An ideal Pb^2+ ion-selective electrode is moved from a 7.25 times 10^-5 M PbNO_3

Question

An ideal Pb^2+ ion-selective electrode is moved from a 7.25 times 10^-5 M PbNO_3 solution to a 0.00326 M PbNO_3 solution at 25 degree 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? Delta E = An ideal ion-selective electrode demonstrates a Nernstian response to the analyte ion. The response (E) can be determined using the following equation, E = constant plusminus 0.5916/z_A log (A_A) where Z_A is the magnitude of the charge on the analyte ion (A), and A is the activity of the analyte ion. The concentration of the analyte ion can be used as an estimate of the activity. The sign of the log term is positive if the analyte is a cation and negative if the analyte is an anion.

Explanation / Answer

Nernest equation is given by

E = Eo(Pb+2/Pb) + (0.059/n) log ( [Pb2+] )    where n = charge of Pb i.e +2

now since we have two cases   E2-E1 = 0.059/n log [Pb2+]2 - 0.059/n log [Pb2+]1

    delta E = ( 0.059/2) log ( 0.00326) - ( 0.059/2) log ( 7.25 x 10^ -5)

       = -0.07336 + 0.122

   = 0.0488 V

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