can oxidation number method implemented on all redox equation? tell it by this example

MnO4(-ve)+I (-ve) gives MnO2 +I2

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In most situations of balancing an equation, you are not told whether the reaction is redox or not. In these circumstances, you can use a procedure called the oxidation number method.

Step 1

The skeleton equation is:

MnO₄^-  +  I^-  → MnO₂  +  I₂

Step 2

Oxidation number of various atoms involved in the reaction.

+7    -2          -1        +4    -2         0

Mn   O₄^-  +  I^-  →  Mn  O₂  +  I₂

Step 3

For Mn oxidation number changes from +7 to +4 so it is reduced. For I oxidation number changes from -1 to 0 so it is oxidised. No change in oxidation number of O.

Step 4

Determine the net increase in oxidation number for the element that is oxidized and the net decrease in oxidation number for the element that is reduced.

For I -1 to 0                         Net change = +1

For Mn +7 to +4                                Net change = -3

Step 5

Determine a ratio of oxidized to reduced atoms that would yield a net increase in oxidation number equal to the net decrease in oxidation number.

I atoms would yield a net increase in oxidation number of +3. (Three electrons would be lost by three I atoms.). 1 Mn atom would yield a net decrease of -3. (One Mn atom would gain three electrons.)

Thus the ratio of I atoms to Mn atoms is 3:1.

Step 6

To get the ratio identified in Step 5, add coefficients to the formulas which contain the elements whose oxidation number is changing.

MnO₄^-  + 3I^-  → MnO₂  +  I₂

Still the equation is not balanced as the charge is different on both sides, so follow step 7.

Step 7

Balance the rest of the equation by inspection.

If we place a 3 in front of the I^- and balance the iodine atoms with a 3/2 in front of the I₂, both the atoms and the charge will be balanced.

MnO₄^-  + 3I^-  → MnO₂  + 3/2I₂     

Multiply by 2 to whole equation and we will get the balanced equation,

 

2MnO₄^-  + 6I^-  → 2MnO₂  + 3I₂

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Topperlearning Team.