How many grams of copper will be deposited from a solution of $C u S O_{4}$ bypassing 0.5 faradays of electricity?
[Atomic weight of $C u = 63.5$ g/mole]

31.75 g

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63.6 g

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15.875 g

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127 g

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Solution

The correct option is C 15.875 g
$m = \frac{E}{F} i t$

Given $i t = 0.5$ faraday

Oxidation number of $C u$ in $C u S O_{4} = 2$

$m = \frac{63.5 \times 0.5}{2}$

$m = 15.875 g$

Hence, the correct option is $C$

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Similar questions

2

C u = 63.5 g / m o l

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

through a cupric salt (Atomic weight of Cu = 63.5)

[NCERT 1975; CPMT 1977, 79]

C u S O_{4}

C u S O_{4}

C u

C u S O_{4}

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