Aluminium oxide may be electrolysed at 1000°C to furnish aluminium metal ( At. Mass = 27 amu 1 Faraday= 96,500 Coulombs ) The cathode reaction is

Al^{3+}+3e^{-}\rightarrow Al^{0} To prepare 5.12 kg of aluminium metal by this method would require

  • Option 1)

    5.49\times 10^{7}C\: of\: electricity

  • Option 2)

    1.83\times 10^{7}C\: of\: electricity

  • Option 3)

    5.49\times 10^{4}C\: of\: electricity

  • Option 4)

    5.49\times 10^{10}C\: of\: electricity

 

Answers (1)

As we learnt in

Faraday's first law of electrolysis -

The mass of any substance deposited or dissolved at any electrode during electrolysis is directly propotional to the quantity of electricity passed through the solution.

- wherein

W\alpha\ Q

W=ZIt

Z = mass of electrolyte get deposited

Z = electro chemical equivalent

 

 

From Faraday's 1st law,

W= Z\times Q\left [ W= weight, Z = electrochemical , Q = quantity \: of \: electricity \right ]

E= Z\times F\: \: \left [ E = equivalent \: weight , F = faraday \right ]

or   W= \frac{E}{F}\times Q

or  Q= \frac{W\times F}{E}

Q= \frac{W\times F}{\frac{A}{n}}

\left [ A= Atomic \: weight ,n=valency \: of \: ion \: \right ]

or \: Q= \frac{n\times w\times f}{A}

or \: Q= \frac{3\times 5.12\times10^{3}\times 96500}{27}= 5.49\times 10^{7}C

 


Option 1)

5.49\times 10^{7}C\: of\: electricity

This option is correct.

Option 2)

1.83\times 10^{7}C\: of\: electricity

This option is incorrect.

Option 3)

5.49\times 10^{4}C\: of\: electricity

This option is incorrect.

Option 4)

5.49\times 10^{10}C\: of\: electricity

This option is incorrect.

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