Balancing of redox reaction by oxidation number method

Redox reaction

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Redox reaction-

 Balance the given  reactions by oxidation no.method-

Ex.1-

   FeCl₃ + H₂S ——-> FeCl₂ + HCl + S

FeCl₃———>  FeCl₂              (Reduction)

Oxidation no. of Fe in FeCl₃ = +3

Oxidation no. of Fe in FeCl₂ = +2

FeCl₃———>  FeCl₂              (Reduction) ———–eq (1)

(Decrease in oxi. no. per Fe atom = 1 unit )

(Decrease in oxi. no. per FeCl₃ molecule = 1 unit )

H₂S ———–> S                      ( oxidation )

Oxidation no. of S in H₂S = -2

Oxidation no. S  in free state = 0

H₂S ———–> S                      ( oxidation ) ————eq (2)

(Increase in oxi. no. per S atom = 2 unit )

(Increase in oxi. no. per H₂S molecule = 2 unit )

To make the increase & decrease in oxidation no. equal, multiply eq. (1) by 2 & add both the equations-

2 FeCl₃———> 2 FeCl₂

2 FeCl₃ + H₂S  ———> 2 FeCl₂ + S

To balance H- atom , 2HCl are added to RHS,

2 FeCl₃ + H₂S  ———> 2 FeCl₂ + S + 2HCl

(Balanced equation)

Ex.2-

CuO + NH₃ ——-> Cu + N₂ + H₂O

CuO ——–>  Cu              (Reduction) 

Oxidation no. of Cu in CuO = +2

Oxidation no. of Cu in free state = 0

CuO ——–>  Cu              (Reduction) ———–eq (1)

(Decrease in oxi. no. per Cu atom = 2 unit )

(Decrease in oxi. no. per CuO molecule = 2 unit )

NH₃ ———–> N₂                     ( oxidation ) 

Oxidation no. of N in NH₃ = -3

Oxidation no. N in N₂ = 0

2NH₃ ———–> N₂           ( oxidation ) ————eq (2)   

(Increase in oxi. no. per N atom = 3 unit )

(Total increase in oxi. no. of  NH₃ = 6 unit )

To make the increase & decrease in oxidation no. equal, multiply eq. (1) by 3 & add both the equations-

3CuO ——–> 3 Cu

3CuO  + 2NH₃ ——–> 3 Cu + N₂

To balance H- atom , 3H₂O are added to RHS,

3CuO  + 2NH₃ ——–> 3 Cu + N₂ + 3H₂O

                                                (Balanced equation)

Ex.3-

K₂Cr₂O₇ + HCl ——->  KCl + CrCl₃ +Cl₂ +H₂O

K₂Cr₂O₇———>  CrCl₃              (Reduction)

Oxidation no. of Cr in K₂Cr₂O₇= +6

Oxidation no. of Cr in CrCl₃ = +3

K₂Cr₂O₇———> 2 CrCl₃  ———–eq (1)

(Decrease in oxi. no. per Cr atom = 3 unit )

(Decrease in oxi. no. per K₂Cr₂O₇  molecule = 6 unit )

HCl ———–> Cl₂                      ( oxidation )

Oxidation no. of Cl in HCl = -1

Oxidation no. Cl in Cl₂ = 0

2HCl ———–> Cl₂ —————eq.(2)

(Increase in oxi. no. per Cl atom = 1 unit )

(Total increase in oxi. no. of HCl  = 2 unit )

To make the increase & decrease in oxidation no. equal, multiply eq. (2) by 3 & add both the equations-

6 HCl ———–> 3 Cl₂

K₂Cr₂O₇ + 6 HCl ———> 2 CrCl₃ +3 Cl₂

To balance K- atom , 2KCl are added to RHS,

K₂Cr₂O₇ + 6 HCl ———> 2 KCl + 2 CrCl₃ +3 Cl₂

To balance O- atom , 7 H₂O are added to RHS,

K₂Cr₂O₇ + 6 HCl ———> 2 KCl + 2 CrCl₃ +3 Cl₂ + 7H₂O

To balance H- atom , 8 HCl are added to LHS,

K₂Cr₂O₇ + 14 HCl ———> 2 KCl + 2 CrCl₃ +3 Cl₂ +7 H₂O

                                                                (Balanced equation)

Ex.4-

Cu + HNO₃ ——-> Cu(NO₃)₂ + NO₂ + H₂O

Cu ——–> Cu(NO₃)₂         (Oxidation)

Oxidation no. of Cu in free state = 0

Oxidation no. of Cu in  Cu(NO₃)₂ = +2

Cu ——–> Cu(NO₃)₂         (Oxidation) ———–eq (1)

