Question

Calculate the mass of Li formed by electrolysis of molten LiCl by a current of 7.7×10⁴ A flowing for a period of 25 h . Assume the electrolytic cell is 79 % efficient.

Express your answer using two significant figures. Answer in terms of gLi

Part B

What is the minimum voltage required to drive the reaction?

Express your answer using two significant figures.

Answer 1

25hr = 25*3600s
7.7x10^4 A = 7.7x10^4 Coulomb /second = 7.7x10^4 C/s
1 Faraday = 96485C
Molar mass of Li: 6.941 g/mol

efficiecy 79%

mass ={(7.7x10^4 C/s)*(25*3600s)*(6.941 g/mol)/(96485C/mol)}*79%
= 3.94x10^5 gram

hence g of LI = 3.94x10^5 gram

Total charge passed through cell

= 7.7 x 10^4 x 25 hour x 3600s/hour = 693 x 10^ 7 .....(assuming 100% efficiency)

for 79%efficiency

total charge = 696 x10 ^ 7 x 0.79 = 547.47 x 10^7

moles of electron =547.47 x 10^7C / 96485 C mol-1

= 56.74 x 10 ^ 3 mol e-1

The resistance is missing the question.

The required volatage can be calculated as V= IR

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