0.02 g equivalent of Ag was deposited in an electrolysis experiment. Find the quantity of charge passed. If the same charge is passed through a gold solution, 1.314 g of gold is deposited. Find the oxidation state of gold (Given atomic mass of Au =197 amu)

Answer

Charge =1930 C ; Oxidation state of the Au=+3

Explanation

(a) Calculation of quantity of charge passed :
The reaction at cathode is :
`Ag^(+)(aq)+underset((96500" C"))underset(1Faraday)(e^(-)) to underset(1g equivalent) (Ag(s))`
1g equivalent of Ag was deposited on passing charge `=96500" C"`.
0.02g equivalent of Ag was deposited on passing charge `=((96500C))/((1g" equiv".))xx(0.02g " equiv".)=1930" C"`
(b) Calculation of oxidation state of gold
1.314g of gold is deposited on passing charge`=1930" C"`
197g of gold is deposited on passing charge `=((1930 C))/((1.314 g))xx(197 g)=289353" C"`
`=(289353)/(96500)=3F`.
`:.` The oxidation state of gold is +3
`Au^(3+)(aq)+underset((3F))(3e^(-)) to underset((197g))(Au(s))`