(Increase in oxi. no. per Cu atom = 2 unit )

HNO₃ ———–> NO₂               (reduction )

Oxidation no. of N in HNO₃ = +5

Oxidation no. N in NO₂ = +4

HNO₃ ———–> NO₂                     (reduction ) ————eq (2)

(Decrease in oxi. no. per N atom = 1 unit )

(Decrease in oxi. no. per HNO₃ molecule= 1 unit )

To make the increase & decrease in oxidation no. equal, multiply eq. (2) by 2 & add both the equations-

2HNO₃ ———–> 2 NO₂

Cu  +  2HNO₃  ——–> Cu(NO₃)₂  + 2NO₂

To balance N- atom , 2HNO₃ are added to LHS,

Cu  +  4 HNO₃  ——–> Cu(NO₃)₂  + 2NO₂

To balance H & O – atoms , 2H₂O are added to RHS,

Cu  +  4 HNO₃  ——–> Cu(NO₃)₂  + 2NO₂ +2H₂O

                                                            (Balanced equation)

Ex.5-

I₂  + NaOH ——-> NaI + NaIO₃ + H₂O

I₂ ——–>  NaI             (Reduction)

Oxidation no. of I in I₂ = 0

Oxidation no. of I in NaI = -1

I₂ ——–> 2 NaI              ———–eq (1)

(Decrease in oxi. no. per I- atom = 1 unit )

(Decrease in oxi. no. per I₂ molecule = 2 unit )

I₂ ——–>  NaIO₃                      ( oxidation )

Oxidation no. of I in I₂ = 0

Oxidation no. I in NaIO₃ = +5

I₂ ——–> 2 NaIO₃                 ————eq (2)

(Increase in oxi. no. per I atom = 5 unit )

( Increase in oxi. no. per I₂ molecule = 10 unit )

To make the increase & decrease in oxidation no. equal, multiply eq. (1) by 5 & add both the equations-

5 I₂ ——–> 10 NaI

6 I₂ ——–> 10 NaI + 2 NaIO₃

To balance Na- atom , 12 NaOH are added to LHS,

6 I₂ +12 NaOH  ——–> 10 NaI + 2 NaIO₃

To balance H- atom , 6 H₂O are added to RHS,

6 I₂ +12 NaOH ——–> 10 NaI + 2 NaIO₃ +6H₂O

                                                                (Balanced equation)

Ex.6-

K₂Cr₂O₇ + FeSO₄ + H₂SO₄—->K₂SO₄ + Cr₂(SO₄)₃ +Fe₂(SO₄)₃  +H₂O

K₂Cr₂O₇———>  Cr₂(SO₄)₃              (Reduction)

Oxidation no. of Cr in K₂Cr₂O₇= +6

Oxidation no. of Cr in Cr₂(SO₄)₃ = +3

K₂Cr₂O₇———>  Cr₂(SO₄)₃     ———–eq (1)

(Decrease in oxi. no. per Cr atom = 3 unit )

(Decrease in oxi. no. per K₂Cr₂O₇  molecule = 6 unit )

FeSO₄  ————->  Fe₂(SO₄)₃                   ( oxidation )

Oxidation no. of Fe in FeSO₄ = +2

Oxidation no. Fe in Fe₂(SO₄)₃   = +3

2 FeSO₄  ————->  Fe₂(SO₄)₃      ————–eq.(2)

(Increase in oxi. no. per Fe- atom = 1 unit )

(Total increase in oxi. no.   = 2 unit )

To make the increase & decrease in oxidation no. equal, multiply eq. (2) by 3 & add both the equations,

6 FeSO₄  ————-> 3 Fe₂(SO₄)₃

K₂Cr₂O₇  + 6 FeSO₄  ————-> Cr₂(SO₄) ₃ + 3 Fe₂(SO₄)₃

To balance K- atom , K₂SO₄ are added to RHS,

K₂Cr₂O₇  + 6 FeSO₄  ————-> K₂SO₄ + Cr₂(SO₄) ₃ + 3 Fe₂(SO₄)₃

To balance  sulphate , 7 H₂SO₄ are added to LHS,

K₂Cr₂O₇  + 6 FeSO₄ + 7H₂SO₄  ————-> K₂SO₄ + Cr₂(SO₄) ₃ + 3 Fe₂(SO₄)₃

To balance H- atom ,  7 H₂O are added to RHS,

K₂Cr₂O₇  + 6 FeSO₄ + 7H₂SO₄  ———-> K₂SO₄ + Cr₂(SO₄) ₃ + 3 Fe₂(SO₄)₃ + 7H₂O

                                                                                                                         (Balanced equation)