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Determine the current (in A) to be passe...

Determine the current (in A) to be passed for 30 min through a `CuSO_(4)` solution to deposit 0.406 g of copper.

Text Solution

Verified by Experts

Equivalent weight of Cu in `CuSO_(4)=(63.5)/(2)=31.75`.
31.75 g of Cu is deposited by 96500 C of electricity.
`therefore" 0.406 g of Cu is deposited by "(96500xx0.406)/(31.75)=1233.98C` of electricity.
`"Putting "Q=1233.98C and t=30xx60=1800s" in the equation "Q=Ixxt" gives I"=(1233.98)/(1800)=0.6855.`
So, required current strength = 0.6855 A.
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