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  4. In a solution of CuSO4 how much time will be, required to precipitate 2 g copper by 0.5 ampere current

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Q. In a solution of $CuSO_4$ how much time will be, required to precipitate $2 g$ copper by $0.5$ ampere current

Electrochemistry

Solution:

$Cu^{3+} + \underset {2\,mole, 2\,F}{2e^-} \rightarrow \underset{1\,mole,63.5\,gm}{Cu}$
To deposit $63.5\, gm$ of copper, electricity needed
$= 2 \times 96500 \,C$
To deposit $2\, gm$ of copper, electricity needed
$ = 2\times \frac{96500}{63.5} \times 2$
$= 6078.74\,C$
$Q = 6078.74\,C$
$ I = 0.5$ ampere
$Q = It$
$t$ (in seconds) $= \frac{Q}{I} = \frac{6078.74}{0.5} $
$ = 12157.48\,s$