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

We know that,
`("Mass of Au deposited")/("Mass of Cu deposited")=("Eq. mass of Au")/("Eq. mass of Cu")`
Eq. mass of `Au=197/3`, Eq. mass of `Cu=63.5/2`
Mass of copper deposited
`=9.85xx63.5/2xx3/197 g=4.7625 g`
Let Z be the electrochemical equivalent of Cu.
`E=Zxx96500`
or `Z=E/96500=63.5/(2xx96500)`
Applying `W=ZxxIxxt`
`t=5` hour `=5xx3600` second
`4.7625=63.5/(2xx96500)xxIxx5xx3600`
or `I=(4.7625xx2xx96500)/(63.5xx5xx3600)=0.804` ampere