Step - 1 : To find out atoms undergoing change in O.N.
\(\overset { +1+7-2 }{ KMnO_{ 4 } } +\overset { +1-2+1 }{ KOH } +\overset { +1-1 }{ { H }_{ 2 }{ O }_{ 2 } } \rightarrow +\overset { +1+6-2 }{ K_{ 2 }MnO_{ 4 } } +\overset { 0 }{ { O }_{ 2 } } +\overset { +1-2 }{ { H }_{ 2 }O } \)
Step - 2 : To find out total increase and decrease in O.N .
\(\overset { +7 }{ KMnO_{ 4 } } \rightarrow \overset { +6 }{ K_{ 2 }MnO_{ 4 } } \) (decrease of 1 unit per atom)
\(\overset { -1 }{ { H }_{ 2 }{ O }_{ 2 } } \rightarrow \overset { 0 }{ { O }_{ 2 } } \) (increase of 1 unit per atom or 2 units per atom)
Total increase = 2
Total decrease = 2
Step - 3 : To balance the total increase and decrease in O.N, multiply KMnO₄ by 2.
2KMnO₄ + KOH + H₂O₂ ⟶ K2MnO₄ + O₂ + H₂O
Step - 4 : To balance all atoms other than 'H' and 'O'
2KMnO₄ + 2KOH + H₂O₂ ⟶ 2K₂MnO₄ + O₂ + H₂O
number of oxygen atoms on LHS = 12
number of oxygen atoms on RHS = 11
Hence, multiply H₂O in RHS by 2. The equation becomes,
2KMnO₄ + 2KOH + H₂O₂ ⟶ 2K₂MnO₄ + O₂ + H₂O
Hydrogen atoms balance by themselves. The balanced equation is
2KMnO₄ + 2KOH + H₂O₂ ⟶ 2K₂MnO₄ + O₂ + 2H₂O

Step - 1 : To find atoms undergoing change in O.N.
\(\overset { +1-1 }{ KBr } +\overset { +4-2 }{ MnO_{ 2 } } +\overset { +1+6-2 }{ { H }_{ 2 }SO_{ 4 } } \rightarrow +\overset { +1+1+6-2 }{ KHSO_{ 4 } } +\overset { +2+6-2 }{ MnSO_{ 4 } } +\overset { -1-2 }{ { H }_{ 2 }O } +\overset { 0 }{ { Br }_{ 3 } } \)
Step - 2 : To find the total increase and decrease
\(\overset { +4 }{ MnO_{ 4 } } \rightarrow \overset { +2 }{ MnSO_{ 4 } } \) (Decrease in O.N. of 2 units per atom)
KBr^-1Br₂ (increase in O.N. of I unit per atom)
Total decrease = 2 x 1 = 2
Total increase = 1 x 2 = 2
Step - 3 : To balance the total increase and decrease in O.N, multiply KBr by 2.
2KBr + MnO₂ + H₂SO₄ ⟶ KHSO₄ + MnSO₄ + H₂O + Br₂
Step - 4 : To balance all atoms other than hydrogen and oxygen atoms
2KBr + MnO₂ + H₂SO₄ ⟶ 2KHSO₄ + MnSO₄+ H₂O + Br₂
Since SO₄^-2 (sulphate) radical does not undergo any change in O.N.
balance them RHS = 3 (SO₄²) radical; RHS = I (SO₄^-2)
Hence multiply H₂SO₄ in LHS by 3. The equation now becomes
2KBr + MnO₂ + 3H₂SO₄ ⟶ 2KHSO₄ + MnSO₄ + H₂O + Br₂
To balance 'O' atoms, multiply H₂O in RHS by 2.
2KBr + MnO₂ + 3H2SO4  ⟶ 2KHSO₄ + MnSO₄ + 2H₂0 + Br₂
This is the balanced equation.

