1.

In the electrolysis of acidulated water, it is desired to obtain hydrogen at the rate of 1 cc per second at NTP condition. What should be the current passed?

Answer»

Solution :`2H^(+)+2E^(-)toH_(2)`
Thus, 1 mole of `H_(2)` i.e., 22400 cc at NTP requires 2F=`2xx96500`coulombs
`therefore`1cc at NTP requires `=(2xx96500)/(22400)xx1=8.616C`
As `Q=Ixxt""thereforeI=(Q)/(t)=(8.616)/(1s)=8.616`ampere


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