Chemistry - Electrochemistry Question with Solution | TestHub

${\mathrm{Al}}_2{\mathrm{O}}_3$ ionises as, $${\mathrm{Al}}_2{\mathrm{O}}_3\rightleftharpoons \underset{\text{ cathode }}{3+}+\begin{array}{r}{\mathrm{AlO}}_3^{3-}\\ \text{ anode }\end{array}$$ At cathode $${\mathrm{Al}}^{3+}+\underset{3\mathrm{F}}{3{\mathrm{e}}^{-}}⟶\stackrel{\mathrm{Al}}{27\mathrm{g}}$$ $\because$ Mass of aluminium deposited by $3\mathrm{F}$ of electricity $=27\mathrm{g}$ $\therefore$ Mass of aluminium deposited by $4.0\times 10^4\times 6\times 3600\mathrm{C}$ of electricity $$\begin{array}{rl}& =\frac{27\times 4.0\times 10^4\times 6\times 3600}{3\mathrm{F}}\mathrm{g}\\ & =\frac{27\times 4.0\times 10^4\times 6\times 3600}{3\times 96500}\mathrm{g}\\ & =8.1\times 10^4\mathrm{g}\end{array}$$