# can oxidation number method implemented on all redox equation? tell it by this example
MnO4(-ve)+I (-ve) gives MnO2 +I2

### Asked by Ria Puri | 14th Sep, 2014, 08:11: PM

### Dear riapuriconstellation@gmail.com

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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:

MnO_{4}^{-} + I^{-} → MnO_{2} + I_{2}

**Step 2**

Oxidation number of various atoms involved in the reaction.

**+7 -2 -1 +4 -2 0**

Mn O_{4}^{-} + I^{-} → Mn O_{2} + I_{2}

**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.

MnO_{4}^{-} + **3**I^{-} → MnO_{2} + I_{2}

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 I_{2}, both the atoms and the charge will be balanced.

MnO_{4}^{-} + 3I^{-} → MnO_{2} + 3/2I_{2}

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

2MnO_{4}^{-} + 6I^{-} → 2MnO_{2} + 3I_{2}

Regards

Topperlearning Team.

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:

MnO_{4}^{-} + I^{-} → MnO_{2} + I_{2}

**Step 2**

Oxidation number of various atoms involved in the reaction.

**+7 -2 -1 +4 -2 0**

Mn O_{4}^{-} + I^{-} → Mn O_{2} + I_{2}

**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.

MnO_{4}^{-} + **3**I^{-} → MnO_{2} + I_{2}

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 I_{2}, both the atoms and the charge will be balanced.

MnO_{4}^{-} + 3I^{-} → MnO_{2} + 3/2I_{2}

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

2MnO_{4}^{-} + 6I^{-} → 2MnO_{2} + 3I_{2}

Regards

Topperlearning Team.

### Answered by Arvind Diwale | 15th Sep, 2014, 01:13: PM

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