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On passing 0.5 mool of electrons through...

On passing `0.5 mool` of electrons through `CuSO_(4)` and `Hg_(2)(NO_(3))_(2)` solutions in series using inert electrodes

A

`0.5 mol ` of `Cu` is deposited

B

`0.5 mol` of `Hg` is deposited

C

`0.125 mol` of `O_(2)` is produced

D

`0.5 mol ` of `O_(2)` is produced

Text Solution

Verified by Experts

The correct Answer is:
b,c
`Cathode : [{:(Cu^(2+),+,H^(o+),,Hg_(2)^(2+)),(Hg_(2).^(2+),+,2e^(-),rarr,2Hg),(implies0.5F,,=, 0.5,mol e Hg):}`
`Anode : [{:(SO_(4)^(2-),,overset(o+)(O)H,,NO_(3)^(c-),),(2overset(o+)(O)H,,rarr,O_(2),2H_(2)O,4e^(-)),(implies4F,,-=,1molimplies, 0.5F,-=(1)/(8)mol e O_(2)):}`
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