How many grams of lead may be formed by the passage of 7,962 C through an electrolytic cell that contains an aqueous Pb(IV) salt.

Question

answer is 4.27. Can you explain

Answer 1

Pb⁴⁺ + 4e- ==> Pb(s)

7962 C x 1 mol e-/96,485 C x 1 mol Pb/4 mol e- x 207 g Pb/mol Pb = 4.27 g Pb