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SS2: CHEMISTRY - 1ST TERM

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  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
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    1 Quiz
  5. Periodicity and Periodic Properties IV | Week 5
    5 Topics
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    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
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    1 Quiz
  8. Oxidation – Reduction Reaction II | Week 9
    3 Topics
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    1 Quiz
  9. Electrode Potential and Electrochemical Cells I | Week 10
    6 Topics
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    1 Quiz
  10. Electrode Potential and Electrochemical Cells II | Week 11
    5 Topics
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    1 Quiz
  11. Electrolysis I | Week 12
    8 Topics
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    1 Quiz
  12. Electrolysis II | Week 13
    8 Topics
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    1 Quiz
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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)

 

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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)