Question

How many coulombs are required to deposit $ 50\,\,g $ of aluminium when the electrode reaction is $ A{{l}^{3+}}+3{{e}^{-}}\xrightarrow{{}}Al $

• A. $ 536111\,\,C $
• B. $ 536.111\,\,C $
• C. $ 96500\,\,C $
• D. $ 38600\,\,C $

Answer 1

Answer: (A)

$ A{{l}^{3+}}+3{{e}^{-}}\xrightarrow{{}}Al $ 1 mole of $ Al $ requires 3 moles of electrons or $ 3\times 96500\,\,C $ 1 mole of $ Al=27\,\,g $ $ 27\,\,g $ of $ Al $ require $ =3\times 96500\,\,C $ $ 50\,\,g $ of $ Al $ require $ =\frac{3\times 96500\times 50}{27} $ $ =536111\,\,C $