Lesson 9: Standard Oxidation Potential (SOP)

The ability of chemicals to gain and lose electrons can be quantified using a list of standard oxidation potentials. You will use such a list to determine which chemicals will spontaneously react with one another, which will react only if they are forced to do so, and what voltages can be generated by combinations of chemicals in voltaic cells.

Standard Oxidation Potential | Standard Reduction Potential | Using the SOP List | The SOP & SRP Lists

 

Standard Oxidation Potential

Over the years chemists have made comparisons and measurements of the ability of chemicals to lose electrons. Those comparisons have been compiled into a standard oxidation potential list. One such list is given in example 16 in your workbook and on the SOP List section a bit farther down on this page. If  you to take a look at the one in your workbook you will be able to refer to it while you read this page.

Comparison with Expectations

This list contains much information, but for now see how the list we made using the periodic table and lab work compares to this standard list. (An abbreviated SOP list is shown here.) Near the top of this list are potassium, calcium, sodium, magnesium, aluminum and zinc. Then go about half-way down the list to find lead. (Lead is listed in a couple places. Skip the lead plus sulfate because we didn't test lead plus sulfate. We tested lead changing to lead ion by itself.) Further on down is copper. We also had iodide ion. it is right below copper; then a little further down is bromide, and then chloride, and then fluoride way down at the bottom.

Abbreviated SOP List K K+ + e-   2.93 Ca Ca2+ + 2e-    2.87 Na Na+ + e-    2.71 Mg Mg2+ + 2e-    2.37 Al Al3+ + 3e-    1.66 Zn Zn2+ + 2e-    0.76 Pb Pb2+ + 2e-    0.13 Cu Cu2+ + 2e-    -0.34 2 I- I2 + 2e-   -0.54 2 Br- Br2 + 2e-   -1.07 2 Cl- Cl2 + 2e-   -1.36 2 F- F2 + 2e-   -2.87

 

So you can see that our rating of the ease of oxidation for these chemicals was not bad. The uncertain order of sodium and calcium is established. (The reason for calcium being higher is primarily that the calcium ions have a higher charge and smaller size than sodium ions and thus form stronger more stable bonds to water molecules.) The uncertain position of aluminum is also established. You can see that the general trend for the ease of losing electrons fits in pretty well with what can be deduced about the ease of oxidation and reduction from the position of an element on the periodic table.

Potentials

This oxidation potential list does more than just list the chemicals in order of their ease of oxidation. On the right hand side of the list there is a numerical rating (or measure) of their ease of oxidation, EMF or E^o. It is measured in volts compared to an arbitrary standard.

That standard, right in the middle of the chart, is hydrogen becoming hydrogen ion and giving off two electrons. That reaction is used as a comparison standard so it's given an arbitrary rating of 0 volts and everything else is compared to that in terms of being easier to oxidize or not as easily oxidized as hydrogen. The chemicals that can be oxidized are listed in the left hand column.

Abbreviated SOP List K K+ + e-  2.93 Ca Ca2+ + 2e- 2.87 Na Na+ + e- 2.71 Mg Mg2+ + 2e- 2.37 Al Al3+ + 3e- 1.66 Zn Zn2+ + 2e- 0.76 Pb Pb2+ + 2e- 0.13 H2 2 H+ + 2e-   0.00 Cu Cu2+ + 2e- -0.34 2 I- I2 + 2e- -0.54   2 Br- Br2 + 2e- -1.07 2 Cl- Cl2 + 2e- -1.36 2 F- F2 + 2e-   -2.87

This standard oxidation potential list is far from a complete list. There are many, many reactions. This is just a sampling.

You should notice in the list in your workbook that some lines include more than one chemical. If the oxidation rating is based on the presence of some other chemical in the reaction, that chemical is listed there, too. Two examples are that lead is more easily oxidized in the presence of sulfate ion and gold is more easily oxidized in the presence of chloride ion. (That is part of the reason why aqua regia can dissolve gold.)

