please clarify if the following are oxidation or reduction reaction and kindly explain the reason.
 
1) CaCO3---CaO+CO2
2)H2S+Cl2---2HCl+S
3) H2S+SO2---2H2O+S

Asked by Sinthuja | 6th Aug, 2014, 09:43: PM

Expert Answer:

1) CaCO3 → CaO + CO2

+2 + 4 –2           +2 –2       +4 –2

 CaCO3 (s)  → CaO(s) + CO2(g)

It is not a redox reaction because the oxidation number of no element changes.

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.

 

Answered by Hanisha Vyas | 8th Aug, 2014, 01:05: PM

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