During electrolysis of a solution of $A g N O_{3}, 9650$ coulomb of charge pass through the electrolytic cell; the mass of silver deposited on the cathode will be:

21.6 g

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108 g

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10.8 g

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1.08 g

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Solution

The correct option is C $10.8 g$
The half cell reaction for the formation of Silver:-

$A g^{+} + 1 e^{-} \to A g_{(s)}$

Formula:-

(i) No. of moles of $A g = \frac{Q (c)}{96500 C m o l^{- 1} e^{-}} \times m o l e r a t i o$

(ii) Mole ratio $= \frac{m o l e s o f A g f o r m e d}{m o l e s o f e l e c t r o n s} \Rightarrow 1$

$∴$ No. of moles of $A g = \frac{9650 C}{96500 C m o l^{- 1} e^{-}} \times 1$

$∴ \frac{W}{m o l a r m a s s} = \frac{9650 C}{96500 C m o l^{- 1} e^{-}} \times 1$

$∴ \frac{W (g)}{108 g m o l^{- 1}} = \frac{9650 C}{96500 C m o l^{- 1} e^{- 1}}$

$∴ W = \frac{9650 \times 108 g m o l^{- 1}}{96500 m o l^{- 1}} = \frac{108 g}{10} = 10.8 g$

$∴$ Mass of silver deposited $= 10.8 g$ .

Hence, the correct option is $C$

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During electrolysis of a solution of AgNO3,9650 coulomb of charge pass through the electrolytic cell; the mass of silver deposited on the cathode will be:

During electrolysis of a solution of $A g N O_{3}, 9650$ coulomb of charge pass through the electrolytic cell; the mass of silver deposited on the cathode will be: