1.

How much gm of Ag is obtained on cathode by passing 9650 Coulomb electricity through aqueous solutionof AgNO_(3) using inert electrode ? (Ag=108" gm "mol^(-1))

Answer»

108 gm
10.8 gm
1.08 gm
32.4 gm

Solution :`9650` COULOMB=`(9650)/(96500)=0.1` FARADAY
`Ag_((aq))^(+)+E^(-) to Ag_((S))`. . . Cathodic REACTION 1 faraday gives 1 mol Ag=108 gm Ag
So, 0.1 Faraday gives 0.1 mol Ag=10.8 gm Ag


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