A-Level Redox, electrode potentials and electrolysis: worked solution

Question

In a small-scale trial the researchers passed a current of 0.150 A for 1200 seconds. Step 1 gave a lithium yield of 90.0%, while Steps 2 and 3 may be taken as fully (100%) efficient. Determine the mass of lithium produced in Step 1.

Charge passed: Q = It = 0.150 x 1200 = 180 C. Moles of electrons = Q/F = 180/96500 = 1.87 x 10⁻³ mol. Li+ + e- -> Li, so 1 mol e- gives 1 mol Li; at 100% this would be 1.87 x 10⁻³ mol Li. Step 1 is only 90.0% efficient: moles Li = 0.900 x 1.87 x 10⁻³ = 1.68 x 10⁻³ mol. Mass = 1.68 x 10⁻³ x 6.94 = 0.0117 g (about 11.7 mg).