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how long (in hours) must a current of 2.32 amps be passed to produce 10.3 g of copper metal from a solution of aqueous copper(ii) chloride?

Sagot :

bharathparasad577

In 3.75 hours a current of 2.32 amps be passed to produce 10.3 g of copper metal from a solution of aqueous copper(ii) chloride.

According to Faraday's first law, the amount of chemical reaction that takes place at any electrode during electrolysis is proportional to the amount of electricity that is passed through the electrolyte.

According to Faraday's first law-

w= zit

w= (E/F) it

w- weight of copper , which is 10.3 grams according to the given question.

i- current = 2.32 ampere

t- time =?

E- equivalent weight of copper = atomic weight/ charge

⇒ 63.5/ 2 = 31.7 g.

F- Faradays constant= 96500C

∴10.3 = (31.7/96500) x 2.32 x t

⇒ t= 13517 sec

⇒t= 13517 / 60x60

∴t= 3.75 hours

Hence, In 3.75 hours a current of 2.32 amps be passed to produce 10.3 g of copper metal from a solution of aqueous copper(ii) chloride.

To learn more about Faraday's law refer-https://brainly.com/question/1640558

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