The amount of Ag plated out if 1.2 g of Zn dissolves is 3.959 g .
Given ,
A current is applied to two electrolytic cells in series .
In the first ,silver is deposited and in the second a zinc electrode is consumed .
the reactions involving are ;
Ag+ (aq) + e = Ag
Zn = Zn2+ (aq) +2e
thus the resultant equation is ,
2Ag+ (aq) +Zn = 2Ag + Zn2+
Thus for every mole of Zn dissolves , there is 2 moles of Ag is formed .
65.38 g of Zn contains = 1 moles
1.2 g of Zn contains = 1.2/65.38 =0.01835 moles
for every 1 mole of Zn dissolves there is 2 moles of Ag formed .
Thus the amount of Ag formed in moles =2(O.01835) =0.0367 Moles
1 mole of Ag contains = 107.86g
0.0367 moles of Ag contains = 107.86 (0.0367) =3.959 g of Ag
Hence ,the amount of Ag plated out if 1.2 g of Zn dissolves is 3.959 g .
Learn more about electrolytic cell here:
brainly.com/question/19854746
#SPJ4
