Question

1(a)       Give the oxidation numbers for the underlined atoms for each of the following:

                        (i)         ClO^-

                        (ii)        OF₂      

                        (iii)       ICl₃

            (b) An aluminium oxalate complex was prepared by dissolving aluminium metal in potassium hydroxide, followed by reaction with oxalic acid.The resulting complex was analysed for oxalate content by titration against potassium permanganate:

                        A 0.1105 g sample of the aluminium oxalate complex was dissolved in deionized water (30 cm³) and bench dilute (2 M) sulphuric acid (10 cm³)added.The resulting solution was heated to about 60 ºC and titrated against standardized potassium manganate (VII) solution.14.35 cm³ of a 0.02 mol dm^-3 KMn0₄ solution was required.

(i)         Write out a balanced equation for the overall redox process.

(ii) Use the analytical data to deduce if the prepared complex was K₃Al(C₂O₄)₃.3H₂O or KAl(C₂O₄)₂(H₂O)₂.2H₂O

                        Molar masses /g mol^-1

                        K, 39.00; Al, 26.98; C, 12.01; O, 16.00; H, 1.0

Answer 1

16)

a)

(i) Let oxidation number of Cl= x

So x + (-2) = -1

x = +1

(ii) let oxidation number of O =x

So x+2(-1) = 0

x = +2

(iii) Let oxidation number of I = x

x + 3(-1) = 0

x = +3

(b)

(i)The reaction for overall redox process will be

2 MnO₄^- (aq) + 5 C₂O₄^2- (aq) + 16 H⁺ (aq) --> 2Mn²⁺ (aq) + 10CO₂ (g) + 8H₂O

(ii) The complex weight = 0.1105 grams

Amount of water = 30 cm^3

The volume of KMnO4 reacted = 14.35 cm^3

Molarity of KMnO4 = 0.02 Mol / L

so normality = 0.02 X 5 = 0.1 geq / L

milliequivalents of KMnO4 used = Normality X Volume (mL) = 0.1 X 14.35 = 1.435 miliequivalents

so these much of gram equivalents must have been used of the complex

Gram equivalents = Mass / Equvalent weight

1.435 X 10^-3 = 0.1105 / equivalent weight

So equivalent weight = 77