(b)(i) Write an equation to represent the reaction of hydrogen sulphide with iron (III) chloride solution.
(ii) Mention one change observed during the reaction in (b)(i) above.
(a) IUPAC name of \(Fe_2(SO_4)_3\)
The compound contains the iron(III) cation, \(Fe^{3+}\), and the sulphate anion, \(SO_4^{2-}\). Because iron shows more than one oxidation state, the oxidation number of the metal must be shown in Roman numerals, and the anion is named in full as the tetraoxosulphate(VI) ion. The IUPAC name is therefore iron(III) tetraoxosulphate(VI) (commonly, iron(III) sulphate).
(b)(i) Reaction of hydrogen sulphide with iron(III) chloride solution
Hydrogen sulphide is a reducing agent. It reduces iron(III) to iron(II), and is itself oxidised, so that its sulphur is deposited as free (elemental) sulphur:
\[2FeCl_3(aq) + H_2S(g) \rightarrow 2FeCl_2(aq) + 2HCl(aq) + S(s)\]
In ionic terms the essential change is \(2Fe^{3+} + H_2S \rightarrow 2Fe^{2+} + 2H^+ + S\), which shows clearly that iron is reduced from the \(+3\) to the \(+2\) state while sulphide sulphur (\(-2\)) is oxidised to sulphur (\(0\)).
(b)(ii) One observed change
The yellow-brown iron(III) chloride solution turns pale green, showing that iron(III) has been reduced to iron(II). At the same time a pale yellow deposit (turbidity) of sulphur forms in the mixture. Either of these observations is acceptable: the colour change from yellow-brown to pale green, or the appearance of the yellow solid sulphur.