Subjective Type

Calculate the number of electrons lost or gained during electrolysis of : $$ (a) \, 2g \,Br^{-} $$ ions , $$ (b)\, 1g \,Cu^{2+} $$ ions .

Solution

(a) Eq. of $$ Br $$ used = $$ \dfrac{2}{80} $$ for $$ 2Br^- \rightarrow Br_2 + 2e $$

$$ \because $$ $$ 1 $$ eq. of an element involve = $$ 1 $$ Faraday charge

$$ = 6.023 \times 10^{23} $$ electrons

$$ \therefore $$ $$ \dfrac{2}{80} $$ eq. of $$ Br^- $$ involve $$ = \dfrac{6.023\times 10^{23} \times 2}{80} $$

$$ = 1.51 \times 10^{22} $$ electrons lost

(b) Similarly , calculate for $$ Cu^{2+} + 2e \rightarrow Cu $$

No. of electrons gained = $$ 1.89 \times 10^{22} $$

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