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  4. When a solution of CuSO4 is electrolysed for 15 minutes with a current of 1 amp, then the mass of Cu deposited at the cathode will be

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Q. When a solution of $CuSO_4$ is electrolysed for $15$ minutes with a current of $1$ amp, then the mass of $Cu$ deposited at the cathode will be

JIPMERJIPMER 2016Electrochemistry

Solution:

$I = 1.0\, A, t = 15\, min = 15 \times 60 = 900\, s$
Quantity of electricity passed = $ I \times t$
$= 1.0 \times 900$ coulombs
$= 900 \,C$
The reaction occurring at the cathode is
$Cu^{2+} + 2e^- \rightarrow Cu$
Thus, $2F $ i.e. $2 \times 96500 \, C\,$ deposit $\, Cu$
$ 1 \, mole = 63.5 \, g$
and $900\, C$ will deposit $Cu = \frac{63.5 }{2 \times 96500} \times 900$
$= 0.26 \, g$
$\approx 0.3 \, g$