Back to Course

SS2: CHEMISTRY - 1ST TERM

0% Complete
0/0 Steps
  1. Periodicity and Periodic Table I | Week 1
    5 Topics
    |
    1 Quiz
  2. Quantum Numbers Orbitals & Electrical Structure | Week 2
    6 Topics
    |
    1 Quiz
  3. Periodicity and Periodic Table II | Week 3
    12 Topics
    |
    1 Quiz
  4. Periodicity and Periodic Properties III | Week 4
    11 Topics
    |
    1 Quiz
  5. Periodicity and Periodic Properties IV | Week 5
    5 Topics
    |
    1 Quiz
  6. Mass-Volume Relationship in Reaction | Week 6
    8 Topics
    |
    1 Quiz
  7. Types of Reactions: Oxidation and Reduction | Week 7 & 8
    7 Topics
    |
    1 Quiz
  8. Oxidation – Reduction Reaction II | Week 9
    3 Topics
    |
    1 Quiz
  9. Electrode Potential and Electrochemical Cells I | Week 10
    6 Topics
    |
    1 Quiz
  10. Electrode Potential and Electrochemical Cells II | Week 11
    5 Topics
    |
    1 Quiz
  11. Electrolysis I | Week 12
    8 Topics
    |
    1 Quiz
  12. Electrolysis II | Week 13
    8 Topics
    |
    1 Quiz

Quizzes




  • Kofa Study
Lesson Progress
0% Complete

Topic Content:

  • Theory Questions & Answers – Oxidation – Reduction Reaction II

Theory Questions – Oxidation – Reduction Reaction II

1. Balance the following simple ionic equation

a. Zn(s) + Ag+(aq) → Zn2+(s) + Ag(aq)

b. Mg(s) + H+(aq) → Mg2+(s) + H2(g)

c. Fe3+(aq) + I–(aq) → Fe2+(s) + I(s)

d. Ca(s) + H+(aq) → Ca2+(s) + H2(g)

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

2. Balance the following complex ionic equation in acidic medium

a. Cr2O72-(aq) + Fe2+(aq) → Cr3+(aq) + Fe3+(aq)

b. MnO4–(aq) + Pb2+(aq) → Pb4+(aq) + Mn2+(aq)

c. Cr2O72-(aq) + SO2(aq) → Cr3+(aq) + SO2-(aq)

d. Cl2O2–(aq) + Sn2+(aq) → Cl–(aq) + Sn4+(aq)

e. MnO4–(aq) + Fe2+(aq) → Mn2+(aq) + Fe3+(aq)

 

You are viewing an excerpt of this Topic. Subscribe Now to get Full Access to ALL this Subject's Topics and Quizzes for this Term!

Click on the button "Subscribe Now" below for Full Access!

Subscribe Now

Note: If you have Already Subscribed and you are seeing this message, it means you are logged out. Please Log In using the Login Button Below to Carry on Studying!


Responses

Your email address will not be published. Required fields are marked *

Evaluation Questions

1. Balance the following simple ionic equation

a. Zn(s) + Ag+(aq) → Zn2+(s) + Ag(aq)

Solution  ⇒ \( \scriptsize \underset{0}{Zn_{(s)}} \: + \: \underset{+1}{Ag^+_{(aq)}} \: \rightarrow \: \underset{+2}{Zn^+_{(s)}} \: + \: \underset{0}{Ag{(aq)}} \)

Zn(s) → Zn2+(aq) (oxidation) equation (I)

Ag+(aq) → Ag(aq) (reduction) equation (II)

Zn(s) → Zn2+(aq) + 2e– × 1

Ag+(aq)  + e– →  Ag(aq)  × 2

Zn(s) → Zn2+(aq) + 2e–

2Ag+(aq)  + 2e– →  2Ag(aq)

⇒ Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(aq)

 

b. Mg(s) + H+(aq) → Mg2+(s) + H2(g)

Solution  ⇒ \( \scriptsize \underset{0}{Mg_{(s)}} \: + \: \underset{+1}{H^+_{(aq)}} \: \rightarrow \: \underset{+2}{Mg^{2+}_{(s)}} \: + \: \underset{0}{H_2{(g)}} \)

Mg(s) → Mg2+(aq) (oxidation) equation (I)

H+(aq) → H2(g) (reduction) equation (II)

Mg(s) → Mg2+(aq) + 2e–

2H+(aq)  + 2e– →  H2(g)

Mg(s) + 2H+(aq)  → Mg2+ +  H2(g)

 

 

c. Fe3+(aq) + I–(aq) → Fe2+(s) + I(s)

Solution  ⇒ \( \scriptsize \underset{+3}{Fe^{3+}_{(aq)}} \: + \: \underset{-1}{I^-_{(aq)}} \: \rightarrow \: \underset{+2}{Fe^{2+}_{(s)}} \: + \: \underset{0}{I_{(s)}} \)

(already balanced)

I–(aq)  → I(s)  (oxidation) equation (I)

Fe3+(aq)  → Fe2+(s) (reduction) equation (II)

I–(aq)  → I(s) + e–

Fe3+(aq) + e– → Fe2+(s)

Fe3+(aq)  + I–(aq) → Fe2+(s)+ I(s)

