Question

$H_{2}S$ 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 $C{u}^{2+}$
• B. reduction of $C{u}^{2+}$
• C. double decomposition
• D. reduction and oxidation

Answer 1

Answer: (C)

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