Chem 102 Discussion 14 KEY
Chem 102 – Discussion 14 – Dr. Naleway
Electrochemistry ANSWER KEY
1) Electrochemical Cells
Half-Cell Reduction Potentials
Volts
I2+ 2e <==> 2 I
0.53
Br2 + 2e <==> 2 Br
1.07
Using the above data construct a voltaic cell and determine the standard potential
and the cell potential at 25oC if the concentations of [I ] and [Br ] are 0.2 M and
0.02 M respectively.
Br2 + 2 I  2 Br + I2
Br2 goes from 0 oxidation state to -1 oxidation state (this is the reduction part of the
cell). 1.07 V
I- goes from -1 oxidation state to 0 oxidation state (this is the oxidation part of the cell).
For the oxidation part, change the sign, since potentials are always given as reductions.
-053 V
Total E°cell = 0.54 V (add reduction and oxidation values together)
Nernst equation:
Ecell = E°cell – 0.0591 log Q
n
n = 2 since two electrons are being transferred in the redox reaction.
Log Q = (0.02)2 / (0.2)2 = log (0.01) = -2
Ecell = 0.54 V - 0.0591/2 (-2) = 0.599 V
What will be the voltage if we raise the temperature to 100 oC?
Use the other form of the Nernst equation for temperatures not at standard temperature
of 298 K.
Ecell = E°cell – RTlnQ
nF
Ecell = 0.54 V – (0.008314)(373) ln (0.02)2/(0.2)2
2 (96.485)
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Chem 102 Discussion 14 KEY
0.54 V – (0.008314)(373)(-4.605) = 0.614 V
2 (96.485)
2) Balance the following reaction in a BASIC environment. Show number of
electrons transferred.
Al(s) + NO2¯ (aq)  AlO2¯ (aq) + NH3 (g)
Al goes from a 0 oxidation state to a +3 oxidation state, which means it has been
oxidized (3e- are transferred)
N goes from a +3 oxidation state to a -3 oxidation state, which means it has been
reduced (6e- are transferred)
Since the number of electrons being oxidized must equal the number of electrons being
reduced, a coefficient of 2 must be placed in front of the Al compounds on each side of
the equation. Then split the reaction up into two half reactions and balance the
remaining atoms by adding OH and H2O, since the environment is basic.
Oxidation half reaction:
2 Al ----> 2 AlO2 + 6 e-
To balance out the -2 total charge on the right side, 2 OH will need to be added to the
left side of the equation. This will balance out the charges on both sides.
2 Al + 2 OH
-
----> 2 AlO2 + 6 e-
Reduction half reaction:
NO2 + 6e- ----> NH3
To balance out the -1 total charge on the left side, 1 OH will need to be added to the
right side of the equation. This will balance out the charges on both sides.
NO2 + 6e- ----> NH3 + OH
Put both equations together and cancel out what is the same on both sides, and then
add H2O to balance the remaining H and O atoms.
2 Al + 2 OH + NO2 + 6e- ----> 2 AlO2 + NH3 + OH + 6e2 Al + OH + NO2 ----> 2 AlO2 + NH3
Finally, add one H2O to the left side to balance out the O and H atoms.
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Chem 102 Discussion 14 KEY
2 Al + H2O + OH + NO2 ----> 2 AlO2 + NH3
Then double check to make sure the equation is atom balanced on each side as well as
charge balanced on each side.
3) The following initial rate data are for the oxidation of arsenate ion by ferric ion
in solution
AsO33- + 2 Fe3+ + 2OH-
 AsO43- + 2 Fe2+ + H2O
If 2 micromoles of arsenate were used up in this reaction over a 2 second period,
what current in amperes would be generated by this reaction?
As in AsO33- has an oxidation state of +3
As in AsO43- has an oxidation state of +5
This means that 2e- are transferred
2 x 10-6 moles arsenate x 2e- transferred = 4 x 10-6 moles of electrons
4 x 10-6 moles e- x 96500 C / mol = 0.386 C
Ampere = C / s = 0.386 C / 2 sec = 0.193 A
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