Balancing Oxidation Reduction Equation

(a) Ferric sulfate is produced by oxidation of ferrous sulfate with acidic potasium permanganate  solution.

(i)Introduction: Potasium permanganate (KMnO₄) is oxidizing agent and ferrous sulfate (FeSO₄) is a reducing agent.

(ii) Half Equation:

Reduction half equation: Mn⁺⁷ ion of  KMnO₄ converts into manganous (Mn⁺²) ion by accepting 5 electrons.  K⁺ ion and water is produced in acidic solution.

KMnO₄  +  8H⁺ +  5e   =   Mn⁺²      +    K⁺       +    4H₂O

Oxidation half equation: Fe²⁺ ion of FeSO₄ converts ferric ion (Fe³⁺) by donating one electrons.

FeSO₄  →  Fe²⁺   + SO₄^2-

Or, Fe⁺² – e = Fe⁺³          

As the number of accepting electrons is equal to the number of donating electrons in Redox reaction, by multiplying this equation by 5 and we  get

5Fe⁺²  -5e = 5Fe⁺³

(iii) Combined equation: We get oxidation-reduction full equation by addition of oxidation half equation and reduction half equation

KMnO₄  +  8H⁺ +  5Fe⁺²    =   Mn⁺²      +    K⁺       +   5Fe⁺³     +  4H₂O

(iv) Balanced full equation: By supplying necessary number of sulfate ions  (SO₄^2-) and we get

KMnO₄  +  4H₂SO₄ +  5FeSO₄    =   MnSO₄   +  1/2 K₂SO₄       +   5/2 Fe₂(SO₄)₃  +  8H₂O

Or, 2KMnO₄  +  8H₂SO₄ +  10FeSO₄    =   2MnSO₄   +   K₂SO₄       +   5Fe₂(SO₄)₃  +  8H₂O

 

(b) Ferric sulfate is produced by oxidation of ferrous sulfate with acidic potassium dichromate.

(i)Introduction: Potassium dichromate (K₂Cr₂O₇) is oxidizing agent and ferrous sulfate (FeSO₄) is a reducing agent.

(ii) Half Equation:

Reduction half equation: Cr⁺⁶  ion of  K₂Cr₂O₇ converts into chromium (Cr⁺³ ) ion by accepting 3 electrons.  K+ ion and water is produced in acidic solution.

K₂Cr₂O₇  +  14H⁺    +   6e   =   2Cr⁺³      +    2K⁺     +   7H₂O

Oxidation half equation: Fe²⁺ ion of FeSO₄ converts ferric ion (Fe³⁺) by donating one electrons.

FeSO₄  →  Fe²⁺   + SO₄^2-

Or, Fe⁺² – e = Fe⁺³          

As the number of accepting electrons is equal to the number of donating electrons in Redox reaction, by multiplying this equation by 6 and we  get

6Fe⁺²  -6e = 6Fe⁺³

(iii) Combined equation: We get oxidation-reduction full equation by addition of oxidation half equation and reduction half equation

K₂Cr₂O₇  +  14H⁺ +  6Fe⁺²    =   Cr⁺³      +    2K⁺       +   6Fe⁺³     +  7H₂O

(iv) Balanced full equation: By supplying necessary number of sulfate ions  (SO₄^2-) and we get

K₂Cr₂O₇   +  7H₂SO₄ +  6FeSO₄    =   Cr₂(SO₄₎₃   +   K₂SO₄       + 6Fe₂(SO₄)₃  +  7H₂O

 

(c) Carbon dioxide is produced by oxidation of acidic potassium permanganate with oxalic acid.

(i)Introduction: Potassium permanganate (KMnO₄) is oxidizing agent and oxalic acid (H₂C₂O₄) is a reducing agent.

(ii) Half Equation:

Reduction half equation: Mn⁺⁷ ion of  KMnO₄ converts into manganous (Mn⁺²) ion by accepting 5 electrons.  K+ ion and water is produced in acidic solution.

KMnO₄  +  8H⁺ +  5e   =   Mn⁺²      +    K⁺       +      4H₂O

Oxidation half equation: C³⁺ ion of H₂C₂O₄ converts  C⁴⁺ ion of CO₂  by donating two electrons.

H₂C₂O₄  (C³⁺)   -2e   =   2CO₂ (C⁴⁺)       +     2H⁺

As the number of accepting electrons is equal to the number of donating electrons in Redox reaction, so by multiplying equation-(i) by 2  and equation-(ii) by 5  and we  get

2KMnO₄  +  16H⁺ +  10e   =   2Mn⁺²   +  2K⁺   +     8H₂O

5H₂C₂O₄  (C³⁺)   –  10e   =   10CO₂ (C ⁴⁺)  +   10H⁺

(iii) Combined equation: We get oxidation-reduction full equation by addition of oxidation half equation and reduction half equation

2KMnO₄  +  6H⁺ +    5H₂C₂O₄ =   2Mn⁺²      +    2K⁺       +  10CO₂  +   8H₂O

(iv) Balanced full equation: By supplying necessary number of sulfate ions  (SO₄^2-) and we get

2KMnO₄  +  3H₂SO₄ +    5H₂C₂O₄  =   2MnSO₄      +    K₂SO₄  +  10CO₂  +   8H₂O

 

(d) Oxygen is produced by oxidized with acidic  potassium permanganate solution to H₂O₂ ?

