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

Moles of Au deposited `=(9.85)/(197)=0.05`
`implies` gram equivalent of Au deposited `=0.5xx3=0.15`
Now, according to faraday's law of electrolysis,if same quaity of electricity is passed through diffrernt cell connected in series, same number of gram equivalent electroytes are discharged at respectively electrodes.
`implies` gram equivalnet of Cu deposited =0.15
`implies` amount of Cu deposited `=0.15xx(63.5)/(2)=4.7625`
Also, coloumbs passed =`0.15xx96500=Ixx5xx60xx60 impliesI=(0.15xx96500)/(5xx3600)=0.80A`