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How many electrons would be required to deposit 6.35 g of copper at the cathode during the electrolysis of an aqueous solution of copper sulphate ? (Atomic mass of copper = 63.5 u, NA=Avogadro’s constant) :

 

Option: 1

\frac{N_{A}}{20}


Option: 2

\frac{N_{A}}{10}


Option: 3

\frac{N_{A}}{5}


Option: 4

\frac{N_{A}}{2}


Answers (1)

best_answer

According to the reaction:

Cu^{2+}+2e^{-}\rightarrow Cu

We require 2 moles of electrons or 2NA? electrons to deposit 1 mol or 63.5 g of Cu.

So, for 6.35 g of Cu, require \frac{2 \mathrm{~N}_{\mathrm{A}}}{10} electrons.

After simplifying. 

\frac{2 \mathrm{~N}_{\mathrm{A}}}{10} =\frac{ \mathrm{~N}_{\mathrm{A}}}{5}

Therefore, Option 3 is correct.

Posted by

SANGALDEEP SINGH

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