In this section all reactions will be redox processes which can occur through an electron transfer mechanism.
In these cases the two half reactions show the actual loss and gain of electrons occurring in the transfer.
Simple examples of this type of reaction are the gain and loss of electrons by metals:
2 Ag+ + Mg 2 Ag + Mg2+
Oxidation: Mg Mg2+ + 2 e-
Reduction: Ag+ + e- Ag
Fe2+ + Ce4+ Fe3+ + Ce3+
Oxidation: Fe2+ Fe3+ + e-
Reduction: Ce4+ + e- Ce3+
With electron transfer processes, it is possible to partially separate the
two half reactions from each other. The electrons can be transferred
from one half reaction to the other through a metal wire:
In fact, this set-up will not work. Although a path for the electron transfer is available, through the wire, the process will quickly stop as charge builds up in the two pots due to the electron transfer.
To allow the transfer to continue, some way of equalizing the charge caused by the electron transfer must be included.
The Salt Bridge.
The way that this is accomplished is to make a connection between the two half reaction solutions such that ions can be transferred between the two thus preserving electrical neutrality. Such an ion-transfer connection is called a salt bridge.
In this case, as electrons are transferred from the magnesium to the silver half reactions, the resulting charge transfer can be neutralized by a movement of NO3- towards the magnesium and of K+ towards the silver.
The metals dipping in to the solutions are called electrodes.
A Voltaic Cell.
A set-up such as the one above is called an electrochemical cell.
If the cell is arranged so that a spontaneous flow of electrons (and ions) occurs, then the cell is called a Voltaic Cell.
Voltaic cells are usually written (by convention) so that the half reaction on the left is the oxidation half reaction, the electrons being produced here and traveling in the wire to the reduction half reaction on the right. Additional labeling is of anode (where oxidation takes place, left compartment), and cathode (where reduction takes place, right compartment).
Which way will electrons flow?
If two half reactions are coupled together to form a voltaic cell, which way will the electrons flow?
Experiment has shown that the various half reactions exert an electron pressure, or voltage, which is dependent on the ion concentrations but not on the amount of solution present. These voltages are tabulated under unit activity (1 M for all ions) against a reference half reaction (because there must be two half reactions to make a full reaction!):
2 H+ + 2 e- H2.
The Hydrogen Reference Electrode.
Because hydrogen is not a metal and cannot physically form an electrode, an inert electrode of platinum metal is used.
Because Magnesium exerts a higher electrical pressure (i.e. voltage) than does hydrogen, the hydrogen electrode in this example becomes the cathode.
By measuring the voltages under unit activity (all ions 1 M), a table of standard reduction potentials can be set up.
Some Standard Reduction Potentials, Eº, in volts at 298 K.
|Reduction Half Reaction||
|Li+ + e- ¾ Li(s)||- 3.05|
|Na+ + e- ¾ Na(s)||-2.71|
|Al3+ + 3 e- ¾ Al(s)||-1.66|
|Zn2+ + 2 e- ¾ Zn(s)||-0.76|
|PbSO4(s) + 2 e- ¾ Pb(s) + SO42-||-0.36|
|Ni 2+ + 2 e- ¾ Ni(s)||-0.25|
|AgI(s) + e- ¾ Ag(s) + I-||-0.15|
|Pb2+ + 2 e- ¾ Pb(s)||-0.13|
|2 H+ + 2 e- ¾ H2(g)||0.00|
|Cu2+ + 2 e- ¾ Cu(s)||0.34|
|Fe3+ + e- ¾ Fe2+||0.77|
|Ag+ + e- ¾ Ag(s)||0.80|
|Br2(l) + 2 e- ¾ 2 Br-||1.07|
|O2(g) + 4 H+ + 4 e- ¾ 2 H2O||1.23|
|MnO2(s) + 4 H+ + 2 e- ¾ Mn2+ + 2 H2O||1.23|
|Cr2O72- + 14 H+ + 6 e- ¾ 2 Cr3+ + 7 H2O||1.33|
|Cl2(g) 2 e- ¾ 2 Cl-||1.36|
|MnO4- + 8 H+ + 5 e- ¾ Mn2+ + 4 H2O||1.52|
|PbO2(s) + SO42- + 4 H+ + 2 e- ¾ PbSO4(s) + 2H2O||1.68|
|F2(g) + 2 e- ¾ 2 F-||2.87|
Reduction Potentials, Oxidation Potentials and Reactions.
Reduction potentials are an indication of the ability of a species to pick up electrons from another species. The reactions indicated are half reactions (the reduction half) and for any real process they must be coupled with an oxidation half reaction.
Oxidation potentials are exactly the opposite of reduction potentials, that is if you reverse the reaction, the voltage is negated.
For a complete reaction, the total voltage for the two half reactions must be positive for a spontaneous process. If a negative voltage is calculated, the reaction is spontaneous in the reverse direction.
From this, by analogy, cell potentials are thermodynamic quantities: they must be positive for a spontaneous process.
Using Standard Reduction Potentials.
