1.

1.234 gm gold is deposited on electrode when 3 ampere current is pass through solution containing AuCl_(4)^(-) ions. So calculate for how long such current should be passed ? (Atomic weight of Au=197 gm/mol)

Answer»

20 min. 8 sec.
10 min. 4 sec.
30 min. 12 sec.
10 min. 40 sec.

Solution :In `AuCl_(4)^(-) AU=(+3)`
`therefore Au_((aq))^(3+)+3e^(-)toAu_((S))`
So, 3F is REQUIRED for 1 mol Au
So, 3F will gives 197 gm Au.
So, for 1.234 gm GOLD required faraday
`=(1.234xx3)/(197)=(1.234xx3)/(197)xx96500` coulomb
Time=`("coulomb")/("ampere")=(1.234xx96500xx3)/(197xx3)`
=604.47 second
`~~10` minute 4 second.


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