Answers section
Questions 5.4.1(c)
- Complete the following table to describe, in words and chemical equations, the reducing property of hydrogen sulphide.
Substance Observations when hydrogen sulphide is reacted with the substance Equation for the reaction SO2(g) NO2(g) Br2(aq) FeCl3(aq) KMnO4(aq) K2Cr2O7(aq) H2O2(aq) HNO3(aq) H2SO4(aq) - State the common observation in all these demonstrations, neglecting the acids.
Answers to Questions 5.4.1(c) 
Substance Observations when hydrogen sulphide is reacted with the substance Equation for the reaction SO2(g) Colourless liquid droplets and yellow specks on the wall of the test tube. 2H2S(g) + SO2(g) → 2H2O(g) + 3S(s) NO2(g) Nitrogen (IV) is decolouried. Yellow specks deposit on the wall of the vessel. H2S(g) + NO2(g) → S(s) + H2O(g) + NO(g) Br2(aq) Bromine water is decolourized, and a dispersed yellow solid is formed. H2S(g) + Br2(aq) → S(s) + 2HBr(aq) FeCl3(aq) Colour of the solution changes from yellow-brown to green; a dispersed yellow solid is formed. H2S(g) + 2FeCl3(aq) → 2FeCl2(aq) + S(s) + HCl(aq) KMnO4(aq) The solution is decolourized, and a dispersed yellow solid is formed. 2MnO4-(aq) + 5H2S(g) + 6H+ → 2Mn2+(aq) + 8H2O(l) + 5S(s) K2Cr2O7(aq) Colour of the solution changes from orange to green, and a dispersed yellow solid is formed. Cr2O72-(aq) + 3H2S(g) + 8H+ → 2Cr3+(aq) + 7H2O(l) + 3S(s) H2O2(aq) A dispersed yellow solid. H2O2(aq) + H2S(g) → 2H2O(l) + S(s) HNO3(aq) A red-brown gas, and dispersed yellow solid 2HNO3(aq) + H2S(g) → S(s) + 2H2O(l) + 2NO2(g) H2SO4(aq) A dispersed yellow solid. H2SO4(aq) + 3H2S(g) → 4S(s) + 4H2O(l) NB: With H2SO4(l), sulphur would be oxidized to SO2 NaOH(aq) (equal moles) A colourless solution. NaOH(aq) + H2S(aq) → NaHS(aq) + H2O(l) NaOH(aq) excess A colourless solution. 2NaOH(aq) + H2S(aq) Na2S(aq) + 2H2O(l) - Yellow solid deposit (of sulphur).