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Question

Write balanced net ionic equation for the following reaction in basic solution:
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
Steps for Balancing redox reactions:
  • Identify the oxidation and reduction halves.
  • Find the oxidising and reducing agent.
  • Find the n-factor of oxidising and reducing agents.
  • Cross multiply the oxidising and reducing agent with the n-factor of each other.
  • Balance the atoms other than oxygen and hydrogen.
  • Balance oxygen atoms.
  • Balance hydrogen atoms.

For basic medium:

If x oxygens are less on one side than the other side add x H2O units to that side. As soon as we add x H2O units for balancing oxygen, we add generally 2x (or whatever suitable) H+ ions on the opposite side to balance hydrogen.Then after that we add equal (equal to number of H+ added) number of OH ions to the both sides and combine H+ and OH ions to form H2O. Then we cancel H2O which is common to both sides and which can be eliminated thus finally OH is left on one side and the equation is balanced. Then we can verify whether the equation is balanced or not by checking the balance of charge on both sides.

formula used for the n-factor calculation,
nf=(|O.S.ProductO.S.Reactant|×number of atoms

Taking the given equation and following the above mentioned steps:
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.

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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