What weight of copper will be deposited by passing $2$ faradays of electricity through a cupric salt?(Atomic weight of $C u = 63.5 g / m o l$ )

2.0 g

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3.175 g

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63.5 g

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127.0 g

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Solution

The correct option is B $63.5 g$
$2$ faradays of electricity flowed.
Charge flowed, $Q = 96500 \times 2 C$
$∴ m = E \times Q = \frac{63.5}{2 \times 96500} \times 96500 \times 2$ $= 63.5 g$

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What weight of copper will be deposited by passing 2 faradays of electricity

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