Whatis the amount of silver deposited by passing 2 A for 20 min through molten silver nitrate ? If the same quantity of electricity is passed through a molten salt solution of gold, 1.6 g of gold is deposited at cathode, then find out its equivalent mass and oxidation state of gold in gold salt. (atomic weight of Au = 197, Ag = 108)

Whatis the amount of silver deposited by passing 2 A for 20 min throug

1.	Whatis the amount of silver deposited by passing 2 A for 20 min through molten silver nitrate ? If the same quantity of electricity is passed through a molten salt solution of gold, 1.6 g of gold is deposited at cathode, then find out its equivalent mass and oxidation state of gold in gold salt. (atomic weight of Au = 197, Ag = 108)
Answer» Solution :Number of gram equivalents `= ("CURRENT "xx" time")/(96500)` `= (2xx 20 xx 60)/(96,500) = 0.0248` Amount of silver deposited `= 0.0248 xx 108 = 2.68g` Let the equilvalent weight of Au be `E_(Au)`. `:. (E_(Au))/(E_(HG)) = (W_(Au))/(W_(Ag))` `rArr (E_(Au))/(108) = (1.6)/(2.68) rArr E_(Au) = 64.47` `:.` Oxidation state of gold `= ("Atomic weight")/("Equivalent weight")` `= (197)/(64.47)` `= 3.05 ~~3` Gold is present in `Au^(+3)` state in the salt