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Balancing Redox Reactions I

    Oxidation-Reduction or "redox" reactions occur when elements in a chemical reaction gain or lose electrons, causing an increase or decrease in oxidation numbers. In order to balance these reactions, we must use what is called the Half Equation Method.

    1. 1. Introduction
    2. 2. Balancing
    3. 3. Step-by-step Examples
    4. 4. More Examples
    5. 5. Practice Problems
    6. 6. References
    7. 7. Outside links

    Introduction

    In a redox reaction, one or more element becomes oxidized, while one or more element becomes reduced. Oxidation is the loss of electrons while reduction is the gain of electrons. An easy way to remember this is to think of the charges; an element's charge gets reduced if they gain electrons (A good acronym to use to remember the difference is L.E.O.= Lose Electron Oxidation & G.E.R.= Gain Electron Reduction). Redox reactions usually occur in two environments: acidic and basic. In order to balance redox equations, understanding oxidation states is necessary.

    Balancing

    To start the balancing, we must first identify which element(s) is being oxidized and which element(s) is being reduced. 

    For example looking at the following reaction in an acidic solution MnO4- (aq) + I- (aq) 

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    Mn2+ (aq) + I2(s)

    We see that Manganese(Mn) goes from a charge of +7 to a charge of +2. Since it gained 5 electrons, it is being reduced. Because it is being reduced, it is considered the oxidizing agent of the reaction.

    Next we see that Iodine (I) goes from a charge of -1 to 0. Thus it lost electrons, and is being oxidized. Iodine (I) is therefore considered to be the reducing agent of this reaction.

    Now we can form our half equations using what we just found out.

    First half is the reduction half: MnO4- (aq)

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     Mn2+ (aq)

    Next we have the oxidation half: I- (aq) 

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    I2 (s)

    Now comes the balancing part. For the reduction half, all we have to balance is the oxygen, and we do that by adding H2O. Since there are 4 oxygen atoms in MnO4- (aq), we must add 4 water molecules so that the oxygen of the MnO4- (aq) balances out with the oxygen in the H2O. Now that we added water, we introduced hydrogen into the reaction.  Thus, we must balance that as well. This can be done by adding hydrogen ions on the other side as shown below:

    Reduction half: MnO4- (aq) + 8H+ (aq) + 5e-

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     Mn2+ (aq) +4H2O (l)

    To balance the charges we must add electrons. On the left side there is a 7+ charge and the right side there is a 2+ charge. To balance the charges you must add 5 electrons to the side with 7+ charge.

    The reduction half is now sufficiently balanced. Now we need to look at the oxidation half.

    The only thing in the oxidation half is iodine, so we can balance it easily by adding another iodine to the left side.

    Oxidation half: 2I- (aq) 

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    I2 (s) + 2e-

    (note that the electrons were also multiplied because we added another iodine. Also,  a general rule of thumb is that the electrons should appear on either sides in both half reactions so they can cancel out.)

    Now that we have balanced both parts of the reaction, we need to balance the whole reaction. To do this, we just need to balance the electrons on each side. If we look at both equations, we see that there are currently 5 electrons on the left from the reduction half, and 2 on the right from the oxidation half. To balance this, we just find the least common multiple of both numbers and multiply each half by a factor that leads to it. In this case, since the least common multiple is 10, we multiply the reduction half by 2 and the oxidation half by 5.

    Reduction half becomes: 2MnO4- (aq) + 16H+ (aq) + 10e-

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    2Mn2+ (aq) + 8H2O (l)

    Oxidation half becomes: 10I- (aq) 

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    5I2 (s) + 10e-.

    Now we just have to add the two parts together and cancel out anything that is on both sides, in this case, only the electrons can be cancelled out.

    2MnO4- (aq) + 16H+ (aq) + 10e-

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    2Mn2+ (aq) + 8H2O (l)

    10I- (aq) 

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    5I2 (s) + 10e-

    _________________________________________________

    10I- (aq) + 2MnO4- (aq) + 16H+ (aq) 

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     5I2 (s) + 2Mn2+ (aq) + 8H2O (l)

    We have balanced our reaction in an acidic solution!

