Find the amount of current required to deposit 0.02kg of metal in a given electrolysis for 120 seconds. [electro chemical equivalent of the metal = 1.3 x 10\(^{-7}\)kgC\(^{-1}\)]
Attached Figure / Diagram
A
7.8 x 10\(^{-4}\)A
B
7.8 x 10\(^{-2}\)A
C
1.3 x 10\(^3\)A
D
1.3 x 10\(^5\)A
Verified Correct Answer: C.
Physical Principles & Numerical Solution
mass of metal deposited during electrolysis = Zit
Given: mass = 0.02kg, i, t = 120s and Z = 1.3 x 10\(^{-7}\)kgC\(^{-1}\)]
i = \(\frac{\text{m}}{\text{Zt}}\)
i = \(\frac{0.02}{1.3 \times 10^{-7} \times 120}\) = 1282.05 ≈ 1.3 x 10\(^3\)A
Source: WAEC / JAMB / NECO Classroom Archive (Record #70695)