18.5.1 CWK - Copper Concentration Cell - Nernst Equation?

Question

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 :)

A.S. · Accepted Answer

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/nernst/ConcCells.htm

feel free to ask any question

swanger · Answer

Nernst Equation Concentration Cell

Anonymous · Answer

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Fe+2(aq) + Zn(s) = Fe(s) + Zn2+ (aq) According to Nernst equation Ecell = E^0cell - RT/2F ln {[Zn2+]/[Fe2+]} {[Zn(s)] = [Fe(s)] = 1} Initially when [Fe2+] = 1.0M, [Zn2+] = 1.0M Thus Ecell = E^0cell Now when [Fe2+] = 2.0M, [Zn2+] = 1.0M, then Ecell = E^0cell - 2.303 RT/2F log (1.0/2.0) = E^0cell + 0.059/2 log 2.0 (At 25^0C 2.303RT/F = 0.059) = E^0cell + 0.059 x 0.3010/2 = E^0cell + 0.0089 Thus when concentration of Fe2+ ion is increased from 1.0 to 2.0M, the potential of the cell will be increased by an amount of 0.0089 V. Fe(s) + Cu2+ (aq) = Fe2+ (aq) + Cu(s) Again according to Nernst equation Ecell = E^0cell - 0.059/2 log{[Fe2+]/[Cu2+]} {[Fe(s) = [Cu(s)] = 1} Here [Fe2+] = 2.0M and [Cu2+] = 1.0M So, Ecell = E^0cell - 0.059/2 x log 2.0 = E^0cell - 0.059 x 0.3010/2 = E^0cell - 0.0089 In this case by increasing the concentration of Fe2+ ion from 1.0 to 2.0 M, the potential of the cell is decreased by 0.0089 V

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18.5.1 CWK - Copper Concentration Cell - Nernst Equation?

3 Answers