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Passing electricity through a dilute sol...

Passing electricity through a dilute solution of sulphuric acid for a certain period of time liberates `168 mL` of gases at `STP`. The quantity of electricity used is

A

`965 C`

B

`9640 C`

C

`96500 C`

D

`96.5 C`

Text Solution

Verified by Experts

The correct Answer is:
A
Overall reaction during the electrolysis of dil. Solution of sulphuric acid is
`2H_(2)O(l) rarr underset(2 Vol.)(2H_(2))(g)+underset(1 Vol.)(O_(2)(g))`
Thus volume of `O_(2)(g)` liberated `= (1)/(3) xx 168 mL = 56 ml O_(2)`
Volume of `H_(2)(g)` liberated `= (2)/(3) xx 168 mL = 112 mL`
`Vol` of `1` eq. of `O_(2)(g)` at `STP` is `5.6 L` or `5600 mL`. It needs `1` faraday of electricity or `96500`, coulombs. Thus, to liberate `56 mL`, of `O_(2)(g)`, we need
`(96500)/(3600) xx 56 = 965 C`
Simultaneously, `112 mL` of `H_(2)(h)` is realesed.
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Knowledge Check

  • On passing electricity through dilute sulphuric acid , the amount of substance liberated at the cathod and the anode is in the ratio

    A
    ` 1 : 8`
    B
    `8 : 1`
    C
    `1 6 : 1`
    D
    `1 : 16`

  • On passing one faraday of electricity through a dilute solution of an acid, the volume of hydrogen obtained at NTP is:

    A
    22400 mL
    B
    1120 mL
    C
    2240 mL
    D
    11200 mL

  • The passage of a constant current through a solution of dilute H_(2)SO_(4) with 'Pt' electrodes liberated 340.5 cm^(3) of a mixture of H_(2) and O_(2) at S.T.P. The quantity of electricity that was passed is:

    A
    96500 C
    B
    965 C
    C
    1930 C
    D
    (1/100) Faraday