BALANCING REDUCTION AND OXIDATION REDOX EQUATIONS

BALANCING REDUCTION AND OXIDATION REDOX EQUATIONS

When balancing redox reactions as written below.

MnO4-   +  Cl- ==> MnO2 +  Cl2  

We begin by checking for the oxidation number value changes of one of the elements for this case, manganese on both sides of the equation. 

Calculating the oxidation number on both sides , it will be clearly seen that manganese changes from +7 to +4  meaning that it's trending towards the negatives. This implies that there is gain of electrons. The oxidation number is seen to decrease. In the case of the chloride ion there is a change from -1 to 0 indicating an increase in oxidation number hence the addition of electrons.

In summary,manganese gains electrons thereby undergoing reduction while chloride ion losses electron to manganese to become chlorine gas.

Writing a balanced redox equation, requires the following

• Divide the equation  break the whole equation into two half equations for each species and 

• balance the electron, 

• balance the water molecules using hydrogen and oxygen count

Step 1: balance electrons difference

Manganese half equation becomes

 MnO4-    ==>  MnO2 

  (+7).              (+4)

The electron change is from +7 to +4. We add 3 electron(e-) to the left hand side to balance up, 

MnO4-   + 3e- ==>  MnO2 

  (+7).    +3e-.             (+4)

for chloride ion 

Cl-  == >Cl2 ,

we balance the electrons loss 

2Cl-  =  Cl2  + 2e- (balanced half equation)

Step 2:balance oxygen difference.

The electrons are now balanced. However there is a difference in the number of oxygen on both sides.

To balance this deficiency,  we add water to both sides with less oxygen. 

MnO4-   +  3e- ==> MnO2 + 2H2O

Then balance with complimentary amount of hydrogen on the other side.

MnO4-   +  3e-  + 4H+==> MnO2 + 2H2O

This leads to an excess amount of hydrogen ions, which we introduce on the other side of the equation to balance both sides.

MnO4-   + 4H- +  5e- = MnO2 + 2H2O (balanced half equation).

 Combining both half equations, before then ensure that the total electron loss equal electron gain.

• 3 electrons was gained by manganese, while

•  2 electrons were lost by chloride ion.

Balance both sides with LCM of 6 in mind.

           MnO4-   + 4H- +  3e- == >  MnO2 + 2H2O (balanced half equation) x 2

2Cl-  =  Cl2  + 2e- (balanced half equation) x 3

Combining both equations

           We have 

2MnO4-   + 8H+ +  6e- = 2MnO2 + 4H2O (balanced half equation) 

6Cl-  =  3Cl2  + 6e- (balanced half equation) 

Since the electrons are now balanced, they will cancel out  so we now combine both half equations: 

           Combining we get the final answer, which is: 

           2MnO4-   + 8H- + 6Cl-  == > 3Cl2  + 2MnO2 + 4H2O