Question

Fundamentals Of Analytical Chemistry 9th Edition - Skoog

Balance the net ionic equation of the following redox reaction using half-reaction method. Show the steps.

Cr(OH)₃ + ClO₃^- --------> CrO₄^2- + Cl^- (basic medium)

Answer 1

Step #1:

Cr(OH)₃ --> CrO₄^2-

ClO₃^- --> Cl^-

Step #2: (Not necessary for this example)

Cr(OH)₃ --> CrO₄^2-

ClO₃^- --> Cl^-

Step #3:

Cr(OH)₃ + H₂O --> CrO₄^2-

ClO₃^- --> Cl^- + 3H₂O

Step #4:

Cr(OH)₃ + H₂O --> CrO₄^2- + 5H⁺

ClO₃^- + 6H⁺ --> Cl^- + 3H₂O

Step #5:

Cr(OH)₃ + H₂O --> CrO₄^2- + 5H⁺ + 3e^-

ClO₃^- + 6H⁺ + 6e^- --> Cl^- + 3H₂O

Step #6:  

2(Cr(OH)₃ + H₂O --> CrO₄^2- + 5H⁺ + 3e^- )

ClO₃^- + 6H⁺ + 6e^- --> Cl^- + 3H₂O

or

2Cr(OH)₃ + 2H₂O --> 2CrO₄^2- + 10H⁺ + 6e^-

ClO₃^- + 6H⁺ + 6e^- --> Cl^- + 3H₂O

Step #7:

2Cr(OH)₃(s) + ClO₃^-(aq)  
     --> 2CrO₄^2-(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^-    
        --> 2CrO₄^2- + 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^-     
    --> 2CrO₄^2- + Cl^- + H₂O + 4H₂O

Step #10:  Cancel or combine the H₂O molecules.

2Cr(OH)₃(s) + ClO₃^-(aq) + 4OH^-(aq)  
--> 2CrO₄^2-(aq) + Cl^-(aq) + 5H₂O(l)