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Standard X

Chemistry

Application of Electrolysis

Some quantity...

Question

Some quantity of electricity was passed through two cells, one containing $H C l$ and the other $A g {N O}_{3}$ solution. What volume of $H_{2}$ will be liberated at NTP, if $1.08 \times 10^{- 4} k g$ of Ag got deposited at the other electrode?
(Molar mass of $A g = 108 g {m o l}^{- 1}$ , $H = 1 g {m o l}^{- 1}$ )

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Solution

$2 e^{-} + A g N O_{3} + H C l ⟶ A g + H_{2} + N O_{3}^{-} + C l^{-}$
By stoichiometry, moles of $A g =$ Moles of $H_{2}$ liberated.
Moles of $A g = \frac{1.08 \times 10^{- 4} \times 10^{3}}{108}$ moles $= 10^{- 3}$ moles.
$∴$ Moles of $H_{2}$ liberated $= 10^{- 3}$ moles.
At NTP, $1$ mole of $H_{2} = 22.414 d m^{3}$
$∴$ $10^{- 3}$ mole of $H_{2} = 22.414 \times 10^{- 3} d m^{3} = 22.414 m^{3}$

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Some quantity of electricity was passed through two cells, one containing HCl and the other AgNO3 solution. What volume of H2 will be liberated at NTP, if 1.08 × 10−4 kg of Ag got deposited at the other electrode?(Molar mass of Ag=108 g mol−1, H=1 g mol−1)

2e−+AgNO3+HCl⟶Ag+H2+NO−3+Cl−By stoichiometry, moles of Ag= Moles of H2 liberated.Moles of Ag=1.08×10−4×103108 moles=10−3 moles.∴ Moles of H2 liberated =10−3 moles.At NTP, 1 mole of H2=22.414dm3∴ 10−3 mole of H2=22.414×10−3dm3=22.414m3