The phrase standard oxidation potential deserves a bit of explanation.

• Standard simply means that these measurements and comparisons are made under standard conditions for redox reactions.  It is under these conditions that the voltage is measured.
  • the concentration of all soluble chemicals is 1 M,
  • the temperature is 25^oC, and
  • the pressure is 1 atm for any gaseous chemicals.
• Oxidation is the type of half-reaction shown here.
• Potential means the voltage which is associated with the tendency for that reaction to occur.

If the concentrations, temperature or pressure are different , then the voltage associated with that particular half-reaction will be different.

 

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Standard Reduction Potential

Many chemists use standard reduction potentials (SRP) instead of standard oxidation potentials (SOP). An example of this is shown in example 17 in your workbook (and on the SRP List section). An abbreviated version is shown below for our immediate use. If you compare these two lists with one another, you will find that they are simply the reverse of one another in every way. Consequently, you need only one of these charts.

For example, if you look at the first reaction on the standard oxidation potential list, you can see that it's the reverse of the last reaction on the reduction potential list and the voltage is just the opposite. That is, if K becomes K⁺ and gives off one electron and has a voltage of 2.93 volts, then K⁺ taking on one electron to become K has a voltage of -2.93 volts.

Abbreviated SRP List F2 + 2e-  2 F-   2.87 Cl2 + 2e-  2 Cl-    1.36 Br2 + 2e-  2 Br-    1.07 I2 + 2e-  2 I-    0.54 Cu2+ + 2e- Cu    0.34 2 H+ + 2e- H2    0.00 Pb Pb2+ + 2e-    -0.13 Zn2+ + 2e- Zn    -0.76 Al3+ + 3e- Al    -1.66 Mg2+ + 2e- Mg   -2.37 Na+ + e- Na   -2.71 Ca2+ + 2e- Ca   -2.87 K+ + e-  K   -2.93

Abbreviated SOP List K K+ + e-   2.93 Ca Ca2+ + 2e-    2.87 Na Na+ + e-    2.71 Mg Mg2+ + 2e-    2.37 Al Al3+ + 3e-    1.66 Zn Zn2+ + 2e-    0.76 Pb Pb2+ + 2e-    0.13 H2 2 H+ + 2e-    0.00 Cu Cu2+ + 2e-    -0.34 2 I- I2 + 2e-   -0.54 2 Br- Br2 + 2e-   -1.07 2 Cl- Cl2 + 2e-   -1.36 2 F- F2 + 2e-   -2.87

Here is one more example. The S.O.P. list shows that lead becoming Pb²⁺ has a voltage of 0.13 volts. If, for some reason, you wanted to know about the tendency of lead ion to gain two electrons, you could simply find lead ion on the right side of the oxidation potential list and note that for it to gain two electrons, it must reverse the reaction as it is written and thus the voltage would be -0.13 volts.

One reason for giving you both the oxidation and the reduction potential list is so that you are aware of the similarity between the two and that you'd be able to use either of these lists, depending on what is available to you when you have to do a calculation or make a comparison.

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Using the SOP List

A standard oxidation potential list can be used to do a variety of things. The most significant of these are discussed below.

Identify Most Easily Oxidized or Reduced Chemicals

You can use it to determine which of many substances has the greatest tendency to be oxidized or reduced. To do that you simply need to look for its position on the list and understand what that position represents.

Chemicals which can be oxidized are written on the left side of the arrows, and the ones which are highest on the left side have the greatest tendency to be oxidized; that is, they are most easily oxidized. In order for one of the chemicals on this list to be reduced, it must be on the right side of the arrows, and those that are furthest down have the greatest tendency to be reduced. Of all the chemicals on this list, potassium has the greatest tendency to be oxidized and fluorine, F2, has the greatest tendency to be reduced. Conversely, fluoride ion has the least tendency to be oxidized and potassium ion has the least tendency to be reduced. Again, you should be able to see the relationships between the position on this potential list and the tendency for the chemicals to be either oxidized or reduced.