    Now you may be wondering, what if it was in a basic solution? The answer is simple in a basic solution, there is OH- ions instead of H+ ions, so all we have to do is add OH- ions to both sides. On the side with the hydrogen, the OH- and the H+ ions will react to form water, which will then cancel out with some of the H2O on the other side. In our case, we need to add 16 OH- ions to both sides because we have 16 hydrogen ions. 

    10I- (aq) + 2MnO4- (aq) + 16H+ (aq) + 16OH- (aq)

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     5I2 (s) + 2Mn2+ (aq) + 8H2O (l) + 16OH- (aq)

    The OH and H react to form H2O.

    10I- (aq) + 2MnO4- (aq) + 16H2O (l) 

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     5I2 (s) + 2Mn2+ (aq) + 8H2O (l) + 16OH- (aq)

    The H2O on the right side cancels out with some on the left.

    10I- (aq) + 2MnO4- (aq) + 8H2O (l) 

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     5I2 (s) + 2Mn2+ (aq) + 16OH- (aq)

    And there you have it! Now you can balance a redox reaction in both acidic and basic solutions.

    Step-by-step Examples

    This is a summary guide that will take you step-by-step through the process.

    To balance a redox reaction, you first take an equation and separate into two half reaction equations specifically oxidation and reduction, and balance them.
    Balance the following in an acidic solution.
    Original Equation:  SO32- (aq) + MnO4- (aq)
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    SO42- (aq) + Mn2+ (aq)
    Step 1:
    Split into two half reaction equations: Oxidation and Reduction
    Oxidation: SO32- (aq)
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     SO42- (aq)  [ oxidation because oxidation state of sulfur increase from +4 to +6]
    >THESE ARE THE SKELETAL HALF EQUATIONS
    Reduction: MnO4+ (aq)
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     Mn2+ (aq) [ Reduction because oxidation state of Mn decreases from +7 to +2]
    Step 2:
    Balance each of the half equations in this order:
    -          Atoms other than H and O
    -          O atoms by adding H2Os with proper coefficient
    -          H atoms by adding H+ with proper coefficient
    The Sulfur atoms and Mn atoms are already balanced, hence we can go onto balancing O atoms
    Oxidation: SO32- (aq) + H2O (l)
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     SO4- (aq)
    Reduction: MnO4- (aq)
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     Mn2+ (aq) + 4H2O (l)
    Then balance out H atoms on each side
    Oxidation: SO32- (aq) + 4H2O (l)
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     SO42- (aq) + 2H+ (aq)
    Reduction: MnO4- (aq) + 8H+ (aq)
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     Mn2+ (aq) + 4H2O (l)
    Step 3:
    Balance the charges of the half reactions by adding electrons
    Oxidation: SO32- (aq) + H2O (l)
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     SO4- (aq) + 2H+  (aq) + 2e-
    Reduction: MnO4- (aq) + 8H+ + 5e-
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     Mn2+ (aq) + 4H2O (l)
    Step 4:
    Obtain the overall redox equation by combining the half reaction, but multiply entire equation by number of electrons in oxidation with reduction equation, and number of electrons in reduction with oxidation equation.
    Oxidation:[ SO32- (aq) + H2O (l)
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     SO4- (aq) + 2H+  (aq) + 2e-] x 5
    Reduction: [ MnO4- (aq) + 8H+ + 5e-
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     Mn2+ (aq) + 4H2O (l) ] x 2
    Overall Reaction:
    Oxidation: 5 SO32- (aq) + 5H2O (l)
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     5SO42- (aq) + 10H+ (aq) + 10e-
    +
    Reduction: 2 MnO4- (aq) + 16H+ (aq) +10e-
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     2 Mn2+ (aq) + 8H2O (l)
    5 SO32- (aq) + 5H2O (l) + 2 MnO4- (aq) + 16H+ (aq) +10e-
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     5SO42- (aq) + 10H+ (aq) + 10e- +2 Mn2+ (aq) + 8H2O (l)
    Step 5:
    Simplify. You can cancel out similar terms on both sides, like the 10e- and waters.
    5 SO32- (aq) + 2 MnO4- (aq) + 6H+ (aq) 
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     5SO42- (aq) + 2Mn2+ (aq) + 3H2O (l)

