Balance Redox Rxn by oxidation No. method

Redox Reaction

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• 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)

I_{2 →} 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 +H₂SO_4→  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

5FeSO_4→  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.

• FeCl₃  +   H₂S  →    FeCl₂+HCl +S

FeCl₃     →      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

2FeCl_{3   →}  2 FeCl₂

H₂S → S

2FeCl₃ +H₂S →  2FeCl₂+S

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

2FeCl₃+H₂S →  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    →  SO_{2       (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_{3  →}   4NO₂

S  → 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