(a)(i) Explain briefly each of the following terms: I. anode; II. cathode.
(ii) Sodium and aluminium are extracted by the electrolysis of molten sodium chloride and alumina respectively. Write balanced equations for the reactions at the anode and cathode during the extraction of: I. sodium; II. aluminium.
(iii) Explain briefly why extraction of aluminium is considered environmentally friendly while that of sodium is not.
(b) Consider the reaction represented by the following equation:
K(_2)Cr(_2)O(_7) + HC1–> KCI + CrCl(_3) + H(_2)O + Cl(_2)
(i) Explain briefly why this reaction is redox.
(ii) Write balanced half equations for the reaction.
(iii) Write the over-all balanced reaction equation.
(c) During the electrolysis of molten Al(_2)O(_3), a current of 6A was passed through the electrolyte for 1 hr. 30 mins. Calculate the mass of aluminium deposited at the cathode.
Explanation
(a)(i) I. Anode – Electrode at which oxidation occurs / electron loss
II. Cathode – Electrode at which reduction occurs / electron gain
(ii) I. Cathodic reaction
Na(^{+_(l)}) + e(^{-}) —> Na(_{s}) .
Anodic reaction
2cl(^{-_(l)}) –> Cl(_{2}) + 2e(^{-})
II. Cathodic reaction
Al(^{3+}_{(l)}) + 3(^{-}) —> Al.
Anodic reaction
2O(^{2-}_{l}) —-> O(_2) + 4e(^{-})
(iii) In the extraction of aluminium, Oxygen is produced whereas in the extraction of sodium, chlorine is produced. Chlorine is an environmental pollutant whereas oxygen is not.
(b)(i) The reaction is a redox because both oxidation and reduction occur simultaneously. The oxidation number of Cr changes from +6 (in K(_2)Cr(_2)O(_7)) to +3 (in CrCl(_3)). Cr is therefore reduced. The oxidation number of Cl changes from –1 (in HCl) to 0 (in Cl(_2)). CI is therefore oxidized.
OR
Removal of hydrogen from HCl is Oxidation. Removal of Oxygen from K(_2)Cr(_2)O(_7) is reduction.
(ii) Reduction Cr(_2)O(_2)(^{2-}) + 14H(^{+}) + 6e(^{-}) (to) 2Cr(^{3+}) + 7H(_2)O Oxidation
2Cl(^{-}) (to) Cl(_2) + 2e(^{-})
(iii) Cr(_2)O(_7)(^{2-}) + 14H(^{+}) + 6Cl(^{-}) –> 2Cr(^{3+}) + 7H(_2)O + 3Cl(_2)
OR
K(_2)Cr(_2)O(_7) + 14HCl (to) 2KCl + CrCl(_3) + H(_2)O + CI(_2)
(c) Q = I x T
6 x (30 +60) x 60
= 32400C
Al(^{3+}) + 3e(^{-}) (to) Al(_{(l)})