Zn + NO₃^- ⟶ Zn + NH₄⁺² in basic medium
Zn ⟶ Zn⁺² + 2e (oxidation) NO₃^- ⟶ NH₄⁺ (Reduction)
Step-1: Balance all atoms other than hydrogen and oxygen.
Zn ⟶ Zn⁺²       NO₃^- ⟶ NH₄⁺
Step-2: For balancing oxygen atom in alkaline medium, add H20 molecules on the side deficient of oxygen atom.
Then balance hydrogen atom, add H20 molecule to the side deficient in hydrogen and equal number of OH- on the other side.
Zn ⟶ Zn⁺²       NO₃^- ⟶ NH₄^-+3H₂O
NO₃^- + 10H₂O ⟶ NH₄⁺ + 3H₂O
NO₃^- + 10H₂O ⟶ NH₄⁺ + 3H₂O + 10OH^-
Step-3: Add electrons to balance charge
Zn ⟶ Zn⁺² +2e
NO₃^- + 7H₂O + 8e ⟶ NH₄⁺ +10OH^-
       Zn ⟶ Zn+2 + 2e x 8
NO3- + 7H2O + 8e ⟶ NH4+ + 10OH-
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8Zn + NO₃^- + 7H₂O ⟶ 8Zn⁺² + NH₄⁺ + 10OH^-
This is the balanced equation.

Cr₂O₇^2- + C₂H₄O ⟶ Cr⁺³ + C₂H₄O₂ In acidic medium.
\(\overset { +6-2 }{ { Cr }_{ 2 }{ O }_{ 7 } } +\overset { +1+1-2 }{ { C }_{ 2 }{ H }_{ 4 }O } \rightarrow \overset { +3+3 }{ Cr } +\overset { 0+1-2 }{ { C }_{ 2 }{ H }_{ 4 }{ O }_{ 2 } } \)
Cr₂O₇^-2 ⟶ Cr⁺³ (Reduction)     C₂H₄O ⟶ C₂H₄O₂  (Oxidation)
Step - 1 : Balance all atoms other than oxygen
Cr₂O₇^-2 ⟶ 2Cr⁺³      C₂H₄O ⟶ C₂H₄O₂
Step - 2 : Balance oxygen atom by adding water molecules on the side deficient of oxygen atom.
Cr₂O₇^-2 ⟶ 2Cr⁺³ + 7H₂O      C₂H₄O + H₂O ⟶ C₂H₄O₂
Step - 3 : To balance hydrogen atoms in acidic medium, add equal number of H" ions on the side deficient of hydrogen atom.
Cr₂O₇^-2 + 14H^-1 ⟶ 2Cr⁺³ + 7H₂O     C₂H₄O + H₂O ⟶ C₂HO₂ + 2H^-1
Step - 4 : Balance the charges by adding sufficient number of electrons
Cr₂O₇^-2 ⟶ 14H⁺ + 6e ⟶ 2Cr⁺³ + 7H₂O ..(1)
C₂H₄O + H₂O ⟶  C₂H₄O₂ + 2H⁺ + Ze  ...(2)
Multiply equation (2) by 3 and add with equation (1)
Cr₂O₇^-2 + 14H⁺ + 6e ⟶ 2Cr⁺³ + 7H₂O
C₂H₄O + H₂O ⟶ C₂H₄O₂ + 2H⁺ +2e x 3
Cr₂O₇^-2 + C₂H₄O + 8H⁺ ⟶ 2Cr⁺³ + 4H₂O + 3C₂H₄O₂

