Steps for Balancing a Redox Equation Using the Method of Half ... - Oakton

Steps for Balancing a Redox Equation Using the Method of Half Reactions 1. Write the equation as a net ionic equation (if it isn't already in that form). (2. Assign oxidation numbers to all "atoms" in each species to figure out what is being oxidized and what is being reduced.* *It turns out that this step is not really necessary; its omission just means you won't know which

half reaction is the oxidation one and which is the reduction one until after you balance each.)

3. Separate the reaction into two half reactions (one oxidation and one reduction). Steps 4-7 provide a procedure for balancing a half reaction. Do these steps for BOTH the oxidation half

reaction and the reduction half reaction (NOTE: sometimes you will not need to "do" one or more of these steps because the atom type will already be balanced, or there won't be any O's or H's). 4. Balance atoms other than O and H (by adding appropriate coefficients). 5. Balance the O atoms by adding H2O's to the side with fewer O atoms (ignore the H's at this point!). 6. Balance the H atoms by adding H+'s to the side with fewer H atoms.. 7. Balance the TOTAL CHARGE (on each side) by adding electrons (e-'s) to the side that has a more

positive total charge [Remember, e-'s are negative!]. *Don't forget to account for coefficients!* 8. Multiply one or both of the half reactions through by an integer, if necessary, to make the

number of electrons lost (in the oxidation half reaction equation) = #e-'s gained (in the reduction half reaction eqn) 9. Add the half reactions; the electrons should cancel at this point (if they don't, then recheck your steps!) 10. IF BASIC CONDITIONS are specified, then for each proton (H+) present, add an OH- to both sides of the

equation. Rewrite the H+ + OH- "pairs" as (an equal number of) H2O molecules and cancel out any H2O's if necessary. 11. IF ACIDIC CONDITIONS you may add an equal number of H2O's as H+'s to write the equation with H3O+'s instead of H+'s. Cancel out any H2O's, if necessary. 12. Calculate the TOTAL CHARGE of both sides as a check.

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Assigning Oxidation Numbers to "Atoms" in chemical species.

1. Atoms in elements are assigned an oxidation number of ZERO. (Note: Ca2+ is not the same as Ca! Ca2+ is an ion, not an element!)

O's in O2 are assigned ZERO; O's in O3 are assigned ZERO; Ca in Ca is assigned ZERO.

2. In compounds OR polyatomic ions, O atoms are typically assigned an oxidation number of ?2, and H atoms are typically assigned an oxidation number of +1. I will not require you to know the exceptions to this rule (peroxides are a common exception). (Note: Don't forget about Rule 1 before assigning oxidation numbers to O and H!!! O2 and H2 have oxidation #'s of zero!)

You can figure out the oxidation numbers of other "atoms" in a species by using Rule number 3: 3. The sum of the oxidation numbers of EACH "ATOM" in a chemical species must equal the net charge

on the species. (Don't forget that if there is more than one atom of the same type in a species, you must sum up the oxidation numbers on ALL of them. BUT that sum is not the oxidation number! Please ask me if you are confused about this.)

Example: in NO3-, each O is assigned an oxidation number of ?2. Since there are 3 O atoms, the sum total of their oxidation numbers is 3 x ?2 = -6. The oxidation number of the single N atom must therefore be the number which when added to ?6 yields the overall charge of ?1. Thus the oxidation number of N in NO3- is +5. (+5 + -6 =1) The oxidation number of O in the above example is ?2; it is NOT ?6.

=> For monatomic species, this rule (Rule 3) reduces to the following "trivial" rule: the oxidation number of any monatomic species is equal to the actual charge on the species. E.g., the oxidation number of Ca in Ca2+ is +2; for K in K+ the oxidation number is +1 (not zero!!!).

Helpful hint: If a compound is ionic, rewrite the formula as the SEPARATED IONS (whose charges you should either know or can figure out). Then assign the oxidation numbers according to rules 2 and 3.

Example: Fe2(CO3)3. You need to rewrite as Fe3+ and CO32- to get all ox. numbers.

Using the method of half reactions to balance a redox equation

Balance: Cl-(aq) + Cr2O72-(aq) Cl2(g) + Cr3+ (in acidic solution). (NOTE: To save space and avoid "clutter", I will leave off state designations until the end)

gains e-`s

Cr2O72- + Cl- Cr3+ + Cl2

+6 -2

-1

+3

0

(acidic)

loses e-`s

Reduction Half Reaction (I show each step so you can see the order of doing things, but

if I were doing this on an exam, I would not show each step separately!)

Cr2O72- 2 Cr3+

(balance all non O's and H's)

Cr2O72- 2 Cr3+ + 7 H2O (balance the O's with H2O)

Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O (balance the H's with H+'s)

+12

+6 need 6 e-`s on left

Cr2O72- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O

Helpful Hint: Always add electrons to the more positive side to bring the charge

"down to" the other side's

(since electrons are negative! )

Oxidation Half Reaction

2 Cl- Cl2 (balance non O's and H's first; no O's or H's to balance!)

-2

0 need 2 e-`s on right

2 Cl- Cl2 + 2 e-

Overall Equation

To make the number of electrons transferred equal, just take the oxidation half reaction and

x 3

multiply it by THREE (to get 6 e-`s total produced--same as the number needed in the

reduction half reaction). Note that this is an "easy" one (that is, you don't need to multiply

BOTH equations by a factor to make the number of electrons equal); make sure that you are

confident that you could do one like we did in class where the least common multiple is neither

of the values of electrons already "given".

6 Cl- 3 Cl2 + 6 e-

Cr2O72- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O

Cr2O72- + 14 H+ + 6 Cl- 3 Cl2 + 2 Cr3+ + 7 H2O

Just to show you how it would look if you were to "convert" the H+'s into H3O+'s like the text chooses to do, at this point you would have to add 14 H2O's to both sides to yield:

Cr2O72- + 14 H3O+ + 6 Cl- 3 Cl2 + 2 Cr3+ + 21 H2O

Adding state designations: Cr2O72-(aq) + 14 H+(aq) + 6 Cl-(aq) 3 Cl2(g) + 2 Cr3+(aq) + 7 H2O(l)

OR Cr2O72-(aq) + 14 H3O+(aq) + 6 Cl-(aq) 3 Cl2(g) + 2 Cr3+(aq) + 21 H2O(l)

Check charge on both sides: (-2) + (+14) + (-6) = +6 (left) and 2(+3) = +6 (right)

OXIDATION - REDUCTION NET - IONIC REACTIONS

Balance the equations using the half-reaction method, and then find the standard cell potential (if you can locate the half reactions in the appendix).

1

Cl2 +

SO2 SO42- + Cl-

(Acidic solution)

2

Cr2O72- +

Sn2+ +

Cl-

Cr3+ +

SnCl4

(Acidic Solution)

Mew net red-ox Reactions

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