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)

0.02 g equivalent of Ag was deposited in an electrolysis experiment. F

1.	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» SOLUTION :(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))`