Question

Calculate the time required for a constant current of 0.898 A to deposit 0.354 g of Tl(III) as Tl(s) on a cathode.

s=?

Calculate the mass of Tl(I) that can be deposited as Tl2O3(s) on an anode at a constant current of 0.898 A over the same amount of time as calculated above.

g=?

Answer 1

Part A)

Tl³⁺ + 3e^-            Tl(s)

96500 coulombs can deposit 1 equivalent Tl

Atomic mass of Tl = 204.3833 u

Therefore, 96500 coulomb can deposit (204.3833)/3 = 68.1277 g Tl

0.354 g of Tl can be deposited by (96500 coulomb x 0.354)/68.1277 = 501.42599 coulomb

Therefore,

501.42599 coulomb = 0.898 A x Time(s)

Time(s) = (501.42599)/0.898 = 558 s

Calculate the time required for a constant current of 0.898 A to deposit 0.354 g of Tl(III) as Tl(s) on a cathode = 558 s

Part B)

Tl⁺     Tl³⁺ + 2e^-

96500 coulomb can produce (204.3833/2) g Tl³⁺

558 x 0.898 coulombs can produce (204.3833/2) x (1/96500) x 558 x 0.898 = 0.530 g

mass of Tl(I) that can be deposited as Tl₂O₃(s) on an anode at a constant current of 0.898 A over a time period of 558 s = 0.530 g