Question

A cell consisting of a saturated calomel electrode and a lead electrode developed a potential of -0.4706 V when immersed in 50.00 mL of a sample. A 5.00 mL addition of standard 0.02 M lead solution caused the potential to shift to -0.4490 V. Calculate the molar concentration of lead in the sample.

Answer 1

let X be the concentration of lead

here saturated calomel electrode acts as a standard electrode having E^* = 0.242 volt

here lead undergoes oxidation and SCE undergoes reduction

At the anode : Pb = Pb²⁺ + 2e^-

At the cathode : 2Hg⁺+ 2e^- = 2Hg

The cell representation is : Pb / Pb²⁺ //KCl / HgCl₂ / Hg

E_{cell =} E_cathode + E_anode

-0.4706 = 0.242 - E _{Pb/ Pb²⁺} ; therefore E _{Pb / Pb ^{2+ =}} 0.7126

hence applying nerst equation : E = E^* + 0.059/n *log (Pb)²⁺

0.7126 = E* + 0.059/2 *LOG (X) ....... 1

Now when the concentration is increased by 0.02 M

Then E_Pb/Pb2+ = 0.6910

therefore , 0.6910 = E + 0.059/2 * log (X+0.02) .......2

divinding 1 and and solving for x we get

X comes to be 0.66 M