The correct option is
D 3Mn(OH)2(s)+2MnO−4(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.Product−O.S.Reactant|×number of atoms Taking the given equation and following the above mentioned steps:
Mn(OH)2(s)+MnO−4(aq)→MnO2(s) Mn oxidation state in
Mn(OH)2=+2 Mn oxidation state in
MnO−4=+7 Mn oxidation state in
MnO2=+4 Cleary,
Mn(OH)2 is undergoing oxidation and
MnO−4 is undergoing reduction.
using the formula of n-factor given above,
nf of
Mn(OH)2=2 nf of
MnO−4=3 Cross multiplying these with
nf of each other.
we get,
3Mn(OH)2(s)+2MnO−4(aq)→MnO2(s) Balancing the Mn element on both sides,
3Mn(OH)2(s)+2MnO−4(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)+2MnO−4(aq)→5MnO2(s)+4H2O Adding
H+ to balance hydrogen,
As LHS of the equation contains 6 hydrogens while RHS contains 8 hydrogens, so adding 2
H+ to LHS.
3Mn(OH)2(s)+2MnO−4(aq)+2H+→5MnO2(s)+2H2O Now adding
OH− to both sides to combine with
H+ and make it
H2O,
3Mn(OH)2(s)+2MnO−4(aq)+2H++2OH−→5MnO2(s)+4H2O+2OH− 3Mn(OH)2(s)+2MnO−4(aq)+2H2O→5MnO2(s)+4H2O+2OH− Eliminating the unrequired common
H2O which is present on the both sides,
3Mn(OH)2(s)+2MnO−4(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.