Example



Example

EXAMPLE Balancing Redox Reactions Using the Half-Reaction Method:     Balance the following redox equation using the half-reaction method.

Cr(OH)3(s) + ClO3-(aq)  -->  CrO42-(aq) + Cl-(aq)   (basic)

Solution: 

Step #1:

Cr(OH)3  -->  CrO42-

ClO3-  -->  Cl-

Step #2: (Not necessary for this example)

Cr(OH)3  -->  CrO42-

ClO3-  -->  Cl-

Step #3:

Cr(OH)3  +  H2O  -->  CrO42-

ClO3-  -->  Cl-  +  3H2O

Step #4:

Cr(OH)3  +  H2O  -->  CrO42-  +  5H+

ClO3-  +  6H+  -->  Cl-  +  3H2O

Step #5:

Cr(OH)3  +  H2O  -->  CrO42-  +  5H+  +  3e-

ClO3-  +  6H+  +  6e-  -->  Cl-  +  3H2O

Step #6:  

2(Cr(OH)3  +  H2O  -->  CrO42-  +  5H+  +  3e- )

ClO3-  +  6H+  +  6e-  -->  Cl-  +  3H2O

or

2Cr(OH)3  +  2H2O  -->  2CrO42-  +  10H+  +  6e-

ClO3-  +  6H+  +  6e-  -->  Cl-  +  3H2O

Step #7:

2Cr(OH)3(s)  +  ClO3-(aq)  
     -->
  2CrO42-(aq)    +  Cl-(aq)    +  H2O(l)  +  4H+(aq)

Step #8: Because there are 4 H+ on the right side of our equation above, we add 4 OH- to each side of the equation.

2Cr(OH)3  +  ClO3-  +   4OH-    
 
        -->
  2CrO42-  +  Cl-  +  H2O  +  4H+  +  4OH-

Step #9: Combine the 4 H+ ions and the 4 OH- ions on the right of the equation to form 4 H2O.

2Cr(OH)3  +  ClO3-   +   4OH-     
 
   
      -->  2CrO42-  +  Cl-  +  H2O  +  4H2O

Step #10:  Cancel or combine the H2O molecules.

2Cr(OH)3(s)  +  ClO3-(aq)  +  4OH-(aq)  
 
 
      -->  2CrO42-(aq)  +  Cl-(aq)  +  5H2O(l)

Step #11: The atoms in our equation balance, and the sum of the charges in each side is -5. Our equation is balanced correctly.

Click here to see a Sample Study Sheet for this task.

Click here to see an exercise that will allow you to try this task yourself. 

Return to the Balancing Redox Equations Page.

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