ISE electrode question Please help:

Question

An ideal Pb2+ ion-selective electrode is moved from a 2.01×10−5Pb(NO3)2 solution to a 1.04×10−3 Pb(NO3)2 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

Nernst equation for a cell @ 298K:

Ecell = Eºcell = 0.0592 / n log Q

n = 2 moles electrons for Pb²⁺ ==> Pb

Eº = standard potential of the cell

Change in E (∆E) = (Eº - 0.0296 log 1.04x10^-3) - (Eº - 0.096 log 2.01x10^-5) = 0.0296 log (2.01x10^-5/1.04x10^-3)

∆E = -0.0507 V = - 50.7 mV

(be sure to check the math)