Question

When a certain amount of current is passed into the AuCl3 solution, which is in an electrolyte cell with Au as the electrode. After 10.0 min, the mass of the cathode increased by 2 g. Calculate the amount of current consumed (given 96.487 C = 1F and Au ≈ 197).

a.
2.4 A


b.
6.2 A


c.
7.2 A


d.
4.9 A


e.
3.2 A

Answer 1

Answer: Cell B: Ag

+

+e

−

⇌Ag at cathode.

1 mole (108 g) of Ag is deposited by 96500 C.

1.45 g of Ag will be deposited by

108

96500×1.45

​

=1295.6C.

Now,

Q=It

1295.6=1.5×t

t=864s

Cell A: Zn

2+

+2e

−

→Zn

2 moles of electrons (2×96500 C of current) produces 1 mole (63.5 g) of zinc.

1295.6 C of electricity will deposit

2×96500

65.3

​

×1295.6=0.438 g of zinc

Cell C: Cu

2+

+2e

−

→Cu

2 moles of electrons (2×96500 C) of current will produce 1 mole (63.5 g) of Cu.

1295.6 C of current will deposit

2×96500

63.5×1295.6

​

=0.426g of copper

Explanation: