If the temperature in the increased as you progressed through your dilutions for

Question

If the temperature in the increased as you progressed through your dilutions for the lab Daniel cell investigation, how would this affect the slope and intercept of your E_cell vs ln(1/[Cu^2+]) plot? Calculate the theoretical potential of a Daniel cell prepared with solutions of [Cu^2+] = 0.50 M and [Zn^2+] = 0.05 M. Instead of using Cu^2+ and a Cu(s) electrode in the cathode half cell, you use a solution of Sn^2+ and a Sn(s) electrode. How would this affect the slope and intercept of your plot?

Explanation / Answer

Daniel cell rxn-

Zn2+ +2e---->Zn , Eocell(anode)=-0.76V

Cu2+ +2e--->Cu ,Eocell(cathode)=+0.34V

Zn +Cu2+ ---->Zn2+ +Cu

Ecell=Eo( cell)-RT/nF ln[Red]/[ox]=Eo( cell)-2.303 RT/nF log [Red]/[ox]=Eo( cell)-2.303 RT/nF log [Red]/[ox]=Eo( cell)-0.0592/n log [Red]/[ox]   where R=8.314J/K.mol,T=298K,F=96485C/mol,n=electrons exchanged by the electrodes

or,Ecell=Eo( cell)-RT/nF ln[Zn2+]/[Cu2+]

So the plot is between Ecell (ordinate ,taken along y-axis) and ln (1/Cu2+) (abscissa,taken along x-axis)

[Zn2+]=1.00M

Eocell=Eo(cathode)-Eo(anode)=0.34-(-0.76)=1.1V

Ecell=1.1V -RT/nF ln(1/[Cu2+])..................(1)

So comparing the eqn (1) with that of linear equation, y=mx+c ,m=slope,c=intercept

slope=-RT/nF

intercept=1.1V

So if T=temperaure increases then intercept does not change but slope becomes more negative or decreases.

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