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DeltaG for the reaction : (4)/(3) Al+...

`DeltaG` for the reaction `:`
`(4)/(3) Al+O_(2)rarr (2)/(3)Al_(2)O_(3)`
is `-772 kJ mol^(-1)` of `O_(2)`.
Calculate the minimum `EMF` in volts required to carry out an electrolysis of `Al_(2)O_(3)`

Text Solution

Verified by Experts

The correct Answer is:
2
`Alrarr Al^(3+)+3e^(-)`
`(4)/(3) mol of Al=(4)/(3)xx3 mol e^(-) =4 mol e^(-)`
`n=4`
`Delta G=-nFE`
`-772xx1000J =-4xx96500xxE`
`:. E=(772xx1000)/(4xx96500)=2.0V`
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DeltaG for the reaction (4)/(3)Al+O_(2)to(2)/(3)Al_(2)O_(3) is -772kJ mol^(-1) of O_(2) . Calculate the minimum EMF in volts requird to carry out and electrolysis of Al_(2)O_(3)

How many coulomb of charge is required for deposition of 2/3 mole of Al in electrolysis of molten Al_(2)O_(3) ?

Knowledge Check

  • On the basis of information available from the reaction (4)/(3)Al+O_(2) rarr (2)/(3)Al_(2)O_(3),DeltaG = -827kJ mol^(-1) of O_(2) , the minimum emf required to carry out of the electrolysis of Al_(2)O_(3) is (F=96,500 C mol^(-1))

    A
    `2.14 V`
    B
    `4.28 V`
    C
    `6.42 V`
    D
    `8.56 V`

  • On the basis of the information available from the reaction 4/3 Al + O_(2) rarr 2/3 Al_(2) O_(3). Delta G^(o) = - 827 " kJmol "^(1) " of " O_(2), The minimum emf required to carry out an electrolysis of Al_(2) O_(3) is :

    A
    `8.56` V
    B
    `2.14` V
    C
    `4.28` V
    D
    `6.42 `V

  • On the bassis of the information available from the reaction 4/3 Al + O_2 rarr 2/3 Al_2 O_3 . Delta G =- 82 7 k J "mol"^(-1) of O_2 the minimum emf required to carry out an electorlysis of Al_2 O_3 is (F= 96500 C "mol"^(-1))

    A
    ` 8.56 V`
    B
    ` 2.14 V`
    C
    ` 4.` 28 V`
    D
    ` 6. 42 V`

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