Download Steps for Balancing a Redox Equation Using the Method of Half

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.
-----------------------------------------------------------------------------------------------------------------------------------------------------------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
(acidic)
0
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)
2+
3+
Cr2O7 + 14 H  2 Cr + 7 H2O (balance the H’s with H+’s)
+12
+6
Cr2O72- + 14 H+ + 6 e-
 need 6 e-‘s on left
 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
2 Cl- 
0
 need 2 e-‘s on right
Cl2 + 2 e-
Overall Equation
x3
To make the number of electrons transferred equal, just take the oxidation half reaction and
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 eCr2O72- + 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:
 3 Cl2(g) + 2 Cr3+(aq) + 7 H2O(l)
OR
2+
Cr2O7 (aq) + 14 H3O (aq) + 6 Cl (aq)  3 Cl2(g) + 2 Cr3+(aq) + 21 H2O(l)
Cr2O72-(aq) + 14 H+(aq) + 6 Cl-(aq)
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).
Cl2 +
1
2
Cr2O72- +
Mew net red-ox Reactions
SO2 
Sn2+ +
SO42- +
Cl- 
Cl-
Cr3+ +
(Acidic solution)
SnCl4
(Acidic Solution)