# 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

Thanks for asking us a question in Ask the Expert section of TopperLearning.com.

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

## Related Videos

- what is the most essential conditions that must be satisfied in a redox reaction
- H2O2+Fe2+=H2O +Fe3+ , give ion exchange method
- Redox Reaction: solve the following equation by ion electron method in acidic medium
- NO3 (-ve)+I (-ve)+H (+) =NO +I2 +H2O
- magnesium reacts with nitric acid to give magnesium nitarate and nitrous oxide gas and liquid water balance this by oxidation number method
- H
_{2}S + KMnO_{4}+ H_{2}SO_{4}→ S+ MnSO_{4}+ KHSO_{4}+ H_{2}O Balance by oxidation number method step by step explain please - Zn + HNO
_{3}= Zn(NO_{3}) + NO_{2}+ H_{2}O balance this equation by oxidation number method step wise - MO4 minus + I gives MnO2+I2
- Solve it
- Define oxidation in terms of oxidation numbers.

### Kindly Sign up for a personalised experience

- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions

#### Sign Up

#### Verify mobile number

Enter the OTP sent to your number

Change