1.

If 50 milli ampere of current is passed through copper coulometer for 60 min, calculate the amount of copper deposited.

Answer»

Solution :ELECTRICAL charge input `=Ixx"t coulombs"`
`=50xx10^(-3)Axx60xx60sec`
`="180 coulombs."`
The chemical reaction is, `Cu^(2+)+2E rarr Cu_((s))`
1 gm atom of COPPER requires 2F CURRENT
`therefore" AMOUNT of copper deposted "=(63.5g.mol^(-1)xx180C)/(2xx96500C)`
`=0.0592gm.`


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