A current of 0.1A was passed for 4hr through a solution of cuprocyanide and 0.3745 g of copper was deposited on the cathode. Calculate the current efficiency for the copper deposition. (Cu GAM 63.5 or Cu-63.5)
A. `79%`
B. `39.5%`
C. `63.25%`
D. `63.5%`

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Correct Answer - A
`m("theoretical")=(63.5xx0.1xx7200)/96500=0.4738` g
`therefore % "efficiency" =0.3745/0.4738xx100=79%`

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