Cr₂O₇^2- + Fe²⁺ ⟶ Cr³⁺ + Fe²⁺ (acidic solution)
Cr₂O₇^2- ⟶ Cr⁺³  (reduction) (The O.N ofCr decreases from +6 to +3)
Fe⁺² ⟶ Fe⁺³ (oxidation) (The O.N of Fe increases from +2 to +3)
Balancing oxidation and reduction half reactions separately.
Cr₂O₇^-2 ⟶ Cr⁺³   Fe⁺² ⟶ Fe⁺³
Step - 1 : Balance' all atoms, other than hydrogen and oxygen.
Cr₂O₇^-2 ⟶ Cr⁺³      Fe⁺² ⟶ Fe³⁺
Step - 2 : Balance oxygen atom by adding equal number of H20 molecules on the side deficient of oxygen.
Cr₂O₇^-2  ⟶ 2Cr⁺³ + 7H₂O     Fe⁺² ⟶ Fe⁺³
Step - 3 : To balance, hydrogen atom in acidic medium, add equal number H⁺ ions on the side deficient of hydrogen atom.
14H⁺ + Cr₂O₇^-2 ⟶ 2Cr⁺³ + 7H₂O    Fe⁺² ⟶ Fe⁺³
Step - 4 : Balance the charges in such a way the charges on either side be the same
6e + Cr₂O₇^-2 + 14H⁺ ⟶ 2Cr⁺³ + 7H₂O    Fe⁺² ⟶ Fe⁺³
Add both equations in such a way the number of electrons taken up by the oxidising agent and the number of electron given by the reducing agent be the same,
Cr₂O₇^-2 + 14H⁺ + be  ⟶ 2Cr⁺³ + 7H₂O
                           Fe⁺² ⟶ Fe⁺³ + e x 6
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Cr₂O₇^-2+14H⁺+6Fe⁺² ⟶ 2Cr⁺³+6Fe⁺³+7H₂O
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The Empirical formula is Na₂SH₂₀O₁₄
Since all hydrogen are present as water of hydration, the number of mole of water present in the compound is 10H₂O.
Hence, the empirical formula Na₂SO₄. 1OH ₂O.
Empirical formula mass of the compound = 2 x 23 + 1 x 32 + 4 x 16 + 10 x 18 = 322.
Molar mass of the compound = 322 g mol^-1.
n = \(\frac{Molar\ mass}{Empirical\ formula\ mass}=\frac{322}{322}=1\)
Hence the molecular formula is Na₂SO₄ 1OH₂O.

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.

Step - 1 : To find atoms undergoing change in O.N.
K₂Cr₂O₇ + HI ⟶ KI + Crl₃ + H₂O + I₂
Step - 2 : To find the total increase and decrease in O.N.
K₂Cr₂O7 + CrI₃  (decrease of 3 unit / atom = Total decrease = 6 units / 2 atom)
HI ⟶ I₂ (increase of 1 unit / atom = total increase 1 x 6 = 6)
Step - 3 : To balance the total increase and decrease in O.N, multiply HI by 6.,
K₂Cr₂O₇ + 6 HI ⟶ KI + Crl₃ + H₂O + I₂
Step - 4 : To balance all atoms other than '0' and 'H'
K₂Cr₂O₇ + 6 HI ⟶ 2KI + 2Crl₃ + H₂O + 3I₂
This makes 14 iodine atoms on RHS. (These iodide ions do not undergo any change in O.N). Hence to balance the iodine atoms add 8HI to LHS.
i.e., K₂Cr₂O₇ + 14 HI ⟶ 2KI + 2CrI₃ + H₂O + 3I₂
These, the oxygen atoms are balanced by making 7H₂O as RHS.
K₂Cr₂O₇ + 14 HI ⟶ 2KI + 2CrI₃ + 7H₂O + 3I₂
The hydrogen atoms are balanced by themselves.
Hence the balanced equation is
K₂Cr₂O₇ + 14HI ⟶ 2KI + 2Crl₃ + 7H₂O + 3I₂

Step - 1 : To find atoms undergoing change in O.N.
\(\overset { +3 }{ { NH }_{ 3 } } +\overset { 0 }{ { F }_{ 2 } } \rightarrow \overset { -1 }{ HF } +\overset { 0 }{ { N }_{ 2 } } \).
Step - 2 : To find the total increase and decrease in O.N.
NH₃ ⟶ N₂    (decrease of 3 units per atom)
F₂ ⟶ HF   (increase in 1 unit per atom)
Total decrease = 6 units (3 x 2)
Total increase =6 units (2 x 3)
Step-3: To balance the increase and decrease in O.N, multiply NH₃ by 2 and F₂ by 3.
2NH₃ + 3F₂ ⟶ HF + N₂
Step-4: To balance all atoms other than oxygen
2NH₃ + 3F₂ ⟶ 6HF + N₂
Hydrogen atom balanced by themselves.
Hence, the balanced equation is 2N₃ + 3F₂ ⟶ 6HF + N₂.

Step - 1 : To find atoms undergoing change in O.N.
\(\overset { -2.5 }{ { C }_{ 4 }{ H }_{ 10 } } +\overset { 0 }{ { O }_{ 2 } } \rightarrow \overset { +4 }{ { CO }_{ 2 } } +\overset { -2 }{ { H }_{ 2 }O } \).
Step - 2 : To find the total increase and decrease in O.N.
C₄H₁₀ ⟶ CO₂ (decrease by 6.5 units per atom)
-O₂ ⟶ H₂O  (increases by 2 units per atom)
Total decrease = 4 x 6.5 = 26 units
Total increase = 2 x 13 = 26
Step - 3 : To balance the total increase and decrease in O.N, multiply C₄H₁₀ by 2 and O₂ by 13.
2C₄H₁₀ + 13O₂ ⟶ CO₂ + H₂O
Step - 4 : To balance all atoms other 'H' and 'O'.
2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + H₂O
Step - 5 : To balance 'O' and 'H' atoms
2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O
hydrogen atoms balanced by themselves.
Hence balanced equations is 2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O.

