Question

$H_2S$ is passed through an acidified solution of copper sulphate and a black precipitate of cupric sulphide is formed. This is due to

• A. oxidation of $Cu^{2+}$
• B. reduction of $Cu^{2+}$
• C. double decomposition
• D. reduction and oxidation

Answer 1

Answer: (C)

$\overset{\text{ +2}}{ Cu} \, \overset{\text{ +6}}{S} \, \overset{\text{ -2}}{ O}_4 + \overset{\text{ +1}}{ H}_2 \, \overset{\text{ -2}}{ S} \rightarrow \overset{\text{ +2}}{ Cu} \,\overset{\text{ -2}}{ S} + \overset{\text{ +1}}{ H}_2 \, \overset{\text{ +6}}{ S} \, \overset{\text{ -2}}{ O}_4 $
Here O.N. of all the atoms remains unaltered, therefore, it is not a redox reaction but is a double decomposition reaction.