(i)Introduction: Potassium permanganate (KMnO₄) is oxidizing agent and hydrogen peroxide (H₂O₂) is a reducing agent.

(ii) Half Equation:

Reduction half equation: Mn⁺⁷ ion of  KMnO₄ converts into manganous (Mn⁺²) ion by accepting 5 electrons.  K+ ion and water is produced in acidic solution.

KMnO₄  +  8H⁺ +  5e   =   Mn⁺²      +    K⁺       +      4H₂O

Oxidation half equation: O^1- ion of H₂O₂   converts  into O₂  by donating two electrons.

H₂O₂    – 2e    =     O₂        +       2H⁺

As the number of accepting electrons is equal to the number of donating electrons in Redox reaction, so by multiplying equation-(i) by 2  and equation-(ii) by 5  and we  get

            2KMnO₄  +  16H⁺ +  10e   =   2Mn⁺²      +    2K⁺       +      8H₂O

5H₂O₂     –  10e   =   5O₂        +     10H⁺

 

(iii) Combined equation: We get oxidation-reduction full equation by addition of oxidation half equation and reduction half equation

2KMnO₄  +  6H⁺ +    5H₂O₂ =   2Mn⁺²      +    2K⁺       +  5O₂  +   8H₂O

(iv) Balanced full equation: By supplying necessary number of sulfate ions  (SO₄^2-) and we get

2KMnO₄  +  3H₂SO₄ +    5H₂O₂   =   2MnSO₄      +    K₂SO₄  +  5O₂  +   8H₂O

 

 

(e) Oxygen is produced by oxidized with acidic  potassium dichromate solution to H₂O₂ .

(i)Introduction: Potassium dichromate (K₂Cr₂O₇) is oxidizing agent and hydrogen peroxide (H₂O₂) is a reducing agent.

(ii) Half Equation:

Reduction half equation: Cr⁺⁶  ion of  K₂Cr₂O₇ converts into chromium (Cr⁺³ ) ion by accepting 3 electrons.  K+ ion and water is produced in acidic solution.

K₂Cr₂O₇  +  14H⁺    +   6e   =   2Cr⁺³      +    2K⁺     +   7H₂O

Oxidation half equation: O^1- ion of H₂O₂   converts  into O₂  by donating two electrons.

H₂O₂    – 2e    =     O₂        +       2H⁺

As the number of accepting electrons is equal to the number of donating electrons in Redox reaction, by multiplying this equation by 3 and we  get

3H₂O₂    –  6e    =     3O₂        +       6H⁺

 

(iii) Combined equation: We get oxidation-reduction full equation by addition of oxidation half equation and reduction half equation

K₂Cr₂O₇  +  8H⁺ +    3H₂O₂   =   2Cr⁺³      +    2K⁺       +  3O₂  +   7H₂O

(iv) Balanced full equation: By supplying necessary number of sulfate ions  (SO₄^2-) and we get

K₂Cr₂O₇  +  3H₂SO₄ +    3H₂O₂   =   Cr₂(SO₄₎₃   +   K₂SO₄       +   3O₂  +   7H₂O

 

 

 

(f) Iodine is produced by oxidized with acidic  potassium dichromate solution to KI?

(i)Introduction: Potassium dichromate (K₂Cr₂O₇) is oxidizing agent and potassium iodide (KI) is a reducing agent.

(ii) Half Equation:

Reduction half equation: Cr⁺⁶  ion of  K₂Cr₂O₇ converts into chromium (Cr⁺³ ) ion by accepting 3 electrons.  K+ ion and water is produced in acidic solution.

K₂Cr₂O₇  +  14H⁺  +   6e   =   2Cr⁺³    +    2K⁺   +  7H₂O

Oxidation half equation: I^1- ion of KI   converts  into I₂  by donating two electrons.

2KI      -2e    =     I₂        +       2K⁺

As the number of accepting electrons is equal to the number of donating electrons in Redox reaction, by multiplying this equation by 3 and we get

6KI    –  6e    =     3I₂        +       6K⁺

(iii) Combined equation: We get oxidation-reduction full equation by addition of oxidation half equation and reduction half equation

K₂Cr₂O₇  +  14H⁺ +    6KI  =   2Cr⁺³      +    8K⁺       +  3I₂  +   7H₂O

(iv) Balanced full equation: By supplying necessary number of sulfate ions  (SO₄^2-) and we get

K₂Cr₂O₇  + 7H₂SO₄  + 6KI   = Cr₂(SO₄)₃   + 4K₂SO₄       +   3I₂  +   7H₂O

 

(g) Carbon dioxide is produced by oxidation of acidic potassium dichromate with oxalic acid.

