Question

An ideal Pb2 ion-selective electrode is moved from a 5.59 × 10-5 M PbNO3 solution to a 0.00449 M PbNO3 solution at 25 °C. By how many millivolts will the potential of the ion-selective electrode change when it is moved from the first solution to the second solution?

Answer 1

Change in concentration of Pb²⁺ ions = 0.00449 - 0.0000559 = 4.43 * 10^-3 M

E⁰_Pb2+/Pb = - 0.13 V

According to Nernst equation,

Eelectrode = E0electrode - (0.05916/n) Log (1/[Mn+]

E⁰_Pb2+/Pb = - 0.13 - (0.05916/2)Log(1/0.00443)

E⁰_Pb2+/Pb = - 0.13 -0.0696

E⁰_Pb2+/Pb = - 0.1996 V

Since the concentration Pb2+ ion is increased the following oxidation reaction can be occured at electrode,

Pb (s) ---------------> Pb²⁺ + 2 e

So,

E⁰_Pb/Pb2+ = + 0.1996 = 0.200 V = 200 milli volts