1.

Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 rnilli ampere current. The time required to liberate 0.01 mole of H_(2) gas at the cathode is (1 Faraday =96500 C mol^(-1) )

Answer»

`9.65 xx10^(4)SEC`
`19.3 XX10^(4)sec`
`28. 95xx10^(4)sec`
`38.6 xx10^(4)sec`

Solution :`H_(2)O hArrH^(*)* OH^(*)`
`H^(*) * c ^(*) * * 1/2 H_(2)`
`therefore0.5` mole of `H_(2)` is liberated by 1 `F = 96500C 0.01 ` mole of `H_(2)` will beliberated by
`=(96500)/(0.5)0.01""=1930C`
`Q=1xx1`
`t * (Q)/(T)* (1930)/(10 * 10^(*3)A)* 19.3* 10^(4)sec`


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