Question

Aluminium oxide may be electrolysed at $ \text{1000}{{\,}^{\text{o}}}\text{C} $ to furnish aluminium metal (atomic mass $ =27\,u;\,1F=96,500\,C $ ).
The cathode reaction is $ A{{l}^{3+}}+3{{e}^{-}}\xrightarrow{{}}A{{l}^{0}} $
To prepare 5.12 kg of aluminium metal by this method would require

• A. $ 5.49\,\,\times \,\,{{10}^{1}}C $ of electricity
• B. $ 5.49\,\,\times \,\,{{10}^{4}}C $ of electricity
• C. $ 1.83\times {{10}^{7}}C $ of electricity
• D. $ 5.49\times {{10}^{7}}C $ of electricity

Answer 1

Answer: (D)

$ A{{l}^{3+}}+3{{e}^{-}}\xrightarrow{{}}Al $
$ w=zQ $
where, $ w= $ amount of metal $ w=5.12\,kg=5.12\times {{10}^{3}}g $
$ z= $ electrochemical equivalent
$ z=\frac{\text{Equivalent}\,\text{weight}}{\text{96500}}=\frac{\text{Atomic}\,\text{mass}}{\text{Electrons}\,\text{ }\!\!\times\!\!\text{ }\,\text{96500}} $
$ z=\frac{27}{3\times 96500} $
$ 5.12\times {{10}^{3}}=\frac{27}{3\times 96500}\times Q $
$ Q=\frac{5.12\times {{10}^{3}}\times 3\times 96500}{27}C $
$ =5.49\times {{10}^{7}}C $