In an electrolysis experiment, current was passed for `5h` through two cells connected in series. The first cell contains a solution of gold and second contains copper sulphate solution. In the first cell, `9.85g ` of gold was deposited. If the oxidation number of gold is `+3`, find the amount of copper deposited at the cathode of the second cell. Also calculate the magnitude of the current in ampere, `(` Atomic weight of `Au` is 1197 and atomic weight of `Cu` is `63.5)`.

Answer

`a. 4.763g`,`b. 0.804A`

Explanation

Equivalent of gold formed `=` Equivalent of `Cu` formed
`:. (W of Au)/( Ew of Au)=(W of Cu)/(Ew of Cu)`
`:' Au^(3+)+3e^(-) rarr Au`
and `Cu^(2+)+2e^(-) rarr Cu`
`:. (9.85)/(197//2)=(W_(Cu))/(63.5//2)`
`W_(Cu)=(9.85xx3xx63.5)/(197xx2)=(1876.425)/(394)=4.763g`
Also, `W=(ZIt)/(96500)`
`:. 4.763=(63.5xxIxx5xx60xx60)/(2xx96500)`
`:. I=(4.763xx2xx96500)/(63.5xx5xx60xx60)=(919259)/(114300)=0.804A`