Step-1: To find atoms undergoing change in O.N.
\(\overset { 0\quad +1\quad -2 }{ { C }_{ 6 }{ H }_{ 12 }{ O }_{ 6 } } +\overset { +1\quad +6\quad -2 }{ H_{ 2 }{ SO }_{ 4 } } \rightarrow \overset { +1-2 }{ { CO }_{ 2 } } +\overset { +4-2 }{ SO_{ 2 } } +\overset { +1-2 }{ { H }_{ 2 }O } \)
Step-2: To find the total increase and decrease in O.N.
C₆H₁₂O₆ ⟶ CO₂ (increase of 4 units per atom)
Total increase = 24 units
H2SO₄ ⟶ SO₂ (decrease of 2 units per atom)
Total decrease = 2 x 12 = 24
Step-3: To balance the total increase and decrease in O.N, multiply H₂SO₄ by 12.
C₆H₁₂O₆ + 12H₂SO₄  ⟶ CO₂ + SO₂ + H₂O
Step-4: Balance all atoms other than 'R' and 'O'
C₆H₁₂O₆ + 12H₂SO₄ ⟶ 6CO₂ + 12SO₂ + H₂O
Step-5: Balance 'O' atoms.
C₆H₁₂O₆+ 12H₂SO₄  ⟶  6CO₂ + 12SO₂ + 18H₂O
Hydrogen atoms balance by themselves.
Hence the balanced equation is
C₆H₁₂O₆ + 12H₂SO₄  ⟶ 6CO₂ + 12SO₂+ 18H₂O.

Step - 1 : To find atoms undergoing change in O.N.
\(\overset { +1+5-2 }{ KIO_{ 3 } } +\overset { +4-2 }{ SO_{ 2 } } +\overset { +1-2 }{ H_{ 2 }O } \rightarrow \overset { +1+1+6-2 }{ KHSO_{ 4 } } +\overset { +1+6-2 }{ H_{ 2 }SO_{ 4 } } +\overset { 0 }{ { I }_{ 2 } } \)
Step - 2 : To find the total decrease and increase in O.N.
KIO₃ ⟶  I₂ (decrease of 5 units per atom)
SO₂ ⟶ H₂SO₄ (increase of 2 units per atom)
Total decrease = 5 x 2 = 10
Total increase = 2 x 5 = 10
Step - 3 : To balance the total decrease and in O.N. increase, multiply KIO₃ by 2 and SO₂ by 5.
2 KIO₃ + 5SO₂ + H₂O ⟶ KHSO₄ + H₂SO₄ + I₂
Step - 4 : To balance all atoms other than 'O' and 'H'
2KIO₃ + 5SO₂ + H2O ⟶ 2KHSO₄ + 3H₂SO₄ + I₂
Step - 5 : To balance 'O' atoms
2 KIO₃ + 5SO₂ + 4H₂O ⟶ 2KHSO₄ + 3H₂SO₄ + I₂
Hydrogen atoms balance by themselves.
Hence, the balanced equations is
2KIO₃ + 5SO₂ + 4H₂O ⟶ 2KHSO₄ + 3H₂SO₄ + I₂

Step-1: To find atoms undergoing change in O.N
\(\overset { 0 }{ P } +\overset { +1\quad +5 }{ HNO_{ 3 } } \rightarrow \overset { +1+5-2 }{ HPO_{ 3 } } +\overset { +1-2 }{ NO } +\overset { +1-2 }{ { H }_{ 2 }O } \)
Step-2: To find total decrease and increase in O.N.
P ⟶ HPO₃ (increase in O.N. of 5 units per atom)
HNO₃ ⟶ NO (decrease in O.N. of3 units per atom)
Total decrease 5 x 3 = 15
Total increase 3 x 5 = 15
Step-3: To balance the total increase and decrease in the equation, by multiplying P by 3 and HNO₃ by 5.
3P + 5HNO3  ⟶ HPO₃ + NO + H₂O
Step-4: To balance all atoms other than 'O' and 'H'
3P + 5HNO₃ ⟶ 3HPO₃ + 5NO + H₂O
Step-5: To balance by oxygen atoms
Oxygen and hydrogen atoms balance by themselves.
Hence the balanced equation is 3P + 5HNO3 ⟶ 3HPO₃ + 5 NO + H₂O

