Redox Reaction source : credit-help.biz

• I2+NaOH NaI+NaIO3 + H2O

I20 →  NaI–1 (Reduction)                                                                   (i)

(Decrease in O NO. Per I-atom =1)

( per I2 molecule= 2)

I2 →   NaIO3                   (oxidation)                                                    (ii)

(in crease in O No. per I- atom =5)

(per I2-molecule =10)

To make the increase & decrease 0 No. in redox rxn multiply eq (1) by 10, but multiply NaI by 10 & I2 by 5 Eq. 2) is multiplied by 2, but I2 by one & NaIO3 by 2.

5 I2 →10NaI                                                  (iii)

I2 → 2NaIO3                                                                                    (iv)

Add both the equations

6I2 →10NaI + 2NaIO3

To balance Na,

6I2 +12NaOH → 10NaI+2NaIO3

To balance H & O-atom

6I2 + 12 NaOH → 10NaI+2NaIO3+6H2O

or

3I2+6NaOH → 5NaI +NaIO3+3H2O

(ii)       KMnO4 +FeSO4 +H2SO4→  K2SO4+MnSO4+Fe2(SO4)3+H2O

FeSO4 →   Fe2(SO4)3                               (Oxidation)                                      (i)

(increase in O No. per Fe atom=1)

(per FeSO4 molecule=1)

KMnO4 →    MnSO4                                        (Reduction)                           (ii)

(Decrease in O No. Per Mn atom =5

per KMnO4 Molecule =5)

To make the increase & decrease in O No. equal in redox rxn multiply eq (1) by (5) but multiply FeSO4 by 5 & Fe2(SO4)3 by 2.5 & then add both equations

5FeSO4→  2.5 Fe2(SO4)3

KMnO4 →  MnSO4

KMnO4 +5FeSO4 →  MnSO4 +2.5Fe2(SO4)3                                                            (iii)

Multiply eq(3) by 2 to round off.

2KMnO4+10FeSO4 →  2MnSO4 +5Fe2(SO4)3

To balance K-atom K2SO4 is added to right hand side (RHS)

2KMnO4+10FeSO4 → K2SO4 +2MnSO4+5Fe2(SO4)3

To balance SO4 add 8H2SO4 to LHS

2KMnO4+10FeSO4+8H2SO4 → K2SO4+2MnSO4+5Fe(SO4)3

To balance O & H-atom, add 8H2O to RHS

2KMnO4+10FeSO4+8H2SO4→  K2SO4+2MnSO4+5Fe2(SO4)3+8H2O balanced eq.

• FeCl+   H2S      FeCl2+HCl +S

FeCl          FeCl2           (Reduction)                          (i)

(Decrease in O No., Per Fe-atom =1, per FeCl3 molecule=1)

H2S       S                    (Oxidation)                                       (ii)

(Increase in O No, Per S-atom = 2,

per H2S molecule=2)

To make the increase & decrease in O No. equal in redox rxn multiply eq (1) by 2 & eq. (2) by 1 & add both equations

2FeCl3     2 FeCl2

H2 S

2FeCl3 +H2  2FeCl2+S

To balance H & Cl atom, add 2 HCl to RHS

2FeCl3+H2→  2FeCl2 + 2HCl +S

(iv)      HNO3+S  SO2+NO2+H2O

HNO3            NO2      (Reduction)                           (i)

(Decrease in O No., Per N-atom =1,

Per HNO3 molecule=1)

S      SO2       (Oxidation)

in crease in O.No., per S atom =4

To make the increase & decrease in O. No. equal in redox rxn, multiply eq(i) by 4 &(2) by 1 & add both equations

4HNO   4NO2

SO2

4HNO3+S   4NO2+SO2

To balance H & O-atoms, add 2H2O on R.H.S.

4HNO3+S →  4NO2+SO2+2H2O

(v)       CuO+NH3   Cu+N2+H2O

CuO        Cu                 (Reduction)                           (i)

(Decrease in O No. per Cu atom =2,

per CuO molecule =2)

NH3      N2                       (Oxidation)                           (ii)

(Increase in O. No. per N-atom=3,

per NH3 molecule=3)

To make the increase & decrease in O.No. equal multiply eq(1) by 3 & eq (2) by 2, while multiplying eq(2) by 2, multiply NH3 by 2 & N2 by 1 & add both equations

3 CuO  3Cu

2NH3  N2

3CuO + 2NH3  3Cu+N2                                      (iii)

To balance H & O -atom, add 3H2O on R.H.S.

3CuO+2NH3  3Cu+N2+3H2O

6          K2Cr2O7 +HCl  KCl + CrCl3 + Cl2 + H2O

K2Cr2O7            CrCl3        (Reduction)               (i)

(decrease in O. No. per Cr atom 3,

per K2Cr2O7 Molecule=6)

HCl     Cl2                 (Oxidation)                                       (ii)

(increase in O.No. per Cl-atom=1,

per HCl molecule=1)

To make the increase & decrease in O No. equal, multiply eq(1) by1, but K2Cr2O7 by 1 & CrCl3 by 2. Multiply eq(2) by 6, but HCl by 6 & Cl2 by 3.

K2Cr2O7 2CrCl3                                                                           (3)

6HCl  3Cl2                                                                                   (4)

Add both the equations,

K2Cr2O7 +6HCl  2 CrCl3+3Cl2

To balance K-atoms

K2Cr2O7+6HCl  2KCl+2CrCl3+3Cl2

To balance Cl-atoms

K2Cr2O7+6HCl+8HCl  2KCl +2CrCl3 + 3Cl2

(=14HCl)

To balance H & O atom

(K2Cr2O7+14HCl  2KCl +2CrCl3+3Cl2+7H2O