Question

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) :-?

Answer 1

Answer: $1.20\times 10^{23}$ electrons.

Explanation:

$CuSO_4\rightarrow Cu^{2+}+SO_4^{2-}$

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

$96500\times 2=193000Coloumb$ of electricity deposits 1 mole or 63.5 g of copper

63.5 g of copper is deposited by 193000 Coloumb

6.35 g of copper is deposited by=$\frac{193000}{63.5}\times 6.35=19300$ Coloumb

1 electron carry charge=$1.6\times 10^{-19}C$

1 mole of electrons contain=$6.023\times 10^{23}$ electrons

Thus  1 mole of electrons carry charge=$\frac{1.6\times 10^{-19}}{1}\times 6.023\times 10^{23}=96500C$

96500 C charge is carried by $6.023\times 10^{23}$ electrons

19300 C charge is carried by $\frac{6.023\times 10^{23}}{96500}\times 19300=1.20\times 10^{23}$ electrons.

Thus $1.20\times 10^{23}$ electrons are required to deposit 6.35 g of copper at the cathode during the electrolysis of an aqueous solution of copper sulphate.

Answer 2

Answer: Na/5 where Na is the avogadro's constant.

Explanation:

We know that a charge of 2F is required to deposit 1mole of copper, then to deposit (6.35/63.5) mole of copper, we need 2F*(0.1)=0.2F

We know that by definition, 1 mole of electrons carry F coloumb of charge. We also know that 1 mole of electrons=Na no of electrons. So Na no of electrons has a charge of F coloumb. So 0.2F is carried by (0.2F)(Na)/F no of electrons, i.e , 0.2 (Na) no of electrons.

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