ICSE Class 9 Answered

+2 + 4 –2           +2 –2       +4 –2

 CaCO_{3 (s)}  → CaO_(s) + CO_2(g)

2) H₂S + Cl₂ → 2HCl + S

(-2)    (0)         (-1)    (0)

 H₂S + Cl₂  → 2HCl +  S

The O.N. of S increases from -2 to 0. So it is undergoing oxidation.

The O.N. of Cl₂ decreases from 0 to -1. So it is undergoing reduction.

Therefore it is a redox reaction.

3) H₂S + SO₂ → 2H₂O + S

  +4       -2           0

 SO₂ + H₂S  → S + H₂O.

Multiply H₂S by 2 to equalize the oxidation numbers on either side of the equation.

 +4       2 x (-2)     0

 SO₂  + 2 H₂S  → S + H₂O.

SO₂+ 2H₂S  → 3S + 2H 3S + 2H₂O

Sulfur is changing from a -2 oxidation number to a +4 oxidation number, so it had to gain six electrons. -2 + (-4) = -6.

Oxygen is changing from a zero oxidation number to a -2, gaining two electrons. 0 + (-2) = -2. Since free oxygen atoms travel in pairs, the actual number of electrons gained is -4.

Therefore it is a redox reaction.