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^{-7}−7kgC−1^{-1}−1]
mass of metal deposited during electrolysis = Zit
Given: mass = 0.02kg, i, t = 120s and Z = 1.3 x 10−7^{-7}−7kgC−1^{-1}−1]
i = mZt\frac{\text{m}}{\text{Zt}}Ztm
i = 0.021.3×10−7×120\frac{0.02}{1.3 \times 10^{-7} \times 120}1.3×10−7×1200.02 = 1282.05 ≈ 1.3 x 103^{3}3A
