Question

After you combine each half-reaction, in the final redox reaction between Mn⁺²_(aq) + Cr₂O₇^2-_(aq) to produce MnO_2(s) and Cr⁺³_(aq) in basic solution, how many water molecules are in the final equation?

	a.	One water on the reactant side of the final equation.
	b.	Three waters on the reactant side of the final equation.
	c.	One water on the product side of the final equation.
	d.	Seven waters on the product side of the final equation.

Answer 1

Mn in Mn+2 has oxidation state of +2

Mn in MnO2 has oxidation state of +4

So, Mn in Mn+2 is oxidised to MnO2

Cr in Cr2O7-2 has oxidation state of +6

Cr in Cr+3 has oxidation state of +3

So, Cr in Cr2O7-2 is reduced to Cr+3

Reduction half cell:

Cr2O7-2 + 6e- --> 2 Cr+3

Oxidation half cell:

Mn+2 --> MnO2 + 2e-

Balance number of electrons to be same in both half reactions

Reduction half cell:

Cr2O7-2 + 6e- --> 2 Cr+3

Oxidation half cell:

3 Mn+2 --> 3 MnO2 + 6e-

Lets combine both the reactions.

Cr2O7-2 + 3 Mn+2 --> 2 Cr+3 + 3 MnO2

Balance Oxygen by adding water

Cr2O7-2 + 3 Mn+2 --> 2 Cr+3 + 3 MnO2 + H2O

Balance Hydrogen by adding H+

Cr2O7-2 + 3 Mn+2 + 2 H+ --> 2 Cr+3 + 3 MnO2 + H2O

Add equal number of OH- on both sides as the number of H+

Cr2O7-2 + 3 Mn+2 + 2 H+ + 2 OH- --> 2 Cr+3 + 3 MnO2 + H2O + 2 OH-

Combine H+ and OH- to form water

Cr2O7-2 + 3 Mn+2 + 2 H2O --> 2 Cr+3 + 3 MnO2 + H2O + 2 OH-

Remove common H2O from both sides

Balanced Eqn is

Cr2O7-2 + 3 Mn+2 + H2O --> 2 Cr+3 + 3 MnO2 + 2 OH-

This is balanced chemical equation in basic medium

Answer: a