When a current is passed through water , acidified with a dilute sulphuric acid , the gases formed at the platinum electrodes are

1 vol. hydrogen ( cathode ) and 2 vol. oxygen (anode)

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2 vol. hydrogen ( cathode ) and 1 vol. oxygen ( anode)

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1 vol. hydrogen (cathode) and 1 vol . oxygen ( anode)

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1 vol . oxygen ( cathode ) and 2 vol . hydrogen ( anode )

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Solution

The correct option is B 2 vol. hydrogen ( cathode ) and 1 vol. oxygen ( anode)
When electric current is passed through acidified water, it decomposes to give Hydrogen gas at cathode and oxygen gas at anode
The reaction is given below :
$2 H_{2} O (l) - - - > 2 H_{2} (g) + O_{2} (g)$
At cathode :
when current is passed from battery, the positive ions move towards cathode. They gain electrons and get neutralized to form H2 gas
$2 H + (a q) + 2 e - - - - - - - - - > H_{2} (g)$
AT Anode:
$4 O H - (a q) - - - - - - - - - - > O_{2} (g) + 2 H_{2} O (l) + 4 e$

Hence option B is correct

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