Question

A current is applied to two electrolytic cells in series. In the first, silver is deposited; in the second, a zinc electrode is consumed.How much Ag is plated out if 1.2 g of Zn dissolves?

Answer 1

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 .

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