Balancing Redox Equations Handout
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Transcript of Balancing Redox Equations Handout
OFB Chapter 12 111/13/09
Chapter 4.11Balancing Redox Equations
This section is intended to be a supplemental resources for students.
A stepwise process to balance oxidation-reduction (redox) reactions is introduced.
Student’s should use these slides to practice balancing redox reactions.
The number of electrons transferred will be used for the Electrochemical cell reactions in Chapter 16
OFB Chapter 12 211/13/09
• This section 4.11deals with Oxidation, Reduction, and electron transfer
• Balancing Redox Reactions• All reactions take place in either
Acid (H+) or Basic (OH-) conditions• Very important to remember the
problem you are working. Is it acidic or basic conditions?
• We will be adding H2O and H+ for acid reactions or H2O and OH- for basic reaction in order to properly balance the reactions
• Practice, Practice, Practice
OFB Chapter 12 311/13/09
Gains ElectronsDecreaseSubstance Reduced
Loses ElectronsIncreaseSubstance Oxidized
Supplies ElectronsIncrease
Reducing Agent, does the reducing
Picks Up electronsDecrease
Oxidizing Agent, does the oxidizing
Gain of ElectronsDecreaseReduction
Loss of ElectronsIncreaseOxidation
Electron Change
Oxidation Number ChangeTerm
The Oxidation States Method
Cl2 Na NaCl+ →
OFB Chapter 12 411/13/09
• Step 1 Write two unbalanced half-equations, one for the species that is oxidized and its product and one for the species that is reduced and its product
• Step 2 Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-equation.
• Step 3 Balance oxygen by adding H 2O to the side deficient in O in each half-equation
• Step 4 Balance hydrogen. – For half-reaction in acidic solution, add H+ on
to the side deficient in hydrogen.
– For a half-reaction in basic solution, add H2O to the side that is deficient in hydrogen and an equal amount of OH- to the other side
OFB Chapter 12 511/13/09
• Step 5 Balance charge by inserting e- (electrons) as a reactant or product in each half-reaction.
• Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication
• Step 7 Check for balance
OFB Chapter 12 611/13/09
Balancing Redox Reactions
• Dithionate ion reacting with chlorous acid in aqueous acidic conditions
S2O62-(aq) +HClO2(aq)
→ SO42-(aq) + Cl2(g)
Will have:
Oxidation half reaction
Reduction half reaction
OFB Chapter 12 711/13/09
• Step 1 Write two unbalanced half-equations, one for the species that is oxidized and its product and one for the species that is reduced and its product
S2O62- → SO4
2-
Dithionate
(we will see eventually that this is the oxidation
reaction)
HClO2 → Cl2
Chlorous acid
(we will see eventually that this is the reduction reaction)
S2O62-(aq) +HClO2(aq)
→ SO42-(aq) + Cl2(g)
OFB Chapter 12 811/13/09
• Step 2 Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-equation.
S2O62- → SO4
2-
HClO2 → Cl2
2
2
OFB Chapter 12 911/13/09
• Step 3 Balance oxygen by adding H 2O to the side deficient in O in each half-equation
S2O62- → 2SO4
2-
2HClO2 → Cl2
(8Os)(6Os)
2H2O +(Need 2 Os)
(4Os)
+ 4H2O(Need 4 Os)
OFB Chapter 12 1011/13/09
• Step 4 Balance hydrogen. – For half-reaction in acidic solution, add H+ on to the
side deficient in hydrogen.
– For a half-reaction in basic solution, add H2O to the
side that is deficient in hydrogen and an equal amount of OH- to the other side
2H2O + S2O62- → 2SO4
2-
2HClO2 → Cl2 + 4H2O
Dithionate ion reacting with chlorous acid in aqueous acidic conditions
Need 4 Hs
Need 6 Hs (2 Hs) (8 Hs)
+ 4H+
6H+ +
(4 Hs)
OFB Chapter 12 1111/13/09
• Step 5 Balance charge by inserting e - (electrons) as a reactant or product in each half-reaction.