(i)Introduction: Potassium dichromate (K₂Cr₂O₇) is oxidizing agent and oxalic acid (H₂C₂O₄) is a reducing agent.

(ii) Half Equation:

Reduction half equation: Cr⁺⁶  ion of  K₂Cr₂O₇ converts into chromium (Cr⁺³ ) ion by accepting 3 electrons.  K⁺ ion and water is produced in acidic solution.

K₂Cr₂O₇  +  14H⁺    +   6e   =   2Cr⁺³    +    2K⁺  +   7H₂O

Oxidation half equation: C³⁺ ion of H₂C₂O₄ converts  C⁴⁺ ion of CO₂  by donating two electrons.

H₂C₂O₄  (C³⁺)   -2e   =   2CO₂ (C⁴⁺)       +     2H⁺

As the number of accepting electrons is equal to the number of donating electrons in Redox reaction, so by multiplying equation-(i) by 1  and equation-(ii) by 3  and we  get

K₂Cr₂O₇  +  14H⁺    +   6e   =   2Cr⁺³      +    2K⁺     +   7H₂O

3H₂C₂O₄  (C³⁺)     –  6e   =   6CO₂ (C ⁴⁺)       +     6H⁺

(iii) Combined equation: We get oxidation-reduction full equation by addition of oxidation half equation and reduction half equation

K₂Cr₂O₇  +  8H⁺  +  3H₂C₂O₄   =   2Cr⁺³      +    2K⁺       +  6CO₂  +   7H₂O

(iv) Balanced full equation: By supplying necessary number of sulfate ions  (SO₄^2-) and we get

K₂Cr₂O₇  +  4H₂SO₄ +  3H₂C₂O₄     =   Cr₂(SO₄₎₃   +   K₂SO₄       +   6CO₂  +   7H₂O

 

(h) Iodine is produced by oxidized with acidic  potassium permanganate solution to KI?

(i)Introduction: Potassium permanganate (KMnO₄) is oxidizing agent and potassium  iodide (KI) is a reducing agent.

(ii) Half Equation:

Reduction half equation: Mn⁺⁷ ion of  KMnO₄ converts into manganous (Mn⁺²) ion by accepting 5 electrons.  K⁺ ion and water is produced in acidic solution.

KMnO₄ +  8H⁺ +  5e   =   Mn⁺²   +  K⁺  +   4H₂O   ——- (i)

Oxidation half equation: I^1- ion of KI   converts  into I₂  by donating two electrons.

2KI      -2e    =     I₂        +       2K⁺    ————–(ii)

As the number of accepting electrons is equal to the number of donating electrons in Redox reaction, so by multiplying equation-(i) by 2  and equation-(ii) by 5  and we  get

            2KMnO₄  +  16H⁺ +  10e   =   2Mn⁺²      +    2K⁺       +    8H₂O

10KI     –  10e   =   5I₂        +     10K⁺

(iii) Combined equation: We get oxidation-reduction full equation by addition of oxidation half equation and reduction half equation

2KMnO₄  +  16H⁺ +    10KI  =   2Mn⁺²     +    12K⁺       +  5I₂  +   8H₂O

(iv) Balanced full equation: By supplying necessary number of sulfate ions  (SO₄^2-) and we get

2KMnO₄  +  8H₂SO₄ +   10KI    =   2MnSO₄      +    K₂SO₄  +  5I₂  +   8H₂O

 

(I) Balance the equation of the REDOX reaction between sodium thiosulfate and Iodine.

Ans:

(i)Introduction:  Sodium thiosulfate is oxidized by Iodine and produces sodium tetra thionate and sodium iodide in the oxidation reduction reaction. Here, Iodine is an oxidizing agent and Sodium thiosulfate (Na₂S₂O₃) is a reducing agent. This oxidation-reduction reaction/equation can be balanced by half equation method or ion electron process.

(ii) Half Equation:

Reduction half equation: I₂ converts into iodide ion (2I^-1) ion by accepting 2 electrons.

I₂      +  2e    =     2I^-1         ——- (i)

Oxidation half equation: Thiosulfate (S₂O₃ ^-2) ion of Na₂S₂O₃  converts  into tetra thionate S₄O₆ ^-2 ion  by oxidation. Here, total charge of S in Na₂S₂O₃ increases from +8 to +10 in tetra thionate S₄O₆ ^-2 ion  by    donating two electrons.

S₂O₃ ^-2     =   S₄O₆ ^-2        + 2e   ————(ii)

(iii) Combined equation: We get oxidation-reduction full equation by addition of oxidation half equation and reduction half equation-

S₂O₃ ^-2     =   S₄O₆ ^-2        + 2e

I₂      +  2e    =     2I^-1        

2S₂O₃ ^-2     +   I₂      =   S₄O₆ ^-2   + 2I^-1        

(iv) Balanced full equation: By supplying necessary number of sodium ions  (Na⁺) and we get

2Na₂S₂O₃    +    I₂      =     Na₂S₄O₆       +   2NaI