[CHM 112] Electrochemistry Lab

[Andrew J. Pounds]

All you need to turn in tomorrow is the report form (not the generalized 
report form with the big box).  Follow the directions on the report form 
and write the equations as indicated.

Here are some pointers:

As you write the balanced cell reactions always treat the cell on the 
RED lead as the cathode (the reduction reaction) and the cell on the 
BLACK lead as the anode (the oxidation reaction).

For example in the first cell on the report form you have Al on the 
black lead and Zn on the red lead.  For the sake of argument I am going 
to assume that the measured cell potential was +0.91 V.

Write the Zn reaction as a reduction...

Zn^2+ + 2e^- --> Zn (s)

and the aluminum reaction as an oxidation...

Al (s) --> Al^3+ + 3e^-

The balanced cell reaction is

3 Zn^2+ + 2Al(s) --> 3Zn (s) + 2 Al^3+

and the cell potential is E_cathode - E_anode where you use the 
REDUCTION potentials for both.  Since the measured E_cell was positive, 
you would conclude that this reaction occurs spontaneously.

Conversely, lets say that when you had the black lead on copper and the 
red lead on aluminum you measured a cell potential of -1.89 V. You still 
assume that the red lead is the cathode (where reduction is taking 
place) and that the black lead is the anode (where oxidation is taking 
place).   You half reactions would therefore be...

for the cathode (as a reduction)

Al^3+ + 3e^- --> Al (s)

and for the anode (as an oxidation)

Cu (s) --> Cu2+ + 2e-.

The balanced cell reaction is thus

2 Al^3+ + 3 Cu(s) --> 2 Al (s) + 3 Cu^2+

and again the E_cell =  E_cathode - E_anode using the reduction 
potentials.  Since you measured a negative cell voltage you would 
conclude that this reaction is not spontaneous as written.

Note -- your measured voltages and your experimentally determined 
voltages will in all likelihood not match each other due to 
contamination issues.

For the concentration cells (the last part) figure out which cell is the 
anode and which cell is the cathode based on the sign of the measured 
voltage and then when you write the overall cell reaction just keep the 
concentration with the components in the balanced chemical reaction.  
For example...

Cu (s) + Cu^2+ (0.01 M) --> Cu (s) + Cu^2+ (0.001 M)

That will make it much easier for you to keep up with what reactant and 
what is the produce when you calculate Q for the Nernst equation.


-- 
Andrew J. Pounds, Ph.D.  (pounds at theochem.mercer.edu)
Associate Professor of Chemistry and Computer Science
Mercer University,  Macon, GA 31207   (478) 301-5627

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