Question

1. In an ionic-concentration corrosion cell involving chromium, which forms a trivalent ion (Cr3+), an electrical current of 7.6 mA is measured. How many at atoms per second are oxidized at the anode?

2 . In an ionic-concentration corrosion cell involving nickel (forming

Ni2+), an electrical current of 3.8 mA is measured. How many Ni atoms per second are oxidized at the anode?

Answer 1

(1)

Current developed, I = 7.6 mA = 7.6*10^-3 A

Charge passed in one second = q = I*t = 7.6*10^-3*1 = 7.6*10^-3 columbs

Charge on one electron = 1.6*10^-19 C

Thus, number of electrons passed = 7.6*10^-3/1.6*10^-19 = 4.75*10¹⁶ electrons

Since one atom of chromium produces 3 electrons on oxidation, so number of atoms oxidised per second = (4.75*10¹⁶)/3 = 1.58*10¹⁶ atoms

(2)

Current developed, I = 3.8 mA = 3.8*10^-3 A

Charge passed in one second = q = I*t = 3.8*10^-3*1 = 3.8*10^-3 columbs

Charge on one electron = 1.6*10^-19 C

Thus, number of electrons passed = 3.8*10^-3/1.6*10^-19 = 2.375*10¹⁶ electrons

Since one atom of Nickel produces 2 electrons on oxidation, so number of atoms oxidised per second = (4.75*10¹⁶)/2 = 1.1875*10¹⁶ atoms