Chemistry Helpers
Oxidation and Reduction Reactions


Oxidation-Reduction Reactions (Redox)
          The four Types of Reactions
          1. Synthesis
          2. Decomposition
          3. Single Replacement
          4. Double Replacement
          * Combustion
Oxidation Numbers: are assigned to atoms or ions as a way to keep track of electron transfers.
Rules for Assigning Oxidation Numbers
1.    The oxidation number for each atom in free elements is always zero (ex: H2, Na, S8)
2.    The oxidation numbers of ions are the same as the charge on the ion. In MgCl2 the Mg+2 ion has an oxidation number of +2. Each of the two Cl- ions has an oxidation number of –1.
3.    Group 1 metals always have a +1 oxidation number.
4.    Group 2 metals always have a +2 oxidation number.
5.    Oxygen has a –2 oxidation number in practically all of its compounds. The exceptions are in peroxides, such as H2O2 and Na2O2 where the oxidation number is –1 and in compounds with fluorine (OF2) in which it is +2.
6.    Hydrogen has an oxidation number of +1 in all its compounds except metal hydrides formed with Group 1 and Group 2 metals, which it is –1.
7.    The sum of the oxidation numbers in a compound must be zero.
8.    The sum of the oxidation numbers in a polyatomic ion must be equal to the charge on the ion. In the CO32- ion, the three oxygens provide a total of –6; the carbon must then be +4.
Finding Oxidation Numbers
Situation 1: Find the oxidation numbers of N and O in N2O.
Solution: Using rule 2, oxygen has an oxidation number of –2. Using rule 7, the total oxidation number of the nitrogen is +2; therefore each N has a +1 oxidation number.
Situation 2: Find the oxidation numbers of K, Mn and O in KmnO4.
Solution: For compounds composed of more than two elements, draw a chart.