Step-1: To find out atoms undergoing change in O.N.
\(\overset { +1+6-2 }{ { K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 } } +\overset { +1+6-2 }{ { H }_{ 2 }SO_{ 4 } } +\overset { +2+6-2 }{ Fe{ SO }_{ 4 } } \rightarrow \overset { +1+6-2 }{ { K }_{ 2 }{ SO }_{ 4 } } +\overset { +3\quad \quad +6-2 }{ { Cr }_{ 2 }(SO_{ 4 })_{ 3 } } +\overset { +3\quad \quad +6-2 }{ { Fe }_{ 3 }({ SO }_{ 4 })_{ 3 } } +\overset { +1-2 }{ { H }_{ 2 }O } \)
Step-2: To find out totel increase and decrease in O.N.
\(\overset { +6 }{ { K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 } } \rightarrow \overset { +3 }{ { Cr }_{ 2 }({ SO }_{ 4 }) } _{ 3 }\)  (Decrease of 3 unit / atom; 6 unit / 2 atoms)
Total decrease = 6 units
\(\overset { +2 }{ Fe_{ 2 }{ SO }_{ 4 } } \rightarrow \overset { +3 }{ { Fe }_{ 2 }({ SO }_{ 4 })_{ 3 } } \)  (increase of 1 unit / atom
Total increase 1 x 6 = 6
Step-3: To balance the total increase and total decrease, multiply FeSO₄ by 6.
K₂Cr2O₇ + 6FeSO₄ +H₂SO₄  ⟶ K₂SO₄ + Fe₂(SO₄)₃ + Cr₂(SO₄)₃ + H₂O
Step-4: Balance all atoms other than 'H' and 'O'.
Since Fe is oxidised (LHS) to Fe⁺³ (RHS) balance Fe₂(SO₄)₃ by multiplying by 3.
Now the equation becomes,
K₂Cr₂O₇ + 6FeSO₄ + H₂SO₄ ⟶ K₂SO₄ + Cr₂(SO₄)₃ + 3Fe₂(SO₄)₃ + H₂O
The sulphur atoms in SO₄-² radical does not undergo any change in O.N.
There are 7, SO₄^-2 radicals in LHS and 13, SO₄-² radicals in RHS.
To balance them, multiply H₂SO₄ in LHS by 7. The equation now becomes.
K₂Cr₂O₇ + 6FeSO₄ + 7H₂SO₄ ⟶ K₂SO₄ + Cr₂(SO₄)₃ + 3Fe₂(SO₄)₃ +H₂O.

Step - 1 : To balance atoms undergoing change in O.N.
\(\overset { 0 }{ Ag } +\overset { +1+5-2 }{ HNO_{ 3 } } \rightarrow +\overset { +1+5-2 }{ AgNO } +\overset { +1-2 }{ { H }_{ 2 }O+ } \overset { +2-2 }{ NO } \)
Step - 2 : To find out total increase and decrease in O.N.
\(\overset { 0 }{ Ag } \rightarrow \overset { +1 }{ Ag(NO_{ 3 })_{ 2 } } \) (increase of 1 unit / atom)
\(\overset { +5 }{ AgNO3 } \rightarrow +\overset { +2 }{ NO } \) (decrease of 3 units / atom)
Total increase = 1 x 3 = 3
Total decrease = 3 x 1 = 3
Step - 3 : To balance the total increase and decrease in O.N, multiply Ag by 3.
3Ag + HNO ⟶ AgNO₃ + H₂O +NO
Step - 4 : To balance all atoms other than hydrogen and oxygen.
3Ag + 4HNO₃ ⟶ 3AgNO₃ + H₂O + NO
Step - 5 : To balance oxygen atoms
3Ag + 4HNO₃ ⟶ 3AgNO₃ + 2H₂O + NO
Hydrogen atoms balanced by themselves.
The balanced equation is 3Ag + 4HNO₃ ⟶ 3AgNO₃ + 2H₂O + NO

Step - 1 : To find out atoms undergoing change in O.N.
\(\overset { 0 }{ Zn } +\overset { +1+5-2 }{ HNO_{ 3 } } \rightarrow +\overset { +2+6-2 }{ Zn(NO_{ 3 })_{ 2 } } +\overset { +1-2 }{ { N }_{ 2 }O } +\overset { +1-2 }{ { H }_{ 2 }O } \)
Step - 2 : To find out the total increase and decrease in O.N.
\(\overset { 0 }{ Zn } \rightarrow \overset { +2 }{ Zn } \) (NO₃)₂ (increase of 2 units per atom)
\(\overset { +5 }{ HNO_{ 3 }\rightarrow } \overset { +1 }{ { N }_{ 2 }O } \) (decrease of 4 units per atom)
Total increase 2 x 4 = 6
Total decrease 4 x 2 = 6
Step - 3 : To balance the total increase and decrease in O.N.
4Zn + 2HNO₃ ⟶ Zn(NO₃)₂ + N₂O + 5H₂O
Step - 4 : Balance the equation
4Zn + 2HNO₃ ⟶ 4Zn(NO₃)₂ + N₂O + 5H₂O
'8' nitrogen atoms, 12 oxygen atoms '8' hydrogen atoms in RHS to be balanced.
For this add 8 HNO₃ to LHS.
4Zn + 10HNO₃ ⟶ 4Zn(NO₃)₂ + N₂O + 5H₂O
The equation is now balanced.