An ideal Pb2+Pb2+ ion-selective electrode is moved from a 5.69×10−55.69×10−5 M Pb(NO3)2Pb(NO3)2 solution to a 3.22×10−33.22×10−3...

An ideal Pb2+Pb2+ ion-selective electrode is moved from a 5.69×10−55.69×10−5 M Pb(NO3)2Pb(NO3)2 solution to a 3.22×10−33.22×10−3 M Pb(NO3)2Pb(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?

Expert Solution

The potential measured by the electrode is given by:

Where R and F are constants (the universal and Faraday's constants, respectively), T is the temperature (298 K in this case), n is the number of electrons involved in the redox process (2 in this due, being this a divalent cation) and E° is the standard potential of the cell. If we consider all of these constants, the temperature, and the change from ln to log, we get:

So the change in the measured potential between the two concentrations will be:

queen_honey_blossom answered 2 years ago

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