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Question

Using Oxidation-number method, find the net ionic equation for the following reaction in basic medium:
Mn(OH)2(s)+MnO4(aq)MnO2(s)

A
6Mn(OH)2(s)+2MnO4(aq)5MnO2(s)+2H2O+OH
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B
4Mn(OH)2(s)+2MnO4(aq)4MnO2(s)+2H2O+2OH
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C
2Mn(OH)2(s)+2MnO4(aq)5MnO2(s)+2H2O+2OH
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D
3Mn(OH)2(s)+2MnO4(aq)5MnO2(s)+2H2O+2OH
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Solution

The correct option is D 3Mn(OH)2(s)+2MnO4(aq)5MnO2(s)+2H2O+2OH
Mn(OH)2(s)+MnO4(aq)MnO2(s)

Mn oxidation state in Mn(OH)2=+2
Mn oxidation state in MnO4=+7
Mn oxidation state in MnO2=+4

Cleary, Mn(OH)2 is undergoing oxidation and MnO4 is undergoing reduction.
Formula used for the n-factor calculation,
nf=(|O.S.ProductO.S.Reactant|×number of atoms
using the formula of n-factor given above,
nf of Mn(OH)2=2
nf of MnO4=3
Cross multiplying these with nf of each other.
we get,
3Mn(OH)2(s)+2MnO4(aq)MnO2(s)

Balancing the Mn element on both sides,
3Mn(OH)2(s)+2MnO4(aq)5MnO2(s)

Adding the H2O to balance the oxygen,
The LHS of the equation contains 14 oxygens while RHS contains only 10 oxygens, so adding 4 H2O to RHS.
3Mn(OH)2(s)+2MnO4(aq)5MnO2(s)+4H2O


Adding H+ to balance hydrogen,

As LHS of the equation contains 6 hydrogens while RHS contains 8 hydrogens, so adding 2H+ to LHS.
3Mn(OH)2(s)+2MnO4(aq)+2H+5MnO2(s)+2H2O

Now adding OH to both sides to combine with H+ and make it H2O,
3Mn(OH)2(s)+2MnO4(aq)+2H++2OH5MnO2(s)+4H2O+2OH
3Mn(OH)2(s)+2MnO4(aq)+2H2O5MnO2(s)+4H2O+2OH

Eliminating the unrequired common H2O which is present on the both sides,

3Mn(OH)2(s)+2MnO4(aq)5MnO2(s)+2H2O+2OH

This is the final balanced equation.

We can also see that charge on both sides is -2. which also indicates that the equation is balanced now.

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