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CBSE Class 12 Chemistry 2017 Delhi Set 1 Solved Paper

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Question : 12 of 26
Marks: +1, -0
(a) Calculate the mass of Ag deposited at cathode when a current of 2 amperes was passed through a solution of AgNO3\mathrm{AgNO}_3 for 15 minutes.
(Given : Molar mass of Ag=108 g mol1,1 F=\mathrm{Ag}=108\ \mathrm{g}\ \mathrm{mol}^{-1}, 1\ \mathrm{F}= 96500 C mol196500\ \mathrm{C}\ \mathrm{mol}^{-1} )
(b) Define fuel cell.
Solution:  
(a)   t=900 s\;t=900\ \mathrm{s}
     Charge   =   Current   ×   Time   \;\;\text{ Charge }\;=\;\text{ Current }\; \times \;\text{ Time }\; =2×900=2 \times 900   =1800 C\;=1800\ \mathrm{C}
According to the reaction
Ag+  (aq)   +eAg(s)\mathrm{Ag}^{+} \;\text{(aq) }\;+ \mathrm{e}^{-} \rightarrow \mathrm{Ag}(\mathrm{s})
We require 1 F1\ \mathrm{F} to deposit 1 mol1\ \mathrm{mol} or 108 g108\ \mathrm{g} of Ag\mathrm{Ag}
For 1800 C1800\ \mathrm{C}, the mass of Ag\mathrm{Ag} deposited will    be   =  108×18001×96500=2.0145 g\;\text{ be }\;=\;\frac{108 \times 1800}{1 \times 96500}=2.0145\ \mathrm{g}
(b) Fuel cell is the name given to the galvanic cells which are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol, etc. directly into electrical energy.
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