2H2O + S2O62-
→ 2SO42- + 4H+
6H+ + 2HClO2
→ Cl2 + 4H2O
-2
-4 +4 Need -2
+ 2e-
+6
Zero charge
Need -6
+ 6e-
Oxidation reaction, electrons are lost on the product side
Reduction reaction, electrons are gained on the reactant side
OFB Chapter 12 1211/13/09
• .Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication
Oxidation Reaction has 2 electrons
Reduction reaction has 6 electrons
Multiply the oxidation reaction by 3. This will give 6 electrons on both sides of the reaction
2H2O + S2O62-
→ 2SO42- + 4H+ + 2e-
3 times
OFB Chapter 12 1311/13/09
• .Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication
3 times
6H2O + 3S2O62-
→ 6SO42- + 12H+ + 6e-
2H2O + S2O62-
→ 2SO42- + 4H+ + 2e-
equals
Now add both Re-Ox reactions
OFB Chapter 12 1411/13/09
• .Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication
6H2O + 3S2O62-
→ 6SO42- + 12H+ + 6e-
6H+ + 2HClO2 + 6e-
→ Cl2 + 4H2O
6H2O + 3S2O62- + 6H+ +
2HClO2 + 6e-
→ 6SO42- + 12H+ + Cl2 +
4H2O + 6e-
2
6
OFB Chapter 12 1511/13/09
• Step 7 Check for balance
2H2O + 3S2O62- + 2HClO2
→ 6SO42- + 6H+ + Cl2
Left side
H 6
O 24
S 6
Cl 2
Right side
H 6
O 24
S 6
Cl 2
If it doesn’t balance, look for simple mistakes first
Omitted subscripts
Omitted superscripts
OFB Chapter 12 1611/13/09
Find the (integer) stoichiometry coefficients for the correctly balanced version of the following reaction
Cr2O72- + H+ + I-
→ Cr+3 + H2O + I3-
OFB Chapter 12 1711/13/09
• Step 1 Write two unbalanced half-equations, one for the species that is oxidized and its product and one for the species that is reduced and its product
Cr2O72- + H+ + I-
→ Cr+3 + H2O + I3-
Cr2O72- → Cr+3
I- → I3-
OFB Chapter 12 1811/13/09
• Step 2 Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-equation.
2
3
Cr2O72- → Cr+3
I- → I3-
OFB Chapter 12 1911/13/09
• Step 3 Balance oxygen by adding H 2O to the side deficient in O in each half-equation
(7Os)
+ 7H2O(Need 7 Os)
(No Oxygen, okay)
Cr2O72- → 2Cr+3
3I- → I3-
OFB Chapter 12 2011/13/09
• Step 4 Balance hydrogen. – For half-reaction in acidic solution, add H+ on to the
side deficient in hydrogen.
– For a half-reaction in basic solution, add H2O to the
side that is deficient in hydrogen and an equal amount of OH- to the other side
Dichromate ion reacting with iodide ion in aqueous acidic conditions
Need 14 Hs
14H+ +
(14 Hs)
Cr2O72-
→ 2Cr+3 + 7H2O
3I- → I3-
(No Hydrogen, okay)
OFB Chapter 12 2111/13/09
• Step 5 Balance charge by inserting e - (electrons) as a reactant or product in each half-reaction.
-2+14
+6
Need -6
+ 2e-
-3 Need -2
6e- +
Oxidation reaction
Reduction reaction
14H+ + Cr2O72-
→ 2Cr+3 + 7H2O
3I- → I3-
-1
OFB Chapter 12 2211/13/09
• .Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication
Reduction reaction has 6 electrons
Oxidation Reaction has 2 electrons
Multiply the oxidation reaction by 3 this will give 6 electrons on both sides of the reaction
3 times
3I- → I3- + 2e-
OFB Chapter 12 2311/13/09
• .Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication
equals
Now add both Re-Ox reactions
3 times
3I- → I3- + 2e-
9I- → 3 I3- + 6 e-
OFB Chapter 12 2411/13/09
• .Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication
6e- + 14H+ + Cr2O72-
→ 2Cr+3 + 7H2O
9I- → 3 I3- + 6 e-
6e- + 14H+ + Cr2O72- + 9I-
→ 2Cr+3 + 7H2O + 3 I3- + 6 e-
equals
OFB Chapter 12 2511/13/09
• Step 7 Check for balance
Left side
H 14
Cr 2
O 7
I 9
Right side
H 14
Cr 2
O 7
I 9
14H+ + Cr2O72- + 9I-
→ 2Cr+3 + 7H2O + 3 I3-
OFB Chapter 12 2611/13/09
Basic SolutionMethod 1
• At Step 4– Instead of adding H+ on the
deficient side as did with the acid solution method
– Add H2O on the H deficient side and add an equal amount of OH- on the other side.
