The concentration cell shown above employs the reaction Cu2+ +2e- -----> Cu E degree = 0.34 V. Both compartments of the cell have copper metal electrodes. The compartment on the the left contains 8.0 x 10^-3 M Cu2+ and the compartment on the right contains 8.5 x 10^-2M Cu2+. The expected potential for this cell at 25degrees C is ______V
In the hint it says E degrees for this cell is 0V
n = 2
last hint is - if the cell volatage you calculate is negative , what does that mean?
I used the Nernst Equation, but still having trouble on how to solve, help please :)
In the hint it says E degrees for this cell is 0V
n = 2
last hint is - if the cell volatage you calculate is negative , what does that mean?
I used the Nernst Equation, but still having trouble on how to solve, help please :)
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reaction at cathode or in the right compartment
Cu2+ + 2e- ------> Cu .......Eo (reduction) = 0.34 V
reaction at anode or in the left compartment :
Cu ----------> Cu2+ + 2e- ......Eo (oxidation) = -0.34 V ( - sign since reaction is reversed)
adding both the reactions ...
Cu2+ + 2e- ---------> Cu
+Cu -----------> Cu2+ + 2e-
--------------------
Cu2+ --------> Cu2+
where Cu2+ in left sides of equation represent conc in cathode or right compartment = 8.5 X 10^-2 M
and Cu2+ in right side of equation represents conc. in anode or left compartment = 8 X 10^-3 M
now Eo(cell) = Eo(reduction) + Eo(oxidation ) = 0.34 + -0.34 = 0
and Ecell) = Eo(cell) - RT/nF ln [ Cu2+] (left) / [Cu2+] (right)
where R = 8.314 J/K/mole
T = 298 K
n = 2
F = 96500 coulombs
E(cell) = 0 - 8.314 X 298 / 2 X 96500 X ln 8 X 10^-3 / 8.5 X 10^-2
E(cell) = -0.013 X ln 0.0941
E(cell) = -0.013 X -2.363 = 0.0307 V
http://www.chem.uiuc.edu/webfunchem/nern…
feel free to ask any question
Cu2+ + 2e- ------> Cu .......Eo (reduction) = 0.34 V
reaction at anode or in the left compartment :
Cu ----------> Cu2+ + 2e- ......Eo (oxidation) = -0.34 V ( - sign since reaction is reversed)
adding both the reactions ...
Cu2+ + 2e- ---------> Cu
+Cu -----------> Cu2+ + 2e-
--------------------
Cu2+ --------> Cu2+
where Cu2+ in left sides of equation represent conc in cathode or right compartment = 8.5 X 10^-2 M
and Cu2+ in right side of equation represents conc. in anode or left compartment = 8 X 10^-3 M
now Eo(cell) = Eo(reduction) + Eo(oxidation ) = 0.34 + -0.34 = 0
and Ecell) = Eo(cell) - RT/nF ln [ Cu2+] (left) / [Cu2+] (right)
where R = 8.314 J/K/mole
T = 298 K
n = 2
F = 96500 coulombs
E(cell) = 0 - 8.314 X 298 / 2 X 96500 X ln 8 X 10^-3 / 8.5 X 10^-2
E(cell) = -0.013 X ln 0.0941
E(cell) = -0.013 X -2.363 = 0.0307 V
http://www.chem.uiuc.edu/webfunchem/nern…
feel free to ask any question