In electrolytic refining of Cu, when 9.65 A of current is passed through an electrolytic cell for 30000 seconds to purify 100 g of an impure Cu metal at the cathode. What is % of impurity in the metal? (Atomic mass of Cu = 63.5 amu)

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Solution

Charge passed = 9.65 x 30000 C
= $\frac{(9.65 \times 30000)}{96500 F}$ = 3 F
So, 3 equivalents = 3/2=1.5 mol of Cu is pure
That is 63.5 x 1.5 = 95.25 gm of pure Cu
% of impure Cu = 100 - 95.25= 4.75%

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Q. In electrolytic refining of Cu, when 9.65 A of current is passed through an electrolytic cell for 30000 seconds to purify 100 g of an impure Cu metal at the cathode. What is % of impurity in the metal? (Atomic mass of Cu = 63.5 amu)

In general, electrolytic refining is a process of purification of impure metal during which an electric current is passed through an electrolytic cell which contains impure metal as anode, a suitable electrolyte and pure metal as cathode.

In the electrolytic refining of Copper (electrolyte is $C u S O_{4}$ solution + dil $H_{2} S O_{4}$ _;Pure Cu cathode and Impure Cu anode), metallic impurities having lower oxidation potential than Cu are deposited as __ sludge.

Q. In the electrolytic refining of Copper (electrolyte is

C u S O_{4}

solution + dil

H_{2} S O_{4}

;Pure Cu cathode and Impure Cu anode), metallic impurities having ___ than Cu are deposited as anode sludge.

Match the column (for electrolytic refining of Cu):
$\begin{matrix} C o l u m n - I & C o l u m n - I I (a) & A n o d e & (p) & T h i n s h e e t s o f p u r e C u (b) & C a t h o d e & (q) & A n a q u e o u s s o l u t i o n o f c o p p e r s u l p h a t e c o n t a i n i n g H_{2} S O_{4} (c) & E l e c t r o l y t e & (r) & A g, A u (d) & A n o d e m u d & (s) & I m p u r e m e t a l o f C u \end{matrix}$

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In electrolytic refining of Cu, when 9.65 A of current is passed through an electrolytic cell for 30000 seconds to purify 100 g of an impure Cu metal at the cathode. What is % of impurity in the metal? (Atomic mass of Cu = 63.5 amu)

Charge passed = 9.65 x 30000 C = (9.65×30000)96500F = 3 F So, 3 equivalents = 3/2=1.5 mol of Cu is pure That is 63.5 x 1.5 = 95.25 gm of pure Cu % of impure Cu = 100 - 95.25= 4.75%

Charge passed = 9.65 x 30000 C
= $\frac{(9.65 \times 30000)}{96500 F}$ = 3 F
So, 3 equivalents = 3/2=1.5 mol of Cu is pure
That is 63.5 x 1.5 = 95.25 gm of pure Cu
% of impure Cu = 100 - 95.25= 4.75%