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  4. A current of 0.5 ampere is passed for 30 minutes through a voltameter containing CuSO4 solution.The weight of Cu deposited will be

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Q. A current of $0.5$ ampere is passed for $30$ minutes through a voltameter containing $CuSO_4$ solution.The weight of $Cu$ deposited will be

AMUAMU 2010Electrochemistry

Solution:

$Q = 1 \times t = 0.5 \times 30 \times 60 = 900 \,C$
$Cu^{2+} + 2e^- \rightarrow Cu$
$2 \times 96500 \,C$ deposit $Cu = 63.5 \,g$
$\therefore 900\, C$ will deposit $Cu$
$ = \frac{63.5}{2\times 96500} \times 900$
$ = 0.296\,g$