Abbreviated SOP List most easily oxidized hardest to reduce K K+ + e-   2.93 Ca Ca2+ + 2e-    2.87 Na Na+ + e-    2.71 Mg Mg2+ + 2e-    2.37 Al Al3+ + 3e-    1.66 Zn Zn2+ + 2e-    0.76 Pb Pb2+ + 2e-    0.13 H2 2 H+ + 2e-    0.00 Cu Cu2+ + 2e-    -0.34 2 I- I2 + 2e-   -0.54 2 Br- Br2 + 2e-   -1.07 2 Cl- Cl2 + 2e-   -1.36 2 F- F2 + 2e-   -2.87 hardest to oxidize most easily reduced

Practice

Take time now to make sure that you understand the importance of the position on this standard oxidation potential list as it relates to the tendency of these chemicals to be oxidized or reduced. After you have done that, check your answers below and continue with this lesson to learn some of the other things we can do with oxidation potential lists.

1. Which has the greater tendency to be oxidized, Al or Cr?
2. Which has the greater tendency to be reduced, I₂ or Fe³⁺?
3. Which has the greater tendency to be oxidized, Fe²⁺ or Sn²⁺?
4. Which has the greater tendency to be reduced, Fe²⁺ or Sn²⁺?

Answers

1. Which has the greater tendency to be oxidized, Al or Cr?  Al is higher on the left.
2. Which has the greater tendency to be reduced, I₂ or Fe³⁺?  Fe³⁺ is lower on the right.
3. Which has the greater tendency to be oxidized, Fe²⁺ or Sn²⁺?   Sn²⁺ is higher on the left.
4. Which has the greater tendency to be reduced, Fe²⁺ or Sn²⁺?  Sn²⁺ is lower on the right.

Identifying Oxidized and Reduced Forms of Chemicals

Using a standard oxidation potential list, you should also be able to identify the oxidized or reduced forms of chemicals. The chemicals listed on the right side are the oxidized form. Remember that the process of losing electrons is oxidation, so potassium can be oxidized. Potassium ion has been oxidized so is in the oxidized state. So you can distinguish between the oxidized and reduced forms of chemicals by whether they are on the left or the right side of the arrows.

There are few chemicals which are on both sides, like the Fe²⁺. About a third of the way down the list at the voltage of .44, Fe²⁺ is the oxidized form when compared to Fe. But if you go further down the list to -.77 volts, Fe²⁺ is the reduced form when compared to Fe³⁺.

Practice

Identify the oxidized and reduced forms of zinc, tin and bromine.

Answers

Element	Oxidized form	Reduced form
zinc	Zn2+	Zn
tin can have more than one oxidized state so multiple answers result	Sn4+ (and Sn2+ when compared to Sn)	Sn (and Sn2+ when compared to Sn4+)
bromine	Br2	Br-

 

Identifying Spontaneous Reactions

To determine which redox reactions will occur spontaneously by using the standard oxidation potential list, you must first realize that in order for a redox reaction to occur, something must lose electrons and something must gain electrons. So, if the chemicals are present on this list, one of the chemicals has to be on the left side of the arrows and the other chemical has to be on the right side of the arrows. In addition, the one on the left must be higher than the one on the right.

In the following examples be sure to locate the chemicals on your larger SOP list, even though the examples on this page will show the chemicals being used. The examples here will show only the chemicals used. You need to get experience finding them in the larger list and to see where they are in the larger context of the entire list.

Examples

Take a simple case off the top: Potassium metal and calcium ion. Those two chemicals will react spontaneously with one another. Potassium can lose electrons and the calcium ion can gain electrons.	K K+ + e-   2.93 Ca Ca2+ + 2e-    2.87
Let's look at a slightly different situation. Go down two more lines to check out sodium ion with magnesium metal. The sodium ion can gain electrons and magnesium metal can lose electrons. However, the tendencies are such that those two will not react spontaneously with one another. Electrons would have to be forced from magnesium metal to sodium ion.	Na Na+ + e-    2.71 Mg Mg2+ + 2e-    2.37
In order to determine that a reaction will occur spontaneously between two chemicals on this list, not only must one be on the left side of the arrows and the other on the right side, but the one that is on the left has to be higher on the list than the one on the right. So K will react with Ca2+ but Mg will not react with Na+.