    Some points to remember as you balance redox reactions:

    • The equation is separated into two half-equations, one for oxidation, and one for reduction.
    • The equation is balanced by adjusting coefficients and adding H2O, H+, and e- in this order:
          1) Balance the atoms in the equation, apart from O and H.
          2) To balance the Oxygen atoms, add the appropriate number of water (H2O) molecules to the other side.
          3) To balance the Hydrogen atoms (including those added in step 2), add H+ ions.
          4) Add up the charges on each side. They must be made equal by adding enough electrons (e-) to the more positive side.
    • The e- on each side must be made equal; if they are not equal, they must be multiplied by appropriate integers to be made the same.
    • The half-equations are added together, cancelling out the electrons to form one balanced equation. Cancel out as much as possible.
    • (If the equation is being balanced in a basic solution, the appropriate number of OH- must be added to turn the remaining H+ into water molecules)
    • The equation can now be checked to make sure it is balanced.

    Next, these steps will be shown in another example: 

    MnO4-(aq) + SO32-(aq) --> MnO2(s) + SO42-(aq)

    First, they are separated into the half-equations: 

    MnO4-(aq) --> MnO2(s) (the reduction, because oxygen is LOST) and

    SO32-(aq) --> SO42-(aq) (the oxidation, because oxygen is GAINED)

    Now, to balance the oxygen atoms, we must add two water molecules to the right side of the first equation, and one water molecule to the left side of the second equation:

    MnO4-(aq) --> MnO2(s) + 2H2O(l)

    H2O(l) + SO32-(aq) --> SO42-(aq)

    To balance the hydrogen atoms (those the original equation as well as those added in the last step), we must add four H+ ions to the left side of the first equation, and two H+ ions to the right side of the second equation.

    4H+ + MnO4-(aq) --> MnO2(s) + 2H2O(l)

    H2O(l) + SO32-(aq) --> SO42-(aq) + 2H+

    Now we must balance the charges. In the first equation, the charge is +3 on the left and 0 on the right, so we must add three electrons to the left side to make the charges the same. In the second equation, the charge is -2 on the left and 0 on the right, so we must add two electrons to the right.

    3e- + 4H+ + MnO4-(aq) --> MnO2(s) + 2H2O(l)

    H2O(l) + SO32-(aq) --> SO42-(aq) + 2H+ + 2e-

    Now we must make the electrons equal eachother, so we multiply each equation by the appropriate number to get the common multiple (in this case, by 2 for the first equation, and by 3 for the second).

    2(3e- + 4H+ + MnO4-(aq) --> MnO2(s) + 2H2O(l))

    3(H2O(l) + SO32-(aq) --> SO42-(aq) + 2H+ + 2e-)

    With the result:

    6e- + 8H+ + 2MnO4-(aq) --> 2MnO2(s) + 4H2O(l)

    3H2O(l) + 3SO32-(aq) --> 3SO42-(aq) + 6H+ + 6e-

    Now we cancel and add the equations together. We can cancel the 6e- because they are on both sides. We can get rid of the 6H+ on both sides as well, turning the 8H+ in the first equation to 2H+. The same method gets rid of the 3H2O(l) on the bottom, leaving us with just one H2O(l) on the top. In the end, the overall reaction should have no electrons remaining. Now we can write one balanced equation:

    2MnO4-(aq) + 2H+ + 3SO32-(aq) --> H2O(l) + 2MnO2(s) + 3SO42-(aq)

    The equation is now balanced in an acidic environment. If necessary, we can balance in a basic environment by adding OH- to turn the H+ into water molecules as follows:

     2MnO4-(aq) + H2O + 3SO32-(aq) --> H2O(l) + 2MnO2(s) + 3SO42-(aq) + 2OH-

    The equation is now balanced in a basic environment.

