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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. <br>
<br>
Here are some pointers:<br>
<br>
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). <br>
<br>
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.<br>
<br>
Write the Zn reaction as a reduction...<br>
<br>
Zn<sup>2+</sup> + 2e<sup>-</sup> --> Zn (s)<br>
<br>
and the aluminum reaction as an oxidation...<br>
<br>
Al (s) --> Al<sup>3+</sup> + 3e<sup>-</sup><br>
<br>
The balanced cell reaction is <br>
<br>
3 Zn<sup>2+</sup> + 2Al(s) --> 3Zn (s) + 2 Al<sup>3+</sup><br>
<br>
and the cell potential is E<sub>cathode</sub> - E<sub>anode</sub>
where you use the REDUCTION potentials for both. Since the measured
E<sub>cell</sub> was positive, you would conclude that this reaction
occurs spontaneously.<br>
<br>
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...<br>
<br>
for the cathode (as a reduction)<br>
<br>
Al<sup>3+</sup> + 3e<sup>-</sup> --> Al (s)<br>
<br>
and for the anode (as an oxidation)<br>
<br>
Cu (s) --> Cu2+ + 2e-.<br>
<br>
The balanced cell reaction is thus<br>
<br>
2 Al<sup>3+</sup> + 3 Cu(s) --> 2 Al (s) + 3 Cu<sup>2+</sup><br>
<br>
and again the E<sub>cell </sub>= E<sub>cathode</sub> - E<sub>anode</sub>
using the reduction potentials. Since you measured a negative cell
voltage you would conclude that this reaction is not spontaneous as
written.<br>
<br>
Note -- your measured voltages and your experimentally determined
voltages will in all likelihood not match each other due to
contamination issues.<br>
<br>
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...<br>
<br>
Cu (s) + Cu<sup>2+</sup> (0.01 M) --> Cu (s) + Cu<sup>2+</sup>
(0.001 M) <br>
<br>
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.<br>
<br>
<br>
<pre class="moz-signature" cols="72">--
Andrew J. Pounds, Ph.D. (<a class="moz-txt-link-abbreviated" href="mailto:pounds@theochem.mercer.edu">pounds@theochem.mercer.edu</a>)
Associate Professor of Chemistry and Computer Science
Mercer University, Macon, GA 31207 (478) 301-5627
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