Home
Class 12
CHEMISTRY
A current of 2A was passed for 1.5 hours...

A current of 2A was passed for 1.5 hours through a solution of `CuSO_(4)` when 1.6 g of copper was deposited. Calculate percentage current efficiency.

Text Solution

Verified by Experts

Amount of current required to deposit 1 mole `Cu(63.5 g) = 2 xx 96500` C
Current required to deposit 1.6 g of copper = `(2 xx 96500 xx 1.6)/63.5 = 4862.99`
Current actually passed through
`=2 x 1.5 xx 60 = 10800`
Current efficiency `=(4862.99)/(10800) xx 100 = 45.03 %`
Promotional Banner

Similar Questions

Explore conceptually related problems

A current of 2A was passed for 1.5 hours through a solution of CuSO_(4) when 1.6g of copper was deposited. Calculate percentage current efficiency.

A current of 2 A was passed for 1 h through a solution of CuSO_4. 0.237g of Cu^(2+) ions was discharged at cathode . The current efficiency is .

A current of 2.68A is passed for one hour through an aqeous solution of CuSO_(4) using copper electrodes. Select the correct statement (s) from the following