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MO4 minus + I gives MnO2+I2

Asked by vaagai2353 | 29 Jun, 2018, 10:29: PM
Step 1

The skeleton equation is:

MnO4-  +  I-  → MnO2  +  I2

Step 2

The 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 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

Answered by Ramandeep | 30 Jun, 2018, 08:25: PM

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