Redox reaction

Redox reaction

source : vinepair.com

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)