(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\(_{(l)}\)