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Electrochemistry question help please!!

An experiment was constructed to determine the passive diffusion of thallium ions Ti+ across the membrane of E.coli bacteria. A suspension of E.coli was prepared and incubated with Ti+ and used as the left half-cell solution in the electrochemical cell constructed as follow:

Ti(s) I Ti+(aq) I H+ (aq) I H2(g) I Pt(s)

(i) Initially the cell potential was +0.297V at 298K. Given that E0(Ti(s), Ti+(aq)) = -0.074V and E0(H2(g), H+(aq)) = 0.000V, calculate the initial concentration of Ti+ added to the solution containing the bacteria. All other species on the right half cell were at their standard state.

(ii) After some time the cell potential stabilised to a new value of +0.302V, calculate the final concentration of Ti+ outside and inside the bacteria, given that the total volume of the solution is 300cm3 and the volume fraction of the bacteria cells was 0.012.
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Original post by sao123
An experiment was constructed to determine the passive diffusion of thallium ions Ti+ across the membrane of E.coli bacteria. A suspension of E.coli was prepared and incubated with Ti+ and used as the left half-cell solution in the electrochemical cell constructed as follow:

Ti(s) I Ti+(aq) I H+ (aq) I H2(g) I Pt(s)

(i) Initially the cell potential was +0.297V at 298K. Given that E0(Ti(s), Ti+(aq)) = -0.074V and E0(H2(g), H+(aq)) = 0.000V, calculate the initial concentration of Ti+ added to the solution containing the bacteria. All other species on the right half cell were at their standard state.

(ii) After some time the cell potential stabilised to a new value of +0.302V, calculate the final concentration of Ti+ outside and inside the bacteria, given that the total volume of the solution is 300cm3 and the volume fraction of the bacteria cells was 0.012.


This is all about the Nernst equation: E = Eo - 0.059 log [Ti+]
(This version of the equation is valid since we are in standard conditions and it's a one electron reaction)

E is the measured cell potential.
Eo = Eo(red) - Eo(ox). The Ti/Ti+ is oxidised since it is on the left hand side, so the H2/H+ is reduced (note this a reference electrode)

You can then work out the initial concentration of Ti+.

After some time, diffusion happens and the concentration of Ti+ changes. Therefore the cell potential changes. Recalculate using the Nernst equation using the new measured cell potential (Eo stays the same though)

This is the conc of Ti+ in solution. You know the volume the bacteria take up (or can work it out!) and therefore calculate the conc of Ti+ in the bacteria.

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