Question

In: Chemistry

You measure the potential (in mV) of two different fluoride containing solutions using a fluoride selective...

You measure the potential (in mV) of two different fluoride containing solutions using a fluoride selective electrode. Solution A contains 0.0642 mol/L of Fluoride and solution B contains 0.0004 mol/L of fluoride. Calculate the difference (εB-εA) in mV. Assume ideal Nernstian response of the electrode. Watch out to obtain the correct sign. Round your answer to an integer (no decimal places)

Solutions

Expert Solution

The Nerst equation is used for determination of electrode potential of an electrochemical cell , with two electrodes, immersed in an electrolyte . Here the fluoride solution is the electrolyte for one of the electrode that is a fluoride selective electrode ,and the other electrode is a reference with fixed half-cell potential.

Nerst equation, for a cell is given by ,

E=Eo+RT/nF ln C(ox)/C(red)

where E=cell potential

Eo=standard potential

R=universal gas constant

n=electrons exchangedin the cell reaction

T=temperature

C(ox)=oxidized species in the cell reaction

C(red)=reduced species in the cell reaction

For the ion selective fluoride electrode,half cell reaction is,

E=Eo-RT/F ln [F-]    , as F- is the oxidized species and the concentration of the reduced species of the electrolyte from the reference electrode (second electrode) is constant,n=1 mol electron exchanged

For solutionA, E(A)=Eo-RT/F ln [F-](A)...........(1)

For solutionA, E(B)=Eo-RT/F ln [F-](B).........(2)

eqn (2)-(1) gives,

E(B)-E(A)=-RT/Fln [F-](A)/[F-](B)=RT/F ln [F-](B)/[F-](A)=[8.314J/Kmol*298K/9.65*10^4 C/mol]*2.303 log (0.0004/0.0642)=0.0592 log (0.0004/0.0642)=0.0592*(-2.205V)=0.130V=130 mV


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