1.

Al_(2)O_(3) is reduced by electrolysis at low potentials and high currents. If 4.0xx10^(4) ampere of current is passed through molten Al_(2)O_(3) for 6 hours, what mass of aluminium is produced ? (Assume 100% current efficiency. At. Mass of Al=27g mol^(-1))

Answer»

`8.1xx10^(4)g`
`2.4xx10^(5)g`
`1.3xx10^(4)g`
`9.0xx10^(3)g`

Solution :`Al^(3+)+3e^(-)toAl`
QUANTITY of electricity PASSED
`=(4.0xx10^(4)A)xx(6xx60xx60s)`
`=864xx10^(6)`coulombs
3F, i.e., `3xx96500C` produce `Al=1` mole, i.e., 27g
`therefore864xx10^(6)C` will produce Al ltbRgt `=(27)/(3xx96500)xx864xx10^(6)g`
`=8.058xx10^(4)g=8.1xx10^(4)g`.


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