OFB Chapter 12 2711/13/09
Find the (integer) stoichiometry coefficients for the correctly balanced version of the following reaction conducted in a basic solution
Ag2S + Cr(OH)3
→ Ag0 + HS- + CrO42-
OFB Chapter 12 2811/13/09
• Step 1 Write two unbalanced half-equations, one for the species that is oxidized and its product and one for the species that is reduced and its product
Ag2S + Cr(OH)3
→ Ag0 + HS- + CrO42-
Ag2S → Ag + HS-
Cr(OH)3 → CrO42-
Reduction reaction
Oxidation State of Ag
+1
0
Oxidation reactionOxidation State of Cr
+3
+6
OFB Chapter 12 2911/13/09
• Step 2 Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-equation.
2Ag2S → Ag + HS-
Cr(OH)3 → CrO42-
Already Balanced
OFB Chapter 12 3011/13/09
• Step 3 Balance oxygen by adding H 2O to the side deficient in O in each half-equation
(4Os)
H2O +
(Oxygen not needed)
Ag2S → 2Ag + HS-
Cr(OH)3 → CrO42-
(3Os)(need 1O)
OFB Chapter 12 3111/13/09
• Step 4 Balance hydrogen. – For half-reaction in acidic solution, add H+ on
to the side deficient in hydrogen.
– For a half-reaction in basic solution, add H2O
to the side that is deficient in hydrogen and an equal amount of OH- to the other side
This problem is in aqueous basic conditions
Needs H
(1 H)
Ag2S
→ 2Ag + HS-
H2O +
+ OH-
(need O)
H2O + Cr(OH)3
→ CrO42-
(3 H)(2 H)
+ 5H2O
5OH- +
Needs H
Add OH-
OFB Chapter 12 3211/13/09
• Step 5 Balance charge by inserting e - (electrons) as a reactant or product in each half-reaction.
-2
Need -2
+ 3e-
-5
Need -3
H2O + Ag2S
→ 2Ag + HS- + OH-
5OH- + H2O + Cr(OH)3
→ CrO42- + 5H2O
-1 -1
2e- +
Reduction reaction
Oxidation reaction
OFB Chapter 12 3311/13/09
• .Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication
Reduction reaction has 2 electrons
Oxidation Reaction has 3 electrons
2e x 3 = 6e
3e x 2 = 6e
OFB Chapter 12 3411/13/09
• .Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication
6e- + 3H2O + 3Ag2S
→ 6Ag + 3HS- + 3OH-
Reduction reaction
3 times Reduction Reaction =
10OH- + 2H2O + 2Cr(OH)3
→ 2CrO42- + 10H2O + 6e-
2 times Oxidation Reaction =
Oxidation reaction
OFB Chapter 12 3511/13/09
• .Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication
Now add both Re-Ox reactions
6e- + 3H2O + 3Ag2S
→ 6Ag + 3HS- + 3OH-
10OH- + 2H2O + 2Cr(OH)3
→ 2CrO42- + 10H2O + 6e-
OFB Chapter 12 3611/13/09
equals
Now add both Re-Ox reactions
6e- + 3H2O + 3Ag2S
→ 6Ag + 3HS- + 3OH-
10OH- + 2H2O + 2Cr(OH)3
→ 2CrO42- + 10H2O + 6e-
6e- + 3H2O + 3Ag2S +
10OH- + 2H2O + 2Cr(OH)3
→ 6Ag + 3HS- + 3OH- +
2CrO42- + 10H2O + 6e-
OFB Chapter 12 3711/13/09
simplify
6e- + 3H2O + 3Ag2S +
10OH- + 2H2O + 2Cr(OH)3
→ 6Ag + 3HS- + 3OH- +
2CrO42- + 10H2O + 6e-
5
7
3Ag2S + 7OH- + 2Cr(OH)3
→
6Ag + 3HS- + 2CrO42- + 5H2O
OFB Chapter 12 3811/13/09
• Step 7 Check for balance
Left side
Ag 6
S 3
O 13
H 13
Cr 2
3Ag2S + 7OH- + 2Cr(OH)3
→ 6Ag + 3HS- + 2CrO4
2- + 5H2O
Right side
Ag 6
S 3
O 13
H 13
Cr 2
OFB Chapter 12 3911/13/09
Find the (integer) stoichiometry
coefficients for the correctly balanced version of the following reaction conducted in a basic solution
Solution
3 Ag2S + 2 Cr(OH)3+ 7 OH-
→ 6 Ag + 3 HS- + 2CrO4
2- + 5H2O
OFB Chapter 12 4011/13/09
Chapter 4.11Redox reactions
• Practice
• We will use the number of electrons transferred again in Chapter 16
OFB Chapter 12 4111/13/09
Balancing Disproportionation Reactions
• Same Steps
• For example
Cl2 → ClO3- + Cl-
• Step 1– Oxidation:
Cl2 → ClO3-
– Reduction:
Cl2 → Cl-