Balancing Oxidation-Reduction (RedOx) Equations *Balancing equation: should obey law of conservation of mass amount of each element should be same in the reactant side and the product side *Balancing redox: charge balance gains and losses of electrons should be equal Half-Reactions *Oxidation and reduction occur simultaneously, however, it is convenient to consider them as separate processes. *Consider the oxidation of Sn2+ by Fe3+: Oxidation: Sn2+ Sn4+ + 2e- Reduction: 2Fe3+ + 2e- 2Fe2+ Net-ionic equation: Sn2+ + 2Fe3+ Sn4+ + 2Fe2+ *The first two equations are the half reactions of the redox reaction.

I. Balancing Redox Reactions Occurring in Acidic Solution

Given the unbalanced equation:

MnO4- + C2O4

2- Mn2+ + CO2

A. Divide the equation into incomplete reactions:

MnO4

- Mn2+

C2O42- CO2

B. Balance each half reaction. How?

Consider MnO4

- Mn2+ First step: Balance the elements other than H and O. *each side contains 1 Mn so we don’t need to put coefficients.

MnO4- Mn2+

Second step: Balance the O atoms by adding H2O to the O deficient side. *You will have to put 4 H2O to the right side of the equation to balance the O atoms.

MnO4- Mn2+ + 4H2O

Third step: Balance H atoms by adding H+ to the H deficient side. *You will have to put 8 H+ to the left side of the equation to balance the O atoms.

8H+ + MnO4- Mn2+ + 4 H2O

Fourth step: Compute for the total charge for each side of the equation. Left side: 8(+1) + 1(-1) = +7 Right side: 1(+2) + 4(0) = +2 Fifth step: Balance the charge by adding e- to the side with the greater over-all positive charge. *The left side has the greater over-all positive charge and you need to add 5e- to balance the charge.

5 e- + 8H+ + MnO4- Mn2+ + 4 H2O

Left side: +7 + 5(-1) = +2 Right side: +2 Consider the other half reaction:

C2O42- CO2

First step: Balance the elements other than H and O. *You need to multiply the right side by 2 to balance the C.

C2O42- 2CO2

Second step: Balance the O atoms by adding H2O to the O deficient side. *The O atoms are already balanced.

C2O42- 2CO2

Third step: Balance H atoms by adding H+ to the H deficient side. *There are no H atoms in each side so there is no need to add H+.

C2O42- 2CO2

Fourth step: Compute for the total charge for each side of the equation. Left side: 1(-2) = -2 Right side: 2(0) = 0 Fifth step: Balance the charge by adding e- to the side with the greater over-all positive charge. *The right side has the greater over-all positive charge and you need to add 2e- to balance the charge.

C2O42- 2CO2 + 2e-

Left side: -2 Right side: 0 + 2(-1) = -2

C. Multiply each half reaction by an integer so that the number of electrons lost in one half cell reaction equals the number gained in the other. You need to multiply the reaction below by 2, so that 10e- were gained

2 x ( 5e- + 8H+ + MnO4- Mn2+ + 4 H2O )

And the reaction below by 5, so that 10e- were lost.

5 x ( C2O42- 2CO2 + 2e- )

So, the resulting equations are:

10e- + 8H+ + 2MnO4- 2Mn2+ + 8H2O

5C2O42- 10CO2 + 10e-

D. Add the two half reactions and simplify where possible canceling species appearing on the both sides of the equation. *You have to cancel the 10e- because it appears on both sides of the equation. [note: no electrons should be present in the final equation.] Balanced equation:

8H+ + 2MnO4- + 5C2O4

2- 2Mn2+ + 8H2O + 10CO2

E. Check the equation to make sure that there are the same number of atoms of each kind and the

same total charge on both sides.

II. Balancing Redox Reactions Occurring in Basic Solution

*Is like balancing a redox reaction in acidic solution except for the second and third step.