    More Examples

    Example 1: Fe(OH)3 + OCl- 

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     FeO42- + Cl- in acidic solution

     
    Step 1:
     
    Reduction: OCl-  
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     Cl-
    Oxidation:   Fe(OH)3 
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     FeO42-
     
    Step 2/3:
     
    Reduction:  2H+ +  OCl- + 2e- 
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     Cl- + H2O
    Oxidation:  Fe(OH)3 + H2O  
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     FeO42- + 3e- + 5H+
     
    Step 4:
     
    Overall Equation:
     
    [ 2H+ +  OCl- + 2e- 
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     Cl- + H2O ] x 3
    [ Fe(OH)3 + H2O  
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     FeO42- + 3e- + 5H+ ] x 2
     
    =
    6H+ + 3OCl- + 6e- 
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     3Cl- +3 H2O
    +
    2Fe(OH)3 +2 H2O  
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     2FeO42- + 6e- + 10H+
     
    6H+ + 3OCl- + 2e- + 2Fe(OH)3 +2 H2O  
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     3Cl- +3 H2O + 2FeO42- + 6e- + 10H+
     
    Step 5:
     
    Simplify:
     
     3OCl- + 2Fe(OH)3  
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     3Cl- + H2O + 2FeO42- + 4H+
     
    Example 2: VO43- + Fe2+ 
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     VO2+ + Fe3+ in acidic solution
    Step 1:
    Oxidation:    Fe2+ 
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     Fe3+
    Reduction:   VO43- 
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     VO2+
     
    Step 2/3:
     
    Oxidation: Fe2+ 
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     Fe3+ + e-
    Reduction: 6H+ + VO43- + e- 
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     VO2+ + 3H2O
     
    Step 4:
     
    Overall Reaction:
     
    Fe2+ 
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     Fe3+ + e-
    +
    6H+ + VO43- + e-
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     VO2+ + 3H2O
    ____________________________
     
    Fe2+ +  6H+ + VO43- + e- 
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     Fe3+ + e- + VO2+ + 3H2O
     
    Step 5:
     
    Simplify:
     
    Fe2+ +  6H+ + VO43-  
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     Fe3+  + VO2+ + 3H2O

    Practice Problems

    Problems:

    Balance the following equations in both acidic and basic environments:

    1) Cr2O72-(aq) + C2H5OH(l) --> Cn3+(aq) + CO2(g)

    2) Fe2+(aq) + MnO4-(aq) --> Fe3+(aq) + Mn2+(aq)

    Solutions:

    1. (Acidic Answer: 2Cr2O7-(aq) + 16H+(aq) + C2H5OH(l) --> 4Cr3+(aq) + 2CO2(g) + 11H2O(l))

        (Basic Answer: 2Cr2O7-(aq) + 5H2O(l) + C2H5OH(l) --> 4Cr3+(aq) + 2CO2(g) + 16OH-(aq))

    2. (Acidic Answer: MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) --> Mn2+(aq) + 5Fe3+(aq) + 4H2O(l))

        (Basic Answer: MnO4-(aq) + 5Fe2+(aq) + 4H2O(l) --> Mn2+(aq) + 5Fe3+(aq) + 8OH-(aq))

    References

    1. Petrucci, Ralph, William Harwood, Geoffrey Herring, and Jeffry Madura. General Chemistry: Principles & Modern Applications. 9th edition. Upper Saddle River, New Jersey: Pearson Prentince Hall, 2007.
    2. Madsen, Dorte. Chapter 5 Lecture. 6 February 2009.
    3. Petrucci, Ralph. General Chemistry. Ninth. Upper Saddle River: Pearson Education, 2007. 157. Print
    4. Petrucci, Ralph. General Chemistry. Ninth. Upper Saddle River: Pearson Education, 2007. 159. Print
    5. Helmenstine, Anne Marie. "How to Balance Redox Reactions - Balancing Redox Reactions." Balancing Redox Reactions - Half-Reaction Method (2009): n. pag. Web. 1 Dec 2009. http://chemistry.about.com/od/generalchemistry/ss/redoxbal.htm
    6. Stanitski, Conrad L. "Chemical Equations." Chemistry Explained Foundations and Applications. 1st. Chemistry Encyclopedia, 2009. Print.
    7. "How to Balance Redox Equations." Youtube. Web. 1 Dec 2009. <http://www.youtube.com/results?searc...balance+redox+>.

     

     

     

     

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