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Faraday's Second Law

9.65 ampere o...

Question

9.65 ampere of current was passed for one hour through Daniel cell. The loss of mass of zinc anode is :

A

11.76g

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B

1.176g

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C

5.88g

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D

2.94g

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Solution

The correct option is A 11.76g
The quantity of electricity passed $= Q (C) = I (A) \times t (s) = 9.65 A \times 3600 s = 34740 C$ .
Molar mass of zinc = 65.4 $\frac{g}{m o l}$
Moles of electrons passed $= \frac{Q (C)}{96500 \frac{C}{m o l e^{-}}} = \frac{34740 C}{96500 \frac{C}{m o l e^{-}}} = \frac{34740}{96500} m o l e^{-}$
The mass of zinc lost $= m o l a r m a s s o f Z n \times m o l e r a t i o \times m o l e s o f e l e c t r o n s p a s s e d = 65.4 \frac{g}{m o l} \times \frac{1 m o l Z n}{2 m o l e^{-}} \times \frac{34740}{96500} m o l e^{-} = 11.76 g .$

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