A constant electric current flows for $4$ hours through two electrolytic cells connected in series. One contain $A g N O_{3}$ solution and second contains $C u C l_{2}$ solution. During this time, $4$ grams of $A g$ are deposited in the first cell.
How many grams of $C u$ are deposited in the second cell?

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Solution

By Faraday's law:
$\Rightarrow \frac{M a s s o f A g}{M a s s o f C u} = \frac{E q . m a s s o f A g}{E g . m a s s o f C u}$

$\Rightarrow$ Eq. mass of $A g = \frac{108}{1} = 108 g$

Eq. mass of $C u = \frac{63.5}{2}$
$\Rightarrow$ Mass of $C u$ deposited $= 4 g \times \frac{63.5}{2} \times \frac{1}{108}$
$= 2.352 g$
Hence, the answer is $2.352 g .$

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A constant electric current flows for 4 hours through two electrolytic cells connected in series. One contain AgNO3 solution and second contains CuCl2 solution. During this time, 4 grams of Ag are deposited in the first cell.How many grams of Cu are deposited in the second cell?

By Faraday's law:⇒MassofAgMassofCu=Eq.massofAgEg.massofCu⇒ Eq. mass of Ag=1081=108g Eq. mass of Cu=63.52⇒ Mass of Cu deposited =4g×63.52×1108 =2.352gHence, the answer is 2.352g.