 

d. Ca(s) + H+(aq) → Ca2+(s) + H2(g)

Answer:

Ca(s) + 2H+(aq) → Ca2+(s) + H2(g)

 

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

Solution: \( \scriptsize \underset{0}{Al_{(s)}} \: + \: \underset{+2}{Fe^{2+}_{(aq)}} \: \rightarrow \: \underset{+3}{Al^{3+}_{(aq)}} \: + \: \underset{0}{Fe_{(s)}} \)

Al(s) → Al3+(aq)   (oxidation) equation (I)

Fe2+(aq) → Fe(s) (reduction) equation (II)

Al(s)  → Al3+(aq) + 3e– × 2

Fe2+(aq)  + 2e–  → Fe(s)  × 3

2Al(s)  → 2Al3+(aq) + 6e–

3Fe2+(aq)  + 6e–  → 3Fe(s)

Net Equation:

2Al(s) + 3Fe2+(aq)   →2Al3+ (aq)+ 3Fe(s)

 

2. Balance the following complex ionic equation in acidic medium

a. Cr2O72-(aq) + Fe2+(aq) → Cr3+(aq) + Fe3+(aq)

Solution:

Reduction half cell:

Cr2O72-(aq) → Cr3+(aq)

Cr2O72-(aq)  + H+(aq) → Cr3+(aq)+ H2O(l)

Cr2O72-(aq)  + 14H+(aq) → 2Cr3+(aq)+ 7H2O(l)

Next balance the number of charges

Total charge on the left-hand side ⇒ (-2) + (+14) = +12

Total charge on the right-hand side = 2(+3) = +6

To balance the charges, we add 6e– to the left-hand side

Cr2O72-(aq) + 14H+(aq) + 6e– → 2Cr3+(aq) + 7H2O(l)

Oxidation half cell:

Fe2+(aq) → Fe3+(aq) + e–

Multiply oxidation half cell by 6

6Fe2+(aq) → 6Fe3+(aq) + 6e–

Net equation:

Cr2O72-(aq)  + 14H+(aq) + 6Fe2+(aq)  → 2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)

 

b. MnO4–(aq) + Pb2+(aq) →  Pb4+(aq) + Mn2+(aq)

Solution

Reduction half cell:

MnO4–(aq) → Mn2+(aq)

MnO4–(aq)  + H+(aq) → Mn2+(aq) + H2O(l)

MnO4–(aq)  + 8H+(aq) → Mn2+(aq) + 4H2O(l)

Next balance the number of charges

Total charge on the left-hand side ⇒ (-1) + (+8) = +7

Total charge on the right-hand side = (+2) = +2

To balance the charges, we add 5e– to the left-hand side

MnO4–(aq)  + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l)

Oxidation half cell:

Pb2+(aq) →  Pb4+(aq) + 2e–

Multiply red. Half ( × 2) and oxd. half ( × 5)

2MnO4–(aq)  + 16H+(aq) + 10e– → 2Mn2+(aq) + 8H2O(l)

5Pb2+(aq) →  5Pb4+(aq) + 10e–

Net equation:

2MnO4–(aq)  + 16H+(aq) + 5Pb2+(aq) → 2Mn2+(aq) + 8H2O(l) + 5Pb4+(aq)

 

c. Cr2O72-(aq) + SO2(aq) → Cr3+(aq) + SO2-(aq)

Solution:

Reduction half cell:

Cr2O72-(aq) → Cr3+(aq)

Cr2O72-(aq) + H+(aq) →  Cr3+(aq) + H2O(l)

Cr2O72-(aq) + 14H+(aq) →  2Cr3+(aq) + 7H2O(l)

Next balance the number of charges

Total charge on the left-hand side ⇒ (-2) + (+14) = +12

Total charge on the right-hand side = 2(+3) = +6

To balance the charges, we add 6e– to the left-hand side

Cr2O72-(aq) + 14H+(aq) + 6e– →  2Cr3+(aq) + 7H2O(l)

Oxidation half cell:

SO2(aq) →  SO2-(aq) + 2e–

Multiply red. Half ( × 1) and oxd. half ( × 3)

Cr2O72-(aq) + 14H+(aq) + 6e– →  2Cr3+(aq) + 7H2O(l)

3SO2(aq) →  3SO2-(aq) + 6e–

Net equation:

Cr2O72-(aq) + 14H+(aq) + 3SO2(aq)  → 2Cr3+(aq) + 7H2O(l)  + 3SO2-(aq)

 

d. Cl2O2–(aq) + Sn2+(aq) → Cl–(aq) + Sn4+(aq)

e. MnO4–(aq) + Fe2+(aq) → Mn2+(aq) + Fe3+(aq)

Report

Harassment or bullying behavior
Contains mature or sensitive content
Contains misleading or false information
Contains abusive or derogatory content
Contains spam, fake content or potential malware

Block Member?

Please confirm you want to block this member.

You will no longer be able to:

  • See blocked member's posts
  • Mention this member in posts
  • Invite this member to groups
  • Message this member
  • Add this member as a connection

Please note: This action will also remove this member from your connections and send a report to the site admin. Please allow a few minutes for this process to complete.

Report

You have already reported this .
error: Alert: Content selection is disabled!!