The electrolysis of a metal salt was carried out by passing a current of 4 amp for 45 minutes. It resulted in the deposition of 2.977g metal. If atomic mass of the metal is `106.4g mol^(-1)`, calculate the charge of metal cation.

Answer

4

Explanation

Let the charge on metal cation be `n^(+)`. The reduction half reaction would then be:
`underset(1"mole")(M^(n+)) + underset(n "mole") (n e^(-)) rarr underset("1 mole= 108.4g")(M)`
The quantity of electricity required to deposit 106.4 g metal= n mole of electron`= n xx 96500C`
`therefore 2.977` g metal will be deposited by `=(n xx 96500 xx 2.977)/(106.4)= n xx 2700C` of electricity
The electricity actually passed `=4 xx 45 xx 60C`
`4 xx 45 xx 60C= n xx 2700C, n= (4 xx 45 xx 60)/(2700)=4`