1 From standard reduction potentials, calculate whether the following processes are spontaneous under standard conditions at 298 K.
a. 2 Ag+ + Cu ¾ Ag + Cu2+
Method: split reaction into its half reactions; find the cell potential.
|1||2 Ag+ + 2 e- ¾ 2 Ag||0.80 v|
|2||Cu ¾ Cu2+ + 2 e-||-0.34 v|
|3||2 Ag+ + Cu ¾ Ag + Cu2+||0.46 v|
Row 1 is the reduction half reaction, the voltage is as it is on the standard reduction potential chart. Note that the amount of reaction taking place does not change the voltage; the voltage changes only with activities.
Row 2 is the oxidation half reaction, the voltage is the negative of the standard reduction potential.
Row 3 is the sum of rows 1 and two.
Since the overall cell potential is positive, the reaction is predicted to be spontaneous.
b. Cl2 + 2 Br- ¾ Br2 + 2 Cl-
|1||Cl2 + 2 e- ¾ 2 Cl-||1.36 v|
|2||2 Br- ¾ Br2 + 2 e-||-1.07 v|
|2||Cl2 + 2 Br- ¾ Br2 + 2 Cl-||0.29 v|
Row 1: the reduction of chlorine.
Row 2: the oxidation of bromine; the oxidation potential is required.
Row 3: the sum of rows 1 and 2.
A positive cell voltage indicates the reaction is spontaneous at 298 K and unit activities.
2. To predict whether a certain oxidizing or reducing agent will be
Can dichromate be used to oxidize chloride ion to chlorine gas?
Note that if dichromate is doing the oxidizing, it is being reduced. In a chemical process the oxidizing agent is reduced.
Note also that the half reactions involved can be taken straight from the table of reduction potentials:
|1||Cr2O72- + 14 H+ + 6 e- ¾ 2 Cr3+ + 7 H2O||1.33 v|
|2||6 Cl- ¾ 3 Cl2 + 6 e-||-1.36 v|
|3||6Cl- + Cr2O72- + 14 H+ ¾ 3 Cl2 + 2 Cr3+ + 7 H2O||-0.03 v|
Row 1: the reduction of the dichromate ion.
Row 2: the oxidation of the chloride ion adjusted so as to produce the 6 electrons used in the dichromate reduction. (The oxidation potential does not change with amount of reaction.)
Row 3: the sum of rows 1 and 2. The equation is balanced.
Since the cell voltage here is negative, the reaction is non-spontaneous in the direction written.
Constructing a Voltaic Cell.
By definition a voltaic cell is one in which a chemical process is occurring spontaneously.
Construct an electrochemical cell using zinc and copper and their respective ions.
Step 1. Look up the reduction potentials and decide which half reaction will be oxidation and which reduction.
2. Set up the cell with the oxidation half reaction on the left, the reduction half reaction on the right (this is the convention).
Zn2+ + 2 e- ¾ Zn(s) -0.76
Cu2+ + 2 e- ¾ Cu(s) 0.34
Because the cell voltage must be positive for a spontaneous process, the half reaction with the lowest reduction potential must be reversed to become the oxidation. Thus the zinc half reaction is the oxidation; the copper half reaction the reduction.
|1||Cu2+ + 2 e- ¾ Cu(s)||0.34 v|
|2||Zn(s) ¾ Zn2+ + 2 e-||0.76 v|
|3||Cu2+ + Zn ¾ Cu(s) + Zn2+||1.10 v|
The Fully Labeled Sketch of the Voltaic Cell:
Cell reaction: Cu2+ + Zn ¾ Cu(s) + Zn2+
Standard Cell voltage: 1.10 v
Shorthand Cell Notation: Zn | Zn2+ || Cu2+ | Cu
Design a voltaic cell
Sketch a voltaic cell which has a voltage of about 2 v.
Method: find a redox pair with a voltage difference of about 2 v.
Create the cell.
Al3+ + 3 e- ¾ Al(s) -1.66
Cu2+ + 2 e- ¾ Cu(s) 0.34
|1||3 Cu2+ + 6 e- ¾ 3 Cu(s)||0.34 v|
|2||2 Al(s) ¾ 2 Al3+ + 6 e-||1.66 v|
|3||3 Cu2+ + 2 Al(s) ¾ Cu(s) + 2 Al3+||2.00 v|
Cell shorthand notation: Al | Al3+ || Cu2+ | Cu
What to do when no elemental metal is involved to form an electrode.
Many half reactions involve ions or gases. How is a voltaic cell set up under such conditions? The answer is to use an inert electrode as it was used for hydrogen. In all cells that you will be asked to sketch, you can use an inert platinum electrode.
For example, sketch the voltaic cell corresponding to the reaction: Fe2+ + Ce4+ ¾ Fe3+ + Ce3+.
|1||Ce4+ + e- ¾ Ce3+.||1.61 v|
|2||Fe2+ ¾ Fe3+ + e-||-0.77 v|
|3||Fe2+ + Ce4+ ¾ Fe3+ + Ce3+.||0.84 v|
Cell shorthand notation: