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

Answer 1

Answer: (C)

double decomposition

Answer 2

{\overset{\text{ +2}}{ { Cu}}$\,$\overset{\text{ +6}}{ { S}}$\,$\overset{\text{ -2}}{ { O}}$_4 +$\overset{\text{ +1}}{ { H}}$_2 \,$\overset{\text{ -2}}{ { S}}$->$\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.