Reactivity in Chemistry

Reduction & Oxidation Reactions

RO8.  Balancing Redox Reactions

In any reaction, it is useful to quantify things.  How much product will there be?  How much of each reactant do we need to add?  What ratios do we need?  There are an awful lot of reactions for which this process is straightforward, but sometimes it can be tricky.  Redox reactions are sometimes on the tricky side (although certainly not always).  For that reason, it's good to have a reliable method for balancing redox reactions: determining the ratios of reactants needed to give the products in the proper amounts.

Suppose, for example, we have a reaction in which silver oxide (Ag2O) reacts with manganese ion (Mn3+) to produce  manganese dioxide and silver.  That's:

 Mn3+  +  Ag2O  -->    MnO2 + Ag

What would the balanced reaction look like? 

The first thing to do is make sure you are working with one half-reaction at a time.  So that's:

Mn3+   -->    MnO2

and

  Ag2O  -->  Ag

We start out by balancing the atoms involved, one at a time.  First we look at the atoms other than hydrogen or oxygen, and we just balance those by adding the right coefficient.

 Mn3+   -->   MnO2

and

 Ag2O  -->    2 Ag

Second, we balance the oxygen atoms by adding water to one side or the other.

Mn3+  +  2 H2O   -->   MnO2 

and

 Ag2O  -->     2 Ag + H2O

Third, we balance any hydrogens by adding protons.

 Mn3+  +  2 H2O -->     4H+   +  MnO2 

and

 Ag2O   + 2 H+   -->   2 Ag + H2O

Fourth, we balance the charge by adding electrons.

 Mn3+  +  2 H2O    -->    4H+   +  MnO2   +  e- 

and

 Ag2O   + 2 H+    + 2 e-     -->     2 Ag + H2O

Fifth, we multiply so that the number of electrons is the same in both reactions.

2 x (Mn3+  +  2 H2O    -->    4H+   +  MnO2   +  e-  )

or

2 Mn3+  +  4 H2O    -->    8 H+   +  2 MnO2   +  2 e-

and

 Ag2O   + 2 H+    + 2 e-     -->     2 Ag + H2O

Sixth, we simply add these two reactions together.  The reaction arrow functions like an equals sign.  The left side adds to the left side, and the right side adds to the right.

2 Mn3+  +  4 H2O  +  Ag2O   + 2 H+    + 2 e-    -->    8 H+   +  2 MnO2   +  2 e-    +  2 Ag + H2O

At that point, gratifyingly, the equation simplifies.  Notice that we have added the same number of electrons to each side; they cancel out.  That's perfect, because it means we have supplied just the right number of electrons from one half reaction to satisfy the other half reaction.

2 Mn3+  +  4 H2O  +  Ag2O   + 2 H+     -->    8 H+   +  2 MnO2     +  2 Ag + H2O

Also if we subtract one water from each side, things get slightly simpler.

2 Mn3+  +  3 H2O  +  Ag2O   + 2 H+     -->    8 H+   +  2 MnO2     +  2 Ag

Subtracting two protons from each side makes it simpler still.

2 Mn3+  +  3 H2O  +  Ag2O       -->    6 H+   +  2 MnO2     +  2 Ag

This method works for any redox reaction, no matter how complicated. 

 

Problem RO8.1.

Balance the following reactions.

a) Cu +  MoO2 -->    Cu2O   +  Mo

b)  NH2OH  +   Ag2O     -->  N2   +    Ag

c) Fe3O4  +   CO    -->  Fe   +   CO2

d)  I2   +   MnO4-    -->   IO3-   +   MnO2

e)  H3Mo7O24   +   S2O32-   -->     Mo   +     SO32-     

 

In the event that the reaction is described as occuring under basic conditions, we can simply "neutralize" our protons at the end, by adding hydroxide to both sides.

2 Mn3+  +  3 H2O  +  Ag2O    +  6 -OH   -->    6 H+    +  6 -OH    +  2 MnO2     +  2 Ag

Which of course means

2 Mn3+  +  3 H2O  +  Ag2O    +  6 -OH   -->    6 H2O    +  2 MnO2     +  2 Ag

  Simplifying to

2 Mn3+    +  Ag2O    +  6 -OH   -->    3 H2O    +  2 MnO2     +  2 Ag

 

Problem RO8.2.

Balance the following reactions under basic conditions.

a)  Fe(OH)2   +   N2H4   -->  Fe2O3   +   NH4+      

b)  MnO4-  +   V3+    -->  HMnO4-     +  VO2+         

 

 

This site is written and maintained by Chris P. Schaller, Ph.D., College of Saint Benedict / Saint John's University (with contributions from other authors as noted).  It is freely available for educational use.

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Structure & Reactivity in Organic, Biological and Inorganic Chemistry by Chris Schaller is licensed under a Creative Commons Attribution-NonCommercial 3.0 Unported License

Send corrections to cschaller@csbsju.edu

This material is based upon work supported by the National Science Foundation under Grant No. 1043566.

Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.

 

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