(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.
(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\(_{(l)}\)
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