Given the unbalanced equation:

Cr(OH)3 + OCl- CrO42- + Cl-

F. Divide the equation into incomplete reactions:

Cr(OH)3 CrO4

2-

OCl- Cl-

G. Balance each half reaction. How?

Consider Cr(OH)3 CrO4

2-

First step: Balance the elements other than H and O. *each side contains equal number of atoms of Cr so we don’t need to put coefficients.

Cr(OH)3 CrO42-

Second step: Balance the O atoms by adding 2OH- to the O deficient side and H2O to the other side. *You will have to put 2OH- to the left side of the equation and H2O to the other side.

2OH- + Cr(OH)3 CrO42- + H2O

Third step: Balance H atoms by adding H2O to the H deficient side and OH- to the other side. *The right side has 3 H atoms greater than the other side so you have to add 3H2O to the left side and 3OH- to the other right side.

3OH- + 2OH- + Cr(OH)3 CrO42- + H2O + 3H2O

By adding similar terms we get,

5OH- + Cr(OH)3 CrO42- + 4H2O

Fourth step: Compute for the total charge for each side of the equation. Left side: 5(-1) + 1(0) = -5 Right side: 1(-2) + 4(0) = -2 Fifth step: Balance the charge by adding e- to the side with the greater over-all positive charge. *The right side has the greater over-all positive charge and you need to add 3e- to balance the charge.

5OH- + Cr(OH)3 CrO42- + 4H2O + 3e-

Left side: -5 Right side: -2 + 3(-1) = -5 Consider the other half reaction:

OCl- Cl-

First step: Balance the elements other than H and O. *each side contains equal number of atoms of Cl so we don’t need to put coefficients.

OCl- Cl-

Second step: Balance the O atoms by adding 2OH- to the O deficient side and H2O to the other side. *You will have to put 2OH- to the right side of the equation and H2O to the other side.

H2O + OCl- Cl- + 2OH-

Third step: Balance H atoms by adding H2O to the H deficient side and OH- to the other side. *H atoms are already balanced.

H2O + OCl- Cl- + 2OH-

Fourth step: Compute for the total charge for each side of the equation. Left side: 1(0) + 1(-1) = -1 Right side: 1(-1) + 2(-1) = -3 Fifth step: Balance the charge by adding e- to the side with the greater over-all positive charge. *The left side has the greater over-all positive charge and you need to add 2e- to balance the charge.

2e- + H2O + OCl- Cl- + 2OH-

Left side: -1 + 2(-1) = -3 Right side: -3

H. Multiply each half reaction by an integer so that the number of electrons lost in one half cell reaction equals the number gained in the other. You need to multiply the reaction below by 2, so that 6e- were gained

2 x ( 5OH- + Cr(OH)3 CrO42- + 4H2O + 3e- )

And the reaction below by 3, so that 6e- were lost.

3 x ( 2e- + H2O + OCl- Cl- + 2OH- )

So, the resulting equations are:

10OH- + 2Cr(OH)3 2CrO42- + 8H2O + 6e-

6e- + 3H2O + 3OCl- 3Cl- + 6OH-

I. Add the two half reactions and simplify where possible canceling species appearing on the both sides of the equation. *You have to cancel the 6e- , 3 H2O and 6OH- because it appears on both sides of the equation. [note: no electrons should be present in the final equation.]

10OH- + 2Cr(OH)3 2CrO42- + 8H2O + 6e-

6e- + 3H2O + 3OCl- 3Cl- + 6OH-

Then,

4OH- [10OH- - 6OH-] + 2Cr(OH)3 + 3OCl- 3Cl- + 2CrO42- + 5H2O [8H2O - 3H2O]

Balanced chemical equation:

4OH- + 2Cr(OH)3 + 3OCl- 3Cl- + 2CrO42- + 5H2O

J. Check the equation to make sure that there are the same number of atoms of each kind and the same total charge on both sides.

MJacinto (061610)