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 197 and atomic weight of `Cu` is `63.5)`.

Explanation

`("Mass Au deposited")/("Mass of Cu deposited")=("Eq mass of Au")/("Eq mass of Cu")`
Eq mass of Au `1=(197)/(3)`
Eq mass of Cu `=(63.5)/(2)`
`=(9.85)/("Mass of Cu deposited") =((197)/(3))/((63.5)/(2))`
Mass of Cu depsited `=9.85 xx (63.5)/(2) (3)/(197)`
=4.7625g
Applying `w=("Eit")/(96500)`
`rArr 4.7625=(63.5//2 xx i xx 5xx 60 xx 60)/(96500)`
`i=(4.7625 xx 2 xx 96500)/(63.5 xx 5 xx 3600)=0`