2007 HCI Prelims Paper 3 FAQs

Explanation for Q4b(i)
CrCl3.6H2O is the general formula for a series of complexes formed.
The complexes can be [Cr(H2O)6]Cl3, [Cr(H2O)5(Cl)]Cl2 . H2O, etc.
Ag+ ions can only precipitate AgCl with the Cl- ions that are NOT bonded to Cr(III).

Explanation for Q5a(i)
Imagine that you have a "half-cell" containing Fe2+ ions (in acidic medium) initially. This is connected to a standard hydrogen electrode via a salt bridge. Now, add Cr2O72- dropwise to Fe2+ so that Fe3+ is formed. Cr2O72- added before equivalence point will change to Cr3+; there is no excess of Cr2O72-. The mixture contains Fe2+ and Fe3+. At X, [Fe2+] = [Fe3+], so we can ESTIMATE the E value to be +0.77 V using "Fe3+ + e- <--> Fe2+, E = +0.77V". After equivalence point (at 16.5 cm3), there is no more Fe2+ (all oxidised to Fe3+). When excess Cr2O72- is added into the mixture, there will be Cr3+ and Cr2O72-. At Y, [Cr3+] = [Cr2O72-], so we can ESTIMATE the E value to be +1.33 V using "Cr2O72- + 14H+ + 6e- <--> 2Cr3+ + 7H2O, E = +1.33V".

Do note that all concentrations (inclusive of H+) must be 1 mol/dm3 to measure the suggested E values at X and Y.