Graphical analysis of Fe /Fe Cells The relationship between cell potential and t

Question

Graphical analysis of Fe /Fe Cells The relationship between cell potential and the concentration of cell components can graphically. The Nern equation Ece E cell (RT/nF) lnQ can be algebraically rearranged i a straight-line equation form of y mx+ b Ecell (RT/nF) lnQ E cell where y cell, m (slope) (RT/nF), x ln (Q), and b (y-intercept Eo cell. For the Fe(IID/Fe(II) cell in Part 4 where 2 Fe3 (aq) cu(s) 2 Fe2 (aq) Cu (aq) the exact form of the Nernst equation becomes: Ecell Since the concentration of the copper ion in this cell was 1.0 M then, (RT/nF) ln (Fe [Fes 12)+ Eo cell cell and 2 (RT/nF) ln (Fe2 /[Fest) Eo cell

Explanation / Answer

Ecell=-2*(RT/nF)*ln(Fe2+/Fe3+)+E0cell

when ln(Fe2+/Fe3+)=0 then Ecell=E0cell

So Fe2+/Fe3+=1

So for 1 M FeSO4 and 1 M FeCl3 ,ln(Fe2+/Fe3+)=0

So E0cell= 0.033

Now from take any value

Ecell=0.085 ,ln(Fe2+/Fe3+)=ln(1/0.1)=2.303

Ecell=-2*(RT/nF)*ln(Fe2+/Fe3+)+E0cell

0.085=-2*8.314*297.15/(n*96500)*2.303+0.033

0.085=-0.118/n+0.033

n=2.269

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