Practice

Here are some additional combinations for practice. (These are also given in example 18 in your workbook.)

1. Will zinc metal and nickel metal react?
2. Will zinc ion react with nickel metal?
3. Will zinc metal react with nickel ion?

Answers

1. Will zinc metal and nickel metal react?  No, they both need to lose electrons. In order for a reaction to occur, one has to lose electrons and one has to gain electrons.
2. Will zinc ion react with nickel metal? No, they are on the opposite sides of the arrows, but the one on the left is lower on the list than the one on the right. So that won't work.
3. Will zinc metal react with nickel ion?  Yes, that will work and when the reaction takes place, the zinc becomes zinc ion and the nickel ion becomes nickel metal.

 

Calculating Voltages

You can also use the standard oxidation potential list to calculate the voltage or tendency of a redox reaction to occur.

Examples

Here we have the reaction of zinc metal and copper ion becoming zinc ion and copper metal.
Because of the relative positions of the chemicals on the oxidation potential list we know that this reaction will occur spontaneously (because zinc is on the left side of the oxidation potential list and copper ion is on the right side and below the zinc).		Zn Zn2+ + 2e-      0.76v Cu Cu2+ + 2e-     -0.34v

 

To figure the potential or voltage for a redox reaction, add the voltages for the two half-reactions. In this case, the oxidation half-reaction is Zn Zn2+ + 2e- and the voltage for this half-reaction on the oxidation potential list is 0.76 volts. The reduction half-reaction is Cu2+ + 2e- Cu . That is the reverse of the way that it is written on the S.O.P. list. Reversing the reaction will also reverse the voltage. Instead of being -0.34 volts, it will be +0.34 volts. You then combine those two voltages, add them together, to get +1.10 volts for the reaction.

Zn Zn2+ + 2e-      0.76v Cu Cu2+ + 2e-      -0.34v Zn Zn2+ + 2e-      0.76v Cu2+ + 2e- Cu     +0.34v ____________________________ Zn + Cu2+ Zn2+ + Cu    1.10v

An alternate approach, but one that accomplishes the same thing, is to leave both half-reactions written as oxidation reactions but subtract the voltage of the lower one because it is going in the reverse direction.

Zn Zn2+ + 2e-      0.76v Cu Cu2+ + 2e-      -(-0.34v) ____________________________ Zn + Cu2+ Zn2+ + Cu    1.10v

 

Next we have zinc metal reacting with iron(II) ion, which is just a little bit below the zinc on the list. In this case the zinc metal becomes zinc ion with a voltage of +0.76. Iron(II) ion becoming iron metal is the reverse of the reaction as written on the standard oxidation potential list. Thus, it will have a voltage of -0.44. The two half-reactions together then would have a voltage of those two added together (+0.76 and  -0.44) giving a voltage of +0.32 volts.

Zn Zn2+ + 2e-      0.76v Fe Fe2+ + 2e-     0.44v Zn Zn2+ + 2e-      0.76v Fe2+ + 2e- Fe     -0.44v ____________________________ Zn + Fe2+ Zn2+ + Fe    +0.32v

 

Positive and Negative Voltages

These two examples are reactions which occur spontaneously. They are reactions with positive voltages. Redox reactions with negative voltages do not occur spontaneously. They must be forced.

For example, if you consider reacting zinc metal with sodium ion, the voltage for that reaction has a negative value and does not occur. So, positive voltages correspond to redox reactions which do occur and negative voltages correspond to redox reactions which do not occur without help.

