Solved: Balance each redox reaction using the

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

Problem 61P Chapter 16

Introductory Chemistry | 5th Edition

  • 2901 Step-by-step solutions solved by professors and subject experts
  • Get 24/7 help from StudySoup virtual teaching assistants
Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo Tro

Introductory Chemistry | 5th Edition

4 5 0 343 Reviews
26
3
Problem 61P

Balance each redox reaction using the half-reaction method.

(a) K(s) + Cr3+ (aq) → Cr(s) + K+(aq)

(b) Mg(s) + Ag+ (aq) → Mg2+ (aq) + Ag(s)

(c) Al(s) + Fe2+ (aq) → Al3+ (aq) + Fe(s)

Step-by-Step Solution:

Problem 61P :

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.

Step 2:

(a) K(s) + Cr3+ (aq) → Cr(s) + K+(aq) :

Step 1:

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

Oxidation state of K (s) = 0, its getting oxidized from  0 to +1.

        Cr3+ is getting reduced from +3 to 0.

Step 2:

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

K(s) →  K+(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 :

        K(s) →  K+(aq) + 1 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  [ K(s) →  K+(aq) + 1 e- ]

1  [ 3e- + Cr3+(aq) → Cr (s) ]

Step 6 :

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

        3 K(s)            →  3 K+(aq) + 3 e-

3e- + Cr3+(aq) → Cr (s)

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

        3 K(s) + Cr3+(aq) → 3 K+(aq) + Cr (s)

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

 3 K(s) + Cr3+(aq) → 3 K+(aq) + Cr (s)

(b) Mg(s) + Ag+ (aq) → Mg2+ (aq) + Ag(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.

        Ag+ is getting reduced from +1 to 0.

Step 2:

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

Mg(s) →  Mg2+(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:

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

Mg(s) →  Mg2+(aq) + 2 e-

1e- + Ag+(aq) → Ag (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.

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

2  [ 1e- + Ag+(aq) → Ag (s) ]

Step 6 :

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

        Mg(s)            →  Mg2+(aq) + 2 e-

2e- + 2Ag+(aq) → 2Ag (s)

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

        Mg (s) + 2Ag+(aq) → Mg2+(aq) + 2Ag (s)

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

 Mg (s) + 2Ag+(aq) → Mg2+(aq) + 2Ag (s)

(c) Al(s) + Fe2+ (aq) → Al3+ (aq) + Fe(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.

        Fe2+ is getting reduced from +2 to 0.

Step 2:

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

Al(s) →  Al3+(aq)

Fe2+(aq) → Fe (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 5

Chapter 16, Problem 61P is Solved
Step 5 of 5

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

Unlock Textbook Solution

Enter your email below to unlock your verified solution to:

Solved: Balance each redox reaction using the

×
Log in to StudySoup
Get Full Access to Introductory Chemistry - 5 Edition - Chapter 16 - Problem 61p

Forgot password? Reset password here

Join StudySoup for FREE
Get Full Access to Introductory Chemistry - 5 Edition - Chapter 16 - Problem 61p
Join with Email
Already have an account? Login here
Reset your password

I don't want to reset my password

Need help? Contact support

Need an Account? Is not associated with an account
Sign up
We're here to help

Having trouble accessing your account? Let us help you, contact support at +1(510) 944-1054 or support@studysoup.com

Got it, thanks!
Password Reset Request Sent An email has been sent to the email address associated to your account. Follow the link in the email to reset your password. If you're having trouble finding our email please check your spam folder
Got it, thanks!
Already have an Account? Is already in use
Log in
Incorrect Password The password used to log in with this account is incorrect
Try Again

Forgot password? Reset it here