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3.12 How much charge is required for the following reductions:

(ii) $1\ mol \ of\ Cu^{2+}\ to\ Cu?$

Answers (1)

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The equation can be written as:-

$Cu^{2+}\ +\ 2e^-\ =\ Cu$

Thus charge required is $=\ 2F$

$= 2(96500) = 193000\ C$

Posted by

Devendra Khairwa

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