to charge by adding electrons to the right side of the oxidation half-reaction and the left side of the reduction half-reaction, making the sum of the charges on both sides of each equation equal.Next, we equal the number of electrons in both half-reactions by multiplying one or both half-reactions by a small whole number.Then we add the two half-reactions together, canceling electrons and other species as necessary.Finally we verify that the reaction is balanced with respect to both mass and charge.

Step 2:

(a) Zn(s) + Sn2+ (aq) → Zn2+ (aq) + Sn(s) :

Step 1:

let’s assign the oxidation states and identify the substances being oxidized and reduced :

Oxidation state of Zn (s) = 0, its getting oxidized from  0 to +2.

        Sn2+ is getting reduced from +2 to 0.

Step 2:

Now, let’s write the oxidation and reduction rates separately :

Zn(s) →  Zn2+(aq)

Sn2+(aq) → Sn (s)

Step 3:

All other elements except H and O are balanced in the reaction and there are no oxygen or

 hydrogen here. So we proceed to next step.  

Step 4:

Now, we balance each half reaction with respect to charge :

Zn(s) →  Zn2+(aq) + 2 e-

2e- + Sn2+(aq) → Sn (s)

Step 5:

Now, we equal the number of electrons in both half-reactions by multiplying one or both half-reactions by a small whole number. Here as we see the number of electrons in both the equations are equal and hence we proceed to next step

Step 6 :

Now, we add the two half-reactions together, canceling electrons and other species as necessary.

        Zn(s)            →  Zn2+(aq)  + 2 e-

2e- + Sn2+(aq) → Sn (s)

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

        Zn(s) +  Sn2+(aq) → Zn2+(aq) + Sn (s)

Hence, the balanced redox reaction using the half-reaction method is :

 Zn(s) +  Sn2+(aq) → Zn2+(aq) + Sn (s)

(b) Mg(s) + Cr3+ (aq) → Mg2+ (aq) + Cr(s) :

Step 1:

let’s assign the oxidation states and identify the substances being oxidized and reduced :

Oxidation state of Mg (s) = 0, its getting oxidized from  0 to +2.

        Cr3+ is getting reduced from +3 to 0.

Step 2:

Now, let’s write the oxidation and reduction rates separately :

Mg(s) →  Mg2+(aq)

Cr3+(aq) → Cr (s)

Step 3:

All other elements except H and O are balanced in the reaction and there are no oxygen or

 hydrogen here. So we proceed to next step.  

Step 4:

Now, we balance each half reaction with respect to charge :

        Mg(s) →  Mg2+(aq)+ 2 e-

3e- + Cr3+(aq) → Cr (s)

Step 5:

Now, we equal the number of electrons in both half-reactions by multiplying one or both half-reactions by a small whole number.

        3  [ Mg(s) →  Mg2+(aq)+ 2 e- ]

2  [ 3e- + Cr3+(aq) → Cr (s) ]

Step 6 :

Now, we add the two half-reactions together, canceling electrons and other species as necessary.

        3 Mg(s)            →  3Mg2+(aq) + 6 e-

6e- + 2 Cr3+(aq) → 2 Cr (s)

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

        3 Mg(s) +  2 Cr3+(aq) →  3Mg2+(aq) + 2 Cr (s)  

Hence, the balanced redox reaction using the half-reaction method is :

 3 Mg(s) +  2 Cr3+(aq) →  3Mg2+(aq) + 2 Cr (s)  

(c) Al(s) + Ag+ (aq) → Al3+ (aq) + Ag(s) :

Step 1:

let’s assign the oxidation states and identify the substances being oxidized and reduced :

Oxidation state of Al (s) = 0, its getting oxidized from  0 to +3.

        Ag+ is getting reduced from +1 to 0.

Step 2:

Now, let’s write the oxidation and reduction rates separately :

Al(s) →  Al3+(aq)

Ag+(aq) → Ag (s)

Step 3:

All other elements except H and O are balanced in the reaction and there are no oxygen or

 hydrogen here. So we proceed to next step.  

Step 4 of 6

Chapter 16, Problem 62P is Solved
Step 5 of 6

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

The full step-by-step solution to problem: 62P from chapter: 16 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM. This full solution covers the following key subjects: reaction, balance, method, half, redox. This expansive textbook survival guide covers 19 chapters, and 2045 solutions. The answer to “Balance each redox reaction using the half-reaction method.(a) Zn(s) + Sn2+ (aq) ? Zn2+ (aq) + Sn(s)(b) Mg(s) + Cr3+ (aq) ? Mg2+ (aq) + Cr(s)(c) Al(s) + Ag+ (aq) ? Al3+ (aq) + Ag(s)” is broken down into a number of easy to follow steps, and 35 words. This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. Since the solution to 62P from 16 chapter was answered, more than 322 students have viewed the full step-by-step answer. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295.

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