Use rule 3 for K, rule 5 for O and rule 7 for Mn.
I.                     Redox Reactions: reactions in which both oxidation and reduction occur.
A.   Oxidation: the loss of electrons by an atom or an ion, increasing the oxidation number.
B.   Reduction: the gain of electrons by an atom or an ion, reducing the oxidation number.
*During Redox there must be an exchange of electrons*
II.                   Redox Reactions:
A.   Electronic equations or equations in which only the atoms/ions being oxidized or reduced are shown.
Ex: 2Al0 + 3Cl20 = 2Al+3Cl3-1
B.   Half Reactions break the electronic equation into an oxidation and reduction half.
Oxidation ½:        Al0 → Al+3 + 3e-
Reduction ½: Cl20 + 2e- → 2Cl-1
C.   Balancing Half Reactions
·        Must conserve the number of electrons!
·        Number of electrons lost equals number of electrons gained
Ø     Oxidizing Agent: causes something else to undergo oxidation by taking on its electrons. This is the substance reduced.
Ø     Reducing Agent: causes something else to undergo reduction by giving its electrons. This is the substance oxidized.
D.   Spectator Ions: ions that are not changed during a redox reaction.
III.                Balancing Simple Redox Reactions:
Remember – overall charge must be conserved!
A.   Identifying Redox Reactions
Once oxidation numbers are assigned, the atom that has shown an increase can be identified as the one that has undergone oxidation. The atom that has a decrease in oxidation number can be identified as the one that has undergone reduction.
*Double Replacement reactions are never redox reactions because NOTHING CHANGES!
Example: HCl + NaOH = NaCl + H2O
IV.               Spontaneous Reactions:
A.   Table J – Reactivity Series of Selected Metals and Nonmetals
Lithium will react more readily with a nonmetal than will any other metal on the list.
Hydrogen (not a metal!!) is shown on the metal side to illustrate what metals will react with acids and which will not. All metals above hydrogen will react spontaneously with acids, while those below hydrogen will not.
B. Spontaneous Reactionsif a reaction proceeds to completion without adding energy sources once it is started, then it is spontaneous.
1.    To predict spontaneous reactions for metals, 2 criteria must exists:
·        You must have something to oxidize and something to reduce.
Mg(s) and Al(s) can only be oxidized
Mg+2 (aq) and Al+3(aq) can only be reduced
*One metal must be neutral, the other an ion!!!
·        The metal undergoing oxidation (the neutral one) must be higher up on Table J than the other.
Cu(s) + Al(s) = cannot be oxidized
2Cr+3 + 3Mn(s) = 3Mn+2 + 2Cr(s)
Cr+3 + Ni(s) = no reaction because Ni is lower than Cr on Table J, must be higher.
2.    To predict spontaneous reactions for nonmetals:
·        Again, there must be something to oxidize and something to reduce:
F2 and Cl2 can only be reduced
F- and Cl- can only be oxidized
·        The one being reduced must be higher on Table J than the ion being reduced.
Cl2 + Br2 = no reaction – can’t both be reduced
Cl2 + 2Br- = 2Cl- + Br2
I2 + Br- = no reaction because I is lower on the table than Br, must be higher
Examples: Which of the following reactions are spontaneous and why?
1.    HCl + Au(s) = AuCl2 + H2
No because Au will not react with H
2.    HCl + Sr(s) = SrCl2 + H2
Yes because Sr is more likely to oxidize than H2
3.    2Na(s) + ZnCl2(aq) = NaCl + Zn
Yes because Na is higher on Table J so it will oxidize while Zn reduces.
V.                 Electrochemistry;
A.   Terminology:
1.    Half-cell: produced when a metal is placed in a solution of a salt of that metal. Provides metals to be oxidized and metal ions to be reduced.
2.    Anion: a negatively charged ion
3.    Cation: a positively charged ion
4.    Electrode:the site at which oxidation or reduction occurs (usually a metal bar)
5.    Anode: the electrode at which oxidation occurs
6.    Cathode: the electrode at which reduction occurs
7.    Electrochemical Cell:a spontaneous redox reaction that creates the flow of electricity (like a battery)
8.    Voltaic Cell:an electrochemical cell in which a spontaneous reaction causes the flow of electrons.
9.    Electrolytic Cell: a redox reaction that is not spontaneous and requires electricity to occur
B.   Electrochemical Cells: 
1.    Voltaic Cells: ½ cells of 2 different metals connected by wires – produces electricity.
·        These are batteries which redox is spontaneous.
·        Electrons flow from the metal being oxidized (reducing agent) to the ion being reduced (oxidizing agent).
·        Salt Bridgefunctions to complete the circuit by allowing ions to flow from reducing side to oxidizing side.
2.    Electrolytic Cells: electricity is used to produce a nonspontaneous redox reaction.
·        Electricity is required! This is not spontaneous!! Requires the use of a Battery.
·        The anode is connected to the Å and the cathode is connected to the q
*The only way to get a group one metal in its elemental form is by electrolysis of their fused salts.
Example: KBr = K(s) + Br2(s)
Memory Aid #1
LEO                      Loss of electrons is oxidation
the lion says          
GER                     Gain of electrons is reduction
Memory Aid #2

The ca-t-ion is the positive ion and the anion is the negative ion.

The cation is attracted to the cathode and the anion is attracted to the anode.

Therefore: cation +, and cathode -

               anion -, and anode +
Memory Aid #3
Think vowels to associate:
  • Oxidation.
  • Increase in oxidation number.
  • Anode is where this occurs.
Think consonants to associate:
  • Reduction.
  • Decrease in oxidation number.
  • Cathode is where this occurs.
Oxidation # balancing
Follow these steps:
1.    Compute oxidation numbers of all species.
2.    Compute the increase and decrease in oxidation numbers and connect them.
3.    Perform a material balance on atoms involved with increase and decrease.
4.    Balance the increase and the decrease.
5.    If in acid: add H+ to balance charge, but if in base: add OH- to balance charge.
6.    Add H2O to balance H atoms.
7.    Check to see if O atoms are balanced.
  • The reaction is in acid if H+ is present.
  • The reaction is in base if OH- is present.
  • It may be appropriate to write a formula equation as a net ionic equation before balancing.
U+4 + MnO4- Mn+2 + UO2+2 in acid.
½ reaction balancing
Steps to follow:
1.    Split the overall reaction into two half reactions each containing one of the species oxidized or reduced. Begin with either half reaction.
2.    Do a material balance on all elements except H and O.
3.    Balance the O atoms by adding H2O’s.
4.    Balance the H atoms by adding H+.
5.    If in acid go to step 6, if in base, eliminate the H+ ions by adding an equal number of OH- ions to both sides of the reaction. The H+ and OH- sum to make water molecules which can now be simplified.
6.    Balance charge by adding electrons to either side.
7.    Repeat steps 2 to 6 for the other half reaction.
8.    Balance the electrons in the two half reactions by multiplying each entire half reaction by the appropriate factor.
9.    Sum the two half reactions.
               U+4 + MnO4- Mn+2 + UO2+2
Balance:                          2H2O + U+4 UO2+2 + 4H+ + 2e- acid.
2x [ 8H+ + 5e- + MnO4- Mn+2 + 4H2O ]
5x [2H2O + U+4 UO2+2 + 4H+ + 2e-]

