Question

When the following equation is balanced properly under basic conditions, what are the coefficients of the species shown? |Cro Cr(OH) MnO2+ Mn(OH)+ (Enter 0 for neither.) (reactant, product, neither) with a coefficient of Water appears in the balanced equation as a How many electrons are transferred in this reaction?

Answer 1

Mn in MnO2 has oxidation state of +4
Mn in Mn(OH)2 has oxidation state of +2
So, Mn in MnO2 is reduced to Mn(OH)2

Cr in Cr(OH)3 has oxidation state of +3
Cr in CrO4-2 has oxidation state of +6
So, Cr in Cr(OH)3 is oxidised to CrO4-2





Reduction half cell:
MnO2 + 2e- --> Mn(OH)2
Oxidation half cell:
Cr(OH)3 --> CrO4-2 + 3e-
Balance number of electrons to be same in both half reactions

Reduction half cell:
3 MnO2 + 6e- --> 3 Mn(OH)2
Oxidation half cell:
2 Cr(OH)3 --> 2 CrO4-2 + 6e-

So, 6 electrons are being transferred

Lets combine both the reactions.
3 MnO2 + 2 Cr(OH)3 --> 3 Mn(OH)2 + 2 CrO4-2
Balance Oxygen by adding water

3 MnO2 + 2 Cr(OH)3 + 2 H2O --> 3 Mn(OH)2 + 2 CrO4-2

Balance Hydrogen by adding H+
3 MnO2 + 2 Cr(OH)3 + 2 H2O --> 3 Mn(OH)2 + 2 CrO4-2 + 4 H+

Add equal number of OH- on both sides as the number of H+
3 MnO2 + 2 Cr(OH)3 + 2 H2O + 4 OH- --> 3 Mn(OH)2 + 2 CrO4-2 + 4 H+ + 4 OH-

Combine H+ and OH- to form water
3 MnO2 + 2 Cr(OH)3 + 2 H2O + 4 OH- --> 3 Mn(OH)2 + 2 CrO4-2 + 4 H2O

Remove common H2O from both sides
Balanced Eqn is
3 MnO2 + 2 Cr(OH)3 + 4 OH- --> 3 Mn(OH)2 + 2 CrO4-2 + 2 H2O
This is balanced chemical equation in basic medium

Answer:
3, 2, 3, 2
Water appear as product with coefficient of 2
6 electrons are being transferred