Calculate the time required for a constant current of 0.800 A to deposit 0.250 g of

(a) Co (II) as the element on a cathode and

(b) as Co3O4 on an anode. Assume 100% current efficiency for both cases.

a) 17 min

b) 4.2 min

Explanation:

The time required can be calculated by:

t = q/i

Where q is the charge, and i is the current. The charge may be calculated by:

q = na*n*F

Where na is the number of moles of the anality, n is the number of moles of electrons, and F is the faraday constant (96485 C/mol).

a) The element will be in the cathode, thus it will reduce:

Co⁺² + 2e⁻ ⇄ Co

The molar mass of Co is 58.93 g/mol, thus:

na = mass/molar mass

na = 0.250/58.93

na = 4.24x10⁻³ mol

n = 2 mol

q = 4.24x10⁻³*2*96485

q = 818.2 C

t = q/i

t = 818.2/0.8

t = 1022.74 s/60

t = 17 min

b) In the anode an oxidation is happening, and n = 2 too. The molar mas of Co3O4 is 240.8 g/mol, thus:

na = 0.250/240.8

na = 1.04x10⁻³ mol

q = 1.04x10⁻³*2*96485

q = 200.3 C

t = 200.3/0.800

t = 250.43 min/60

t = 4.2 min