              2H2O + 5U+4 + 2MnO4- 2Mn+2 + 5UO2+2 + 4H+
The Standard Reduction Potential Table (aka E° Table)
The table is used to:
1.    Predict which spontaneous reactions are likely to occur (total is +)
2.    Predict which non-spontaneous reactions can be forced to occur by an applied current.
3.    Calculate voltages produced by spontaneous reactions or voltages required to force non-spontaneous reactions.
These four selected reactions are listed in the same order as they are on the E° table.
Cu+2 + 2e- Cu(s)          E° = +0.34 V
Pb+2 + 2e- Pb(s)          E° = -0.13 V
Zn+2 + 2e- Zn(s)          E° = -0.76 V
Mg+2 + 2e- Mg(s)         E° = -2.37 V

A general rule of thumb: a positive net E° value indicates that a net reaction is likely spontaneous while a negative net E° value indicates that a net reaction is definitely non-spontaneous.
a.     Pb(s) will react with Cu+2 but not with Zn+2or Mg+2.
b.    Zn(s) will react with Cu+2 and with Pb+2 but not with Mg+2.
c.     Zn+2 will react with Mg(s) but not with Mg+2.
How to Correctly Handle E° Values
1.    Use the E° value as written for the half reaction occurring left to right (ie. the reduction half reaction).
2.    Use the opposite of the E° value for the half reaction occurring right to left (ie. the oxidation half reaction).
3.    When a half reaction is doubled or tripled for the sake of balancing electrons, the E° value is unaffected. This is so because the E° value is a description of energy carried per electron and is therefore unaffected by the number of electrons.
The Hydrogen Reference Cell
The Hydrogen half reaction has an E° value of 0.00V. This value is arbitrarily chosen and all other E° values are measured in cells made by joining the half cell of the species to be measured with a Hydrogen half cell. The E° value of any half cell cannot be measured independently, therefore one half reaction must be assigned a value of zero.

Write Your Questions Below

When sulfur changes from an oxidation state of -2 in sulfides to +6 in the sulfate ion, the sulfur must A. gain two electrons B. lose eight electrons C. lose six electrons D. gain four electrons
November 20, 2013

 Hi Christina, in your question sulfur goes from a negtive to a positive number. This can only be caused by a loss of electrons since they are negative. You can never lose protons or you have changed the element present (remember the atomic number is the number of protons an atom has). Therefore, sulfur lost 8 electrons.

January 9, 2014 -  Replied By Expert

Copper ( II ) Nitrate &ammonium hydroxide; What is the reaction and is this a chemical or physical change?
February 21, 2012

Cu(NO3)2 + NH4OH → Cu(OH)2 + NH4(NO3)

This is a double replacement reaction. Which results in a chemical change.

*Please use your refence table F Solubility Guidelines to determine if these reactions will occur or not.

February 21, 2012 -  Replied By Expert

calcium nitrate & ammonium hydroxide; what is the reaction and is it a chemical or physical reaction?
February 21, 2012

Ca(NO3)2 + NH4OH → Ca(OH)2 + NH4NO3

This reaction will not go to completion because both compounds would be soluble and one needs to be insoluble.  **Again please refer to reference table F**

February 21, 2012 -  Replied By Expert

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