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

`965 C`

`9640 C`

`96500 C`

`96.5 C`

Text Solution

Verified by Experts

The correct Answer is:

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

` 1 : 8`

`8 : 1`

`1 6 : 1`

`1 : 16`

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

22400 mL

1120 mL

2240 mL

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:

96500 C

965 C

1930 C

(1/100) Faraday