Two electrolytic cells containing silver nitrate solution and copper sulphate solution are connected in series. A steady currrent of 2.5 ampere was passed through them till 1.078 g of Ag were deposited. How long did the current flow? What weight of copper will be deposited? (At mass of Ag=107.8, Cu=63.5)

Two electrolytic cells containing silver nitrate solution and copper s

1.	Two electrolytic cells containing silver nitrate solution and copper sulphate solution are connected in series. A steady currrent of 2.5 ampere was passed through them till 1.078 g of Ag were deposited. How long did the current flow? What weight of copper will be deposited? (At mass of Ag=107.8, Cu=63.5)
Answer» Solution :For DEPOSITION of Ag, reaction is: `Ag^(+) +e^(-)toAg ` Thus, 1 mole, i.e., `107.8g` Ag is DEPOSITED by 1F=96500C `therefore1078`g Ag will be deposited by `(96500)/(1078)xx1*078C=965C` Now `Q=Ixxt ""thereforet=(Q)/(I)=(965)/(2.5)=385s=6"min "26S` For deposition of CU, the reaction is `Cu^(2+)+2etoCu` Thus, `2F=2xx96500C` deposit Cu=1 mole=63.5 g `therefore965C` will deposit `Cu=(63.5)/(2xx96500)xx965=0.3175g`