1. In an ionic-concentration corrosion cell involving chromium, which forms a tr
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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? 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?
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?
2 . In an ionic-concentration corrosion cell involving nickel (forming
Explanation / Answer
(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
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