Question

In an electrolytic cell, both a zinc anode and cathode were used in a Zn²⁺(aq) solution. The cell operated for 3 minutes and 35 seconds at 325 milliamperes.

a. Calculate the theoretical mass, in milligrams, of zinc deposited. answer ( 24 mg )

b. The actual mass of zinc deposited was 20.5 mg. Calculate the current efficiency. answer (85%)

Answer 1

current =325 milliamps =325*10^-3 amps

number of coulumbs= 215 second *325*10-3 = 69.875coulumbs

96500 coulumbs =1 mole of electrons

moles of electrosn = 69.875/96500

Zinc loosses two electrons per mole Zn ---------> Zn+2 +2e-, atomic weight of Zinc= 65.38

mass of zinc deposited = (69.875/96500)* (1 mole zinc/2 moles of electrons) =(11.375/96500)*(1/2)*65.38/1 =0.0236 gm =24

2. Efficiency = 100* actual mass of zinc/ theoretial mass of zinc = 100* 20.5/24=85.4%