Step-1: To find atoms undergoing change in O.N.
\(\overset { +2-2 }{ CuO } +\overset { -3+1 }{ { NH }_{ 3 } } \rightarrow \overset { 0 }{ Cu } +\overset { 0 }{ { N }_{ 2 } } +\overset { +1-2 }{ { H }_{ 2 }O } \)
Step-2: To find total increase and decrease in O.N.
CuO ➝ Cu (decrease of 2 unit per atom)
NH3 ➝ N2 (increase of3 unit per atom)
Total decrease = 2 x 3 = 6
Total increase = 3 x 2 = 6
Step-3: To balance the total decrease and increase in O.N, multiply two by 3 and NH3 by 2.
3CuO + 2NH₃ ⟶ Cu + N₂ + H₂O
Step-4: To balance all atoms other than 'H' and 'O'
3CuO + 2NH₃ ⟶ 3Cu + N₂ + H₂O
Step-5: To balance 'O' atoms
3CuO + 2NH₃ ⟶ 3Cu + N₂ + 3H₂O
hydrogen atoms balance by themselves.
Hence the final equation is 
3CuO + 2NH₃ ⟶ 3Cu + N₂ + 3H₂O

When a strip of metallic copper in sliver nitrate solution ~aken in a beaker and after some time, the solution slowly turns blue. This is due to the formation of Cu²⁺ ions, i.e. copper replaces silver from silver nitrate. The reaction is,

It indicates that between copper and silver, copper has the tendency to release electrons and silver to accept electrons. This type of metal displacement reactions are known as competitive electron transfer reactions.

The given equation shows that 1 mole of A reacts with 1 mole of B₂ and I atom of A reacts with 1 molecule of B₂
(i) B is the limiting reagent because 200 molecules of B₂ will react with 200 atoms of A and 100 atoms of A will be left in excess.
(ii) A is limiting reagent because 2 moles of A will react with 2 moles of B and 1 mole of B will be left in excess.
(iii) Both will react completely because it is a stoichiometric mixture. No limiting reagent.
(iv) 2.5 moles of B will react with 2.5 moles of A. Hence B is the limiting reagent.
(v) 2.5 moles of A will react with 2.5 moles of B. Hence A is the. limiting reagent.

(i) In metal displacement reactions, we learnt that zinc replaces copper from copper sulphate solution. Let us examine whether .the reverse reaction takes place or not. As discussed earlier, place a metallic copper strip in zinc sulphate solution. If copper replaces zinc from zinc sulphate solution, Cu²⁺ ions would be released into the solution and the colour of the solution would change to blue. But no such change is observed. Therefore, we conclude that among zinc and copper, zinc has more tendency to release electrons and copper to accept the electrons.

(ii) Let us extend the reaction to copper metal and silver nitrate solution. Place a strip of metallic copper in silver nitrate solution taken in a beaker. After some time, the solution slowly turns blue. This is due to the formation of' Cu²⁺ ions, i.e. copper replaces silver from silver nitrate. The reaction is
(iii) It indicates that between copper and silver, copper has the tendency to release electrons and silver to accept electrons.

(iv) From the above experimental observations, we can conclude that among the three metals, namely, zinc, copper and silver, the electron releasing tendency is in the following order
Zinc> Copper> Silver.

Displacement reaction: Redox reactions in which an ion (or an atom) in a compound is replaced by an ion (or atom) of another element are called displacement reactions. They are further classified into
(i) metal displacement reactions
(ii) non-metal displacement reactions.

(i) Metal displacement reactions :
Place a zinc metal strip in an aqueous copper sulphate solution taken in a beaker. Observe the solution, the intensity of the blue colour of the solution slowly reduced and finally disappeared.
The zinc metal strip became coated with brownish metallic copper. This is due to the following metal displacement reaction.

(ii) Non-metal displacement:

Decomposition reaction: Redox reactions in which a compound breaks down into two or more components are called decomposition reactions. These reactions are opposite to combination reactions. In these reactions, the oxidation number of the different elements in the same substance is changed.

Combination reaction: Redox reactions in which two substances combine to form a single compound are called combination reaction.

Disproportionation reaction : In some redox reactions, the same compound can undergo both oxidation and reduction. In such reactions, the oxidation state of one and the same element is both increased and decreased. These reactions are called disproportionation reactions.

At 273 K and l atm pressure 1 mole of a gas occupies a volume of 22.4L
Therefore, number of moles of oxygen, that occupies a volume of 224 ml at 273 k and 3 atm.
\(=\frac{1 \text { mole }}{273 \mathrm{~K} \times 1 \mathrm{~atm} \times 22.4 \mathrm{~L}} \times 0.224 \mathrm{~L} \times 273 \mathrm{~K} \times 3 \mathrm{~atm}\)
= 0.03 mole
1 mole of oxygen contain 6.022 x 10²³ molecures
0.03 mole of oxygen contains = 6.022 x 10²³ x 0.03
= 1.8 x 10²² molecules of oxygen.