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# Balancing Redox Reactions by half-reactions by cuiliqing

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```									Balancing Redox Reactions

The Half-reaction Method
Balancing Redox Equations
• Many redox reactions occur in acidic or
basic solution.
• H1+(aq) and OH1-(aq) ions, and H2O
molecules often play an important role in
these types of reactions
• The method of half-reactions can be used
to balance a redox reaction given in the
form of a net-ionic equation
The Half-reaction Method
Eg1) Balance the following reaction which
occurs in an acidic solution.

I2(aq) + OCl1-(aq)  IO31-(aq) + Cl1-(aq)

STEP 1: Divide the skeleton equation into
half reactions
I2  IO31-    (oxidation half)
OCl1-  Cl1- (reduction half)
STEP 2: Balance atoms other than H
&O
I2  2IO31-

OCl1-  Cl1-
STEP 3: Balance “O” atoms using H2O

6H2O + I2  2IO31-

OCl1-  Cl1- + H2O
STEP 4: Balance “H” atoms using H1+ ions

6H2O + I2  2IO31- + 12H1+

2H1+ + OCl1-  Cl1- + H2O
STEP 5: Balance the charges on each side
using electrons (THIS IS THE TRICKIEST
STEP!)

6H2O + I2  2IO31- + 12H1+ + 10e-

2e- + 2H1+ + OCl1-  Cl1- + H2O

Note: Look at the total sum of charges on each
side of each half-reaction for this step!
STEP 6: Balance the # of electrons
between the two half-reactions and add
them together algebraically.

1 x( 6H2O + I2  2IO31- + 12H1+ + 10e- )

5 x( 2e- + 2H1+ + OCl1-  Cl1- + H2O )
6H2O + I2  2IO31- + 12H1+ + 10e-

10e- + 10H1+ + 5OCl1-  5Cl1- + 5 H2O
1                       2
6H2O + I2  2IO31- + 12H1+ + 10e-

10e- + 10H1+ + 5OCl1-  5Cl1- + 5 H2O

1H2O + I2  2IO31- + 2H1+
5OCl1-  5Cl1-

H2O + I2 + 5OCl1-  2IO31- + 5Cl1- + 2H1+

It’s balanced!
For basic solutions… 3 more steps!

STEP 1  6 are the same as acidic solutions,
then…
STEP 7: Add the same # of OH1- to both sides to
neutralize the H1+ ions
STEP 8: Combine OH1- and H1+ to form H2O
STEP 9: Cancel any H2O appearing on both sides
of the equation.
Eg.2 Balance the following redox
reaction if it occurs in a basic
solution

MnO41- + C2O42-  MnO2 + CO32-
Homework!

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