current (in amperes) and the time (in seconds). Thus,
= 0.53 x 95 x 60
= 3020 C
= 3.13 x 10-2 moles
Ag. his tells us that one mole of electrons will reduce one mole (197 g) of silver ions.
What mass of metallic silver will be deposited from silver nitrate solution when a current of 0.53 A is passed through it for 95 min ?
Solution:
| The total charge, Q, passed through is
the product of the current (in amperes) and the time (in seconds). Thus, |
Q = It = 0.53 x 95 x 60 = 3020 C | |
| Amount of electrons | = 3020/9.65x104 = 3.13 x 10-2 moles | |
| The half-reaction of the reduction of silver is: Ag+ + e- Ag. his tells us that one mole of electrons will reduce one mole (197 g) of silver ions. | ||
| Hence the mass of silver produced will be | 197 x 3.13 x 10-2 = 6.2 g. | |
