Concentration cells - Mistake in solutions manual for 20.74

[Zellmer, Robert]

There's a mistake in the solutions manual for 20.74.   This is a
question about concentration cells.  Part (c) has an error.  I'm
going to explain Part (a) and conc. cells in general first.

Part (a) asks which compartment is the cathode.

A conc. cell occurs with the same half reactions for both the
anode and cathode and depends on a difference in conc. in
the two half cells.

For a conc. cell the reaction always goes in the direction of conc.
to dilute for the reaction. The reaction runs in the direction to equalize
the conc. of involved substances in solution until E_cell is zero.
The E_cell has to be positive.  The E^o will be zero, since the two
half cells are the same.

Here's the Nernst eqn for a conc. cell:

                    0.0592 V
E = E^o -  (--------------)*log(Q)
                          n

Since the half-reactions for the anode and cathode are the same the
E^o = 0 and we get,

             0.0592 V
E = -  (--------------)*log(Q)
                    n

A conc. cell will be a voltaic cell (spontaneous, E>0) only if log(Q) is
negative since there's a negative sign in the eqn.  For this to occur
Q < 1 (log of a number < 1 is negative).  This means the conc. soln
will be on the reactant side and the dilute soln will be on the product
side.  That way you get the following for Q,

         dilute
Q = ----------
         conc

Since the conc. of the dilute soln is less than the conc of the conc.
soln the Q < 1.   That means the more dilute soln will always be in
the numerator.

For this question you're given the following half reaction:

AgCl(s)  +  e-   ---->   Ag(s)  +  Cl^-(aq)           E^o = +0.22V

You use this as both the reduction and oxidation reaction.  Then
you add them together to get the overall reaction,

AgCl(s)  +  e-   ---->   Ag(s)  +  Cl^-( aq, dilute)          E^o = +0.22V

Ag(s)  +  Cl^-( aq, conc)   ---->   AgCl(s)  +  e-          E^o = +0.22V
--------------------------------------------------------------
Cl^- (aq, conc)   ---->   Cl^- (aq, dilute)

Based on this the reduction reaction is the more dilute soln.  That
would be the cathode in this case.

Part (b):  E^o(cell) = E^o(reduction) - E^o(oxidation), where these
are both reduction potentials from a table of red. pot.

Since, E^o(reduction) = E^o(oxidation),

E^o(cell) = 0 (as I mentioned above).


Part (c) has an error.  They have the following eqn. for E(cell),

             0.0592 V          [Cl^-, dilute]
E = -  (--------------)*log(-----------------)
                    n                [Cl^-, conc]

             0.0592 V           0.0150
E = -  (--------------)*log(-----------)
                    1                  2.55

             0.0592 V
E = -  (--------------)*(-2.2304) = + 0.13204 V = + 0.132 V
                    1

The solutions manual has it set up correctly but made a mistake
with the sign for E(cell).

d)  As the cell runs the [Cl^-] in the anode compartment dec.
while the [Cl^-] in the cathode compartment inc.  As this happens
E(cell) decreases.  This continues until the [Cl-] in both the
anode and cathode compartments equalize, at which point
Q=1 and E(cell)=0.


This is different than what is shown in Section 20.6 and shown in
lecture.  I used a metal electrode with the cation of the electrode
in solution, such as Cu(s) and Cu^2+(aq) in each half cell.  In this
type of conc. cell the dilute solution would be the anode and the
conc. solution would be the cathode.  The reaction quotient, Q,
is still [dilute]/[conc].

Dr. Zellmer
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