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  • In the electrolysis constant current was passed for 4 hour through 2 cells connected in series. The first cell contains a solution of gold salt & the second cell contains copper sulphate soluti

In the electrolysis constant current was passed for 4 hour through 2 cells connected in series. The first cell contains a solution of gold salt & the second cell contains copper sulphate solution 8.85g of gold was deposited in the first cell. If the oxidation number of gold is +3 , Calculate the magnitude of current in ampere.

Option: 1

0.90 A


Option: 2

0.46 A
 


Option: 3

0.85 A


Option: 4

0.78 A


Answers (1)

best_answer

\mathrm{\frac{Mass ~of ~Au ~deposited}{Mass ~of ~Cu ~deposited} = \frac{Eq ~Mass ~of ~Au }{Eq~Mass ~of ~Cu }}

\mathrm{\frac{8.85}{\text { Mass of Cu deposited }}} = \frac{197 / 3}{63.5 / 2}

\mathrm{\text { Mass of Cu deposited }=4.27 \mathrm{~g}}

\mathrm{\text { Applying } \omega=\frac{\text { zit }}{96500}}

\mathrm{4.27=\frac{63.5/2 \times i \times 4 \times 60 \times 60}{96500}}

\mathrm{ i=\frac{4.27 \times 96500}{31.75 \times 4 \times 3600} \\ }

\mathrm{ i=0.90 \mathrm{~A} }

Posted by

Ritika Jonwal

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