Na Na+ + e-      2.71v Zn Zn2+ + 2e-      0.76v 2 Na+ + 2 e- 2 Na      -2.71v Zn Zn2+ + 2e-      0.76v ______________________________ Zn + 2 Na Zn2+ + 2 Na   -1.95v

Incidentally, you do not have to take into account how many electrons are given off when you calculate the voltage. That has already been taken care of and worked into the value.

Practice

Determine the voltages for the following reactions or combinations of chemicals.

 

Answers

Here are the voltages for the following reactions and combinations of chemicals.

Al with Cr3+  0.92v

Pb with Ag+  0.93v

2 I- + Br2 I2 + 2 Br-  0.53v

2 Cr + 3 Cu2+ 2 Cr3+ + 3 Cu  1.08v

If you did not get these values, first check to see that you properly dealt with the negative voltages that some of these half-reactions have. If you are still having troubles check with an instructor.

 

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The SOP & SRP Lists

Standard Oxidation Potentials
Oxidation Half-Reactions	EMF or Eo
K K+ + e- 2.93 Ca Ca2+ + 2 e- 2.87 Na Na+ + e- 2.71 Mg Mg2+ + 2 e- 2.37 Al Al3+ + 3 e- 1.66 H2 + 2 OH- 2 H2O + 2 e- 0.83 Zn Zn2+ + 2 e- 0.76 Cr Cr3+ + 3 e- 0.74 Fe Fe2+ + 2 e- 0.44 Cd Cd2+ + 2 e- 0.40 Pb + SO42- PbSO4 + 2 e- 0.36 Ni Ni2+ + 2 e- 0.25 Sn Sn2+ + 2 e- 0.14 Pb Pb2+ + 2 e- 0.13 H2 2 H+ + 2 e- 0.00 Sn2+ Sn4+ + 2 e- -0.15 Cu Cu2+ + 2 e- -0.34 2 I- I2 + 2 e- -0.54 Fe2+ Fe3+ + e- -0.77 Ag Ag+ + e- -0.80 NO + 2 H2O NO3- + 4 H+ + 3 e- -0.96 Au + 4 Cl- AuCl4- + 3 e- -1.00 2 Br- Br2 + 2 e- -1.07 2 H2O O2 + 4 H+ + 4 e- -1.23 2 Cl- Cl2 + 2 e- -1.36 Au Au3+ + 3 e- -1.50 Mn2+ + 4 H2O MnO4- + 8 H+ + 5 e- -1.51 2 F- F2 + 2 e- -2.87

Standard Reduction Potentials
Reduction Half-Reactions	EMF or Eo
F2 + 2 e- 2 F- 2.87 MnO4- + 8 H+ + 5 e- Mn2+ + 4 H2O 1.51 Au3+ + 3 e- Au 1.50 Cl2 + 2 e- 2 Cl- 1.36 O2 + 4 H+ + 4 e- 2 H2O 1.23 Br2 + 2 e- 2 Br- 1.07 AuCl4- + 3 e- Au + 4 Cl- 1.00 NO3- + 4 H+ + 3 e- NO + 2 H2O 0.96 Ag+ + e- Ag 0.80 Fe3+ + e- Fe2+ 0.77 I2 + 2 e- 2 I- 0.54 Cu2+ + 2 e- Cu 0.34 Sn4+ + 2 e- Sn2+ 0.15 2 H+ + 2 e- H2 0.00 Pb2+ + 2 e- Pb -0.13 Sn2+ + 2 e- Sn -0.14 Ni2+ + 2 e- Ni -0.25 PbSO4 + 2 e- Pb + SO42- -0.36 Cd2+ + 2 e- Cd -0.40 Fe2+ + 2 e- Fe -0.44 Cr3+ + 3 e- Cr -0.74 Zn2+ + 2 e- Zn -0.76 2H2O + 2 e- H2 + 2 OH- -0.83 Al3+ + 3 e- Al -1.66 Mg2+ + 2 e- Mg -2.37 Na+ + e- Na -2.71 Ca2+ + 2 e- Ca -2.87 K+ + e- K -2.93

 

 

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