1.

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

Answer»

`9.65xx10^(4)`sec
`19.3xx10^(4)`sec
`28.95xx10^(4)sec`
`38.6xx10^(4)`sec

Solution :Faraday's LAW
Equivalent of `H_2` PRODUCED `=(1xxt(sec))/(9)`
`0.01xx2=(10xx10^(-3)t)/(96500)=96500xx2=t`
`19.3xx10^(4)sec=1`.


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