(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${}_{(l)}^{3+}$ + 3${}^{-}$ —> Al.
Anodic reaction
2O${}_l^{2-}$ ----> 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${}_{(l)}$