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

in  the answer of this question 

, Q = n×w×f/A

= 3×5.12×10³×96500/27

from where this 10^3 coming

Answers (1)

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

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