Question

Al2O3, is reduced by electrolysis at low potentials
and high currents. If 4.0 104 amperes of current
is passed through molten AI,O, for 6 hours, what
mass of aluminium is produced? (Assume 100%
current efficiency, At mass of Al = 27 g mol-1)​

Answer 1

Answer:

$Al_{2} O_{3}$ ionises as

$Al_{2} O_{3}$⇆$Al_{cathode} ^{+}$+$AlO_{3} ^{3-}$

At cathode

Mass of aluminium deposited by 3 F of electricity=27

$4.0*10^{4} *6*3600C$ of electricity

$=\frac{27*4.0*10^{4} *6*3600}{3F}$

$=\frac{27*4.0*10^{4}*6*3600 }{3*96500} g$

$=8.1*10^{4} g$

Answer 2

Answer: 8.1 x 10^4

Explanation:

Look at the attachment below: