REDUCTION AND OXIDATION REACTIONS


How are redox reactions different?

Oxidation states.

Is it a redox reaction?

Half reactions.

Reduction or oxidation?

Practice with assigning oxidation states.

Balancing redox reactions.


 

HOW ARE REDOX REACTIONS DIFFERENT?

Redox is the term used to label reactions in which the acceptance of an electron (reduction) by a material is matched with the donation of an electron (oxidation). A large number of the reactions already mentioned in the Reactions chapter are redox reactions.

Synthesis reactions are also redox reactions if there is an exchange of electrons to make an ionic bond. If chlorine gas is added to sodium metal to make sodium chloride, the sodium has donated an electron and the chlorine has accepted an electron to become a chloride ion or an attached chlorine.

If a compound divides into elements in a decomposition, a decomposition reaction could be a redox reaction. The electrolysis of water is a redox reaction. With a direct electric current through it, water can be separated into oxygen and hydrogen. H2O ==> H2 + O2 The oxygen and hydrogen in the water are attached by a covalent bond that breaks to make the element oxygen and the element hydrogen. Learning more about the conditions for redox reactions will show that the electrolysis of water is a redox reaction.

A single replacement reaction is always a redox reaction because it involves an element that becomes incorporated into a compound and an element in the compound being released as a free element.

A double replacement reaction usually is not a redox reaction.

 

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OXIDATION STATES

Before we go any further into redox, we must understand oxidation states. The idea of oxidation state began with whether or not a metal was attached to an oxygen. Unattached (free) atoms have an oxidation state of zero. Since oxygen almost always takes in two electrons when it is not a free element, the combined form of oxygen (oxide) has an oxidation state of minus two. The exception to a combined oxygen taking two electrons is the peroxide configuration. Peroxide can be represented by -O-O- where the each dash is a covalent bond and each O is an oxygen atom. Peroxide can be written as a symbol, (O2)2-. The over-simplified way of showing this is that each oxygen atom has a negative one oxidation state, but that is not really so because the peroxides do not come in individual oxygen atoms. Peroxides are not as stable as oxides, and there are very many fewer peroxides in nature than oxides. H2O2 is hydrogen peroxide.

Hydrogen in compound always has an oxidation state of plus one, except as a hydride. A hydride is a compound of a metal and hydrogen. The hydrogen atoms in a hydride have the oxidation state of -1. Hydrides react with water, so there are no hydrides found in nature. The formula XH or XH2 or XH3 or even XH4 where X is a metal is the general chemical formula for hydride.

The rules for oxidation state are in some ways arbitrary and unnatural, but here they are:

 

1. Any free (unattached) element with no charge has the oxidation state of zero. Diatomic gases such as O2 and H2 are also in this category.

2. All compounds have a net oxidation state of zero. The oxidation state of all of the atoms add up to zero.

3. Any ion has the oxidation state that is the charge of that ion. Polyatomic ions (radicals) have an oxidation state for the whole ion that is the charge on that ion. The ions of elements in Group I, II, and VII (halogens) and some other elements only have one likely oxidation state.

4. Oxygen in compound has an oxidation state of minus two, except for oxygen as peroxide, which is minus one.

5. Hydrogen in compound has an oxidation state of plus one, except for hydrogen as hydride, which is minus one.

6. In radicals or small covalent molecules, the element with the greatest electronegativity has its natural ion charge as its oxidation state.

KNOW THIS

 

Now would be a good time to try the oxidation state problems beginning the practice page at the end of this chapter. Problems 1-30 are good examples for practice of assigning oxidation states.

 

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IS IT A REDOX REACTION?

A redox reaction will have at least one type of atom releasing electrons and another type of atom accepting electrons. How can you most easily tell if a reaction is redox? Label every atom on both the reactant and product side of the equation with its oxidation number. If there is a change in oxidation number from one side of the equation to the other of the same species of atom, it is a redox reaction. Each complete equation must have at least one atom species losing electrons and at least one atom species gaining electrons. The loss and gain of electrons will be reflected in the changes of oxidation number.

Lets take the following equation: K2(Cr2O7) + KOH ==> 2 K2(CrO4) + H2O Is it a redox equation or not? Potassium dichromate and potassium hydroxide make potassium chromate and water. Some of the atoms are easy. All of the oxygens in compound have an oxidation state of minus two. All of the hydrogens have an oxidation state of plus one. Potassium is a group one element, so it should have an oxidation state of plus one in the compounds. That seems to make sense because dichromate and chromate ions have a charge of minus two and there are two potassium atoms in each compound. Hydroxide ion has a charge of minus one and it has one potassium. But what about the chromium atoms? We can do a little primitive math on the material either from the starting point of the compound or the ion to find the oxidation state of chromium in that compound. The entire compound must have a net oxidation state of zero, so the oxidation numbers of two potassiums one chromium and four oxygens must equal to zero. 2 K + Cr + 4 O = 0 We know the oxidation state of everything else but the chromium. 2(+1) + Cr + 4 (-2) = 0 and Cr = +6.  Or we could do it from the point of view of the chromate ion. Cr + 4 O = -2 The oxygens are minus two each. Cr + 4 (-2) = -2 Either way Cr = +6. Now the dichromate; 2 K + 2 Cr + 7 O = 0 and 2 (+1) + 2 Cr + 7 (-2) = 0. Then 2 Cr = +12 and Cr = +6. You can do the math for the dichromate ion to see for yourself that the chromium does not change from one side of this equation to the other. As suspicious-appearing as the equation might have seemed to you, it is not a redox reaction.

Consider copper metal in silver nitrate solution becomes silver metal and copper II nitrate. The oxygens do not change. Oxygen in compound is negative two on both sides. The nitrogen cannot change. It does not move out of the nitrate ion where it has an oxidation state of plus five. (Is that right?) The other two have to change because they both are elements with a zero oxidation state on one side and in compound on the other. Silver goes from plus one to zero and Copper goes from zero to plus two.
AgNO3 + Cu ==> Cu(NO3)2 + Ag (not a balanced reaction)

Think of this on a number line. The copper is oxidized because its oxidation number goes up from zero to plus two. The silver is reduced because its oxidation number reduces from plus one to zero.

 

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HALF REACTIONS

Consider the reaction: AgNO3 + Cu ==> Cu(NO3)2 + Ag (not a balanced reaction)

Half reactions are either an oxidation or a reduction. Only the species of atom that is involved in a change is in a half reaction. In the above reaction, silver goes from plus one to zero oxidation state, but to account for everything, the electrons must be placed into the half reaction. e- + Ag+ ==> Ag0 (Reduction) Notice that the half reaction must be balanced in charge also and that the only way to balance it is to add electrons to the more positive side. The other half reaction is that of copper. Cu0 ==> Cu+2 + 2 e- (Oxidation)

This time the material is oxidized and the electrons must appear on the product side. We must double the silver half reaction to cancel out the electrons from right to left. The two half reactions can be added together to make one reaction, thus.

2( e- + Ag+ ==> Ag0)
Cu0 ==> Cu+2 + 2 e-     and the total reaction is:
Cu0 + 2Ag+ ==> Cu+2 + 2Ag0

In the complete reaction the number of electrons lost must equal the number of electrons gained. The number of electrons used in the reduction half reaction must equal the number of electrons produced in the oxidation half reaction. The entire half reactions must be multiplied by numbers that will equalize the numbers of electrons, and the final complete balanced chemical reaction must show these number relationships.

One of the important bits of information from adding the half reactions in this case is that the entire chemical equation will have to have two silver atoms for every copper atom in the reaction for the reaction to balance electrically. This type of information from the half reactions is sometimes the easiest or only way to balance a chemical equation. The redox balancing problems beginning with number 31 at the end of the chapter are good help for your further understanding.

From doing this math on a number of materials, you will find that it is possible to get some strange-looking oxidation states, to include some fractional ones. The oxidation state math works on fractional oxidation states also, even though fractional charges are not possible.

 

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REDUCTION OR OXIDATION?

A reduction of a material is the gain of electrons. An oxidation of a material is the loss of electrons. This system comes from the observation that materials combine with oxygen in varying amounts. For instance, an iron bar oxidizes (combines with oxygen) to become rust. We say that the iron has oxidized. The iron has gone from an oxidation state of zero to (usually) either iron II or iron III. This may be difficult to remember. The easier way to tell if a half reaction is a reduction or oxidation is to plot the changing ion into the number line. If the oxidation state of the ion goes up the number line, it is an oxidation. If it goes down the number line, it is a reduction. Based on the KIS principle (Keep It Simple), remember only one rule for this.

Someone, in a fit of perversity, decided that we needed more description for the process. A material that becomes oxidized is a reducing agent, and a material that becomes reduced is an oxidizing agent.

 

COMING ATTRACTIONS

ELECTROLYSIS
Water, aluminum, copper

ELECTROPLATING
chromium, gold,

ELECTRIC CELLS
carbon zinc , etc electrical potential and voltages

 

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REDOX PROBLEMS
For each element in the following materials list the number of the rule you use to assign oxidation state of that element and list the oxidation state you have found it to be.
 
.
Summary of Oxidation State Rules    
1. Free element O.S. = 02. Compound total O.S. = 03. Ion O.S. = charge
4. Oxygen O.S. = -25. Hydrogen O.S. = +16. Electronegativity rules
  .
MATERIALRULESOXIDATION STATES
 
1. NaClNa 3; Cl 3Na = +1 , Cl = -1
2. KMnO4 K 3; O 4; Mn 3,4 or 2,3,4K = +1, O = -2, Mn = +7
3. diamondC 1C = 0
4. CO2 O 4; C 4,2 C = +4, O = -2
5. COO 4; C 4,2C = +2, O = -2
6. KCN K 3; N 6; C 3,6, 2 K = +1, C = +2, N=-3
7. Na4Fe(CN)6Na 3; N 6; C 6,3; Fe 3,6,2 Na= +1, N= -3, C= +2, Fe= +2
8. Fe2O3 O 4; Fe 4,2 O = -2, Fe = +3
9. Fe3O4 O 4; Fe 4,2O = -2, Fe = +8/3
10. (ClO4)- O 4; Cl 4,3O = -2, Cl = +7
11. (ClO3)- O 4; Cl 4,3 O = -2, Cl = +5
12. (ClO2)- O 4; Cl 4,3 O = -2, Cl = +3
13. (ClO)- O 4; Cl 4,3 O = -2, Cl = +1
14. Cl- Cl 3 Cl = -1
15. Cl2 Cl 1 Cl = 0
16. P2O5 O 4; P 4,2 O = -2, P = +5
17. P4O6 O 4; P 4,2O = -2, P = +3
18. H3PO4 H 5; O 4; P 5,4,2 or 4,3H = +1, O = -2, P = +5
19. Mg3N2Mg 3; N 3,2 Mg = +2, N = -3
20. MgH2Mg 3; H 3,2 Mg = +2, H = -1 (hydride!)
21. NH3H 5; N 5,2H = +1, N = -3
22. N2H4H 5; N 5,2H = +1, N = -2
23. (NH4)+ H 5; N 5,2 H = +1, N = -3
24. N2 N 1N = 0
25. (NO3)- O 4; N 4,3O = -2, N = +5
26. (NO2)- O 4; N 4,3 O = -2, N = +3
27. NO2 O 4; N 4,2 O = -2, N = +4
28. NO O 4; N 4,2O = -2, N = +2
29. N2O O 4; N 4,2O = -2, N = +1
30. Na2O2 Na 3; O 2 Na = +1, O = -1 (peroxide!)

 

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B. For the each word reaction, write the chemical equation without balancing it, write the oxidation state of each element above that element, and write the two half reactions, labeling which is oxidation and which is reduction. You can check your work by balancing the complete reaction using the numbers from the half reaction addition. If you have a problem with an example, check first with the completed balanced equation in the answer section.
 
31. Hydrogen gas burns in oxygen to make water.
 
32. Mercuric oxide, a red powder, is put into a test tube and warmed. Liquid mercury forms on the sides and in the bottom of the tube and oxygen gas escapes from the test tube.
 
33. Potassium chlorate is heated in a test tube. Oxygen gas is made and potassium chloride is left in the bottom of the tube.
 
34. Hydrochloric acid is poured onto zinc metal to make zinc chloride and hydrogen gas.
 
35. A copper wire is put into silver nitrate. Silver metal appears and the solution turns blue from copper II nitrate.
 
36. Magnetite, an ore of iron, is smelted in large hot furnaces by blowing carbon monoxide gas through the ore. The result is liquid (molten) iron and carbon dioxide bubbles.
 
37. Lead metal and lead IV oxide in sulfuric acid produce lead II sulfate and water. This is the reaction in a common lead-acid car battery.
 
38. Methane gas burns in oxygen to make water vapor and carbon dioxide.
 
39. Octane burns with oxygen to make carbon dioxide and water.
 
40. Concentrated nitric acid is put on copper wire. Water and copper II nitrate in the water solution is produced, along with a brownish gas, nitrogen monoxide or nitric oxide, NO.
 
41. Potassium dichromate and hydrochloric acid in solution will make chlorine gas, water, chromium III chloride and potassium chloride. (The soluble salts, of course, remain in the water solution.)
 
 
42. Potassium permanganate solution when added to potassium cyanide in water solution will make manganese IV oxide and potassium hydroxide and water and potassium cyanate (KOCN).
 
43. In a sulfuric acid solution potassium permanganate will titrate with oxalic acid to produce manganese II sulfate, carbon dioxide, water, and potassium sulfate in solution.
 

 

ANSWERS TO REDOX EQUATIONS
 

 

31. 0 0+1 -22( H0 ==>H+1 +e-) Oxidation
H2+O2 ==>H2O2e- + O0 ==> O-2 Reduction
 
Balanced equation   2 H2 + O2 ==> 2 H2O

 

32.+2 -2 0  02e- + Hg+2 ==> Hg0 Reduction
HgO ==>O2+HgO-2 ==> O0 + 2e- Oxidation
 
Balanced equation   2 HgO ==> O2 + 2 Hg

 

33.+1+5-2+1 -106e- + Cl+5 ==> Cl-1 Reduction
KClO3 ==> KCl+O23( O-2 ==> O0 + 2e-) Oxidation
 
Balanced equation   2 KClO3 ==> 2 KCl + 3 O2

 

34.+1 -1 0+2 -1 02( e- + H+1 ==> H0)  Reduction
HCl+Zn==>ZnCl2+H2Zn0 ==> Zn+2 + e-  Oxidation
 
Balanced equation   2 HCl + Zn ==> ZnCl2 + H2

 

35.+1+5- 2 0+2+5 -2 02( e- + Ag+1 ==> Ag0  Reduction
AgNO3+Cu==>Cu(NO3)2+AgCu0 ==> Cu+2 + 2 e-  Oxidation
 
Balanced chemical equation   2 AgNO3 + Cu ==> Cu(NO3)2 + 2 Ag
 

 

36.+8/3-2+2-2 0+4-24 ( C+2 ==> C+4 + 2 e-) Oxidation
Fe3O4+CO ==> Fe+CO23 ( 8/3 e- + Fe+8/3 ==> Fe0) Reduction
 
Balanced chemical reaction   Fe3O4 + 4 CO ==> 3 Fe + 4 CO2

 

37. 0+4-2+1+6-2+2+6-2+1 -2Pb0 ==> Pb+2 + 2e- Oxidation
Pb+PbO2+H2SO4 ==>PbSO4 +H2O2e- + Pb+4 ==> Pb+2 Reduction
 
Balanced equation   Pb + PbO2 + 2 H2SO4 ==> 2 PbSO4 + 2 H2O - lead oxidizes and reduces.

 

38.-4+1 0+1 -2+4-2C-4 ==> C+4 +8e- Oxidation
CH4 +O2 ==>H2O+CO24( 2e- + O0 ==> O-2 ) Reduction
 
Balanced equation   CH4 + 2 O2 ==> 2 H2O + CO2

 

39. -9/4 +1 0+4-2+1-225 ( 2e- + O0 ==> O-2) Reduction
C8H18+O2 ==>CO2 +H2O2 ( C-9/4 ==> C+4 + 25/4 e-) Oxidation
 
Balanced equation   2 C8H18 + 25 O2 ==> 16 CO2 + 18 H2O

 

40.0+1 +5 -2+2-2 +2 +5 -2+1-22( 3e- + N+5 ==> N+2) Reduction
Cu +H(NO3) ==> NO+Cu(NO3)2 +H2O3( Cu0 ==>Cu+2 + 2e-) Oxidation
 
Balanced equation   3 Cu + 8 HNO3 ==> 4 H2O + 2 NO + 3 Cu(NO3)2

 

41.+1+6-2+1-1 0+3 -1+1 -2+1-13( Cl-1 ==> Cl0 + 1e-) Oxidation
K2(Cr2O7)+HCl ==>Cl2+CrCl3+H2O+KCl3 e- + Cr+6 ==> Cr+3 Reduction
 
Balanced equation   K2(Cr2O7) + 14 HCl ==> 3 Cl2 + 7 H2O + 2 CrCl3 + 2 KCl

 

42.+1 +7 -2+1+2-3+1-2+4  -2+1-2+1+1 -2 +4 -3
KMnO4+K(CN)+H2O ==>MnO2+K(OH)+K(OCN)
 
3( C+2 ==> C+4 +2e-) Oxidation2(3e- + Mn+7 ==> Mn+4) Reduction
 
Balanced equation   2 KMnO4 + 3 KCN + H2O ==> 2 MnO2 + 2 KOH + 3 K(OCN)

 

43. +1 +7 -2 +1+3 -2 +1 +6 -2 +2+6 -2 +4-2 +1-2 +1 +6-2
KMnO4 +H2C2O2 +H2SO4   ==> MnSO4+ CO2+ H2O+ K2SO4
 
5e- + Mn+7 ==> Mn+2 Reduction5( C+3 ==> C+4 + e-) Oxidation
 
Balanced   2 KMnO4 + 5 H2C2O4 + 3 H2SO4 ==> 2 MnSO4 + 10 CO2 + 8 H2O + K2SO4

 


 

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