Transcript 20.3 Describing Redox Equations

Balancing Redox Reactions
Chapter 20: Day 2
1
2
Review of Terminology
for Redox Reactions
• OXIDATION—loss of electron(s) by a
species; increase in oxidation number.
• REDUCTION—gain of electron(s);
decrease in oxidation number.
• OXIDIZING AGENT—electron acceptor;
species is reduced.
• REDUCING AGENT—electron donor;
species is oxidized.
20.3 Describing Redox Equations >
CHEMISTRY
& YOU
Why does cut fruit turn brown?
Some fruits, including apples, turn
brown when you cut them. What is
happening on the surface of the fruit?
Oxygen in air reacts with
chemicals on the surface of
the cut fruit. The oxygen
oxidizes the chemicals in
the fruit, causing a redox
reaction and therefore the
color change.
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If changes in oxidation number occur,
the reaction is a redox reaction.
20.3 Describing Redox Equations >
• The element whose oxidation
number increases is oxidized
• The element whose oxidation
number decreases is reduced
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20.3 Describing Redox Equations >
Identifying Redox Reactions
Use the change in oxidation number to
identify whether each reaction is a redox
reaction
Cl2(g) + 2NaBr(aq) → 2NaCl(aq) + Br2(aq)
2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)
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20.3 Describing Redox Equations >
a.
0
Sample Problem 20.5
Assign oxidation numbers.
+1 –1
+1 –1
0
Cl2(g) + 2NaBr(aq) → 2NaCl(aq) + Br2(aq)
The chlorine is reduced;
The bromide ion is oxidized;
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20.3 Describing Redox Equations >
2
Solve
+1 –2 +1
Sample Problem 20.5
ASSIGN OXIDATION NUMBERS.
+1 +6 –2
+1 +6 –1
+1 –2
2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)
• NO change in oxidation number.
• This is not a redox reaction.
This is an
acid-base
(neutralization)
reaction.
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20.3 Describing Redox Equations >
Which of the following are redox
reactions?
A. NH3 + HCl → NH4Cl
B. SO3 + H2O → H2SO4
C. NaOH + HCl → NaCl + H2O
D. H2S + NHO3 → H
H2SO4 + NO2 + H2O
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TRANSFER REACTIONS
Atom/Group transfer (not)
HCl + H2O ---> Cl- + H3O+
Redox: Electron transfer
Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s)
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OXIDATION-REDUCTION
REACTIONS
Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s)
Why 2?
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Balancing Equations
Cu + Ag+ --give--> Cu2+ + Ag
Need to Balance BOTH mass and
CHARGE
Step 1: Divide into half-reactions:
one for oxidation and the other
for reduction.
Ox
Red
Cu ---> Cu2+
Ag+ ---> Ag
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Balancing Equations
Step 2:
Step 3:
Balance each for mass.
Already done in this case.
Balance each half-reaction for
charge by adding electrons.
Ox Cu ---> Cu2+ + 2eRed
Ag+ + e- ---> Ag
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Balancing Equations
Multiply each half-reaction by a
factor to have the electrons lost equal to
number gained
Step 4:
Cu ---> Cu2+ + 2e2 Ag+ + 2 e- ---> 2 Ag
Step 5:
Add to give the overall equation.
Cu + 2 Ag+
---> Cu2+ + 2Ag
The equation is now balanced for
BOTH charge and mass.
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Use the half-reaction method to
balance the following redox equation.
FeCl3 + H2S → FeCl2 + HCl + S
Oxidation: H2S → 2H+ + S + 2e–
Reduction: 2Fe3+ + 2e– → 2Fe2+
2FeCl3 + H2S → 2FeCl2 + 2HCl + S
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20.3 Describing Redox Equations >
Balancing Redox Equations
What are two different methods for
balancing a redox equation?
Two different methods for
balancing redox equations are the
oxidation-number-change
method and the half-reaction
method.
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20.3 Describing Redox Equations >
Using Oxidation-Number Changes
In the oxidation-numberchange method, you balance
a redox equation by
comparing the increases and
decreases in oxidation
numbers.
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20.3 Describing Redox Equations >
Using Oxidation-Number Changes
Step 1: Assign oxidation numbers to all the
atoms in the equation.
• Write the numbers above the atoms.
• The oxidation number is stated per atom.
+3 –2
+2 –2
0
+4 –2
Fe2O3(s) + CO(g) → Fe(s) + CO2(g)
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20.3 Describing Redox Equations >
Using Oxidation-Number Changes
Step 2: Identify which atoms are oxidized and which
are reduced.
• Iron is reduced. +3 to 0
• Carbon is oxidized. +2 to +4
+3 –2
+2 –2
0
+4 –2
Fe2O3(s) + CO(g) → Fe(s) + CO2(g)
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20.3 Describing
Redox Equations >
Using
Oxidation-Number
Changes
Step 3: Use one line to connect the atoms that
undergo oxidation and another such line to
connect those that undergo reduction.
+2 (oxidation)
+3 –2
+2 –2
0
+4 –2
Fe2O3(s) + CO(g) → Fe(s) + CO2(g)
–3 (reduction)
• Write the oxidation-number change at the midpoint
of each line.
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20.3 Describing Redox Equations >
Using Oxidation-Number Changes
Step 4: Make the total increase in oxidation number
equal to the total decrease in oxidation
number by using appropriate coefficients.
3 × (+2) = +6
+3 –2
+2 –2
0
+4 –2
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
2 × (–3) = –6
• The oxidation-number increase should be
multiplied by 3 and the decrease by 2.
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20.3 Describing Redox Equations >
Using Oxidation-Number Changes
Step 5: Finally, make sure the equation is balanced for
both atoms and charge.
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
• If necessary, finish balancing the
equation by inspection.
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Balancing Redox Equations by
Oxidation-Number Change
20.3 Describing Redox Equations >
Balance this redox equation by using the oxidation-number-change method.
2 × (-1) = -2
0
0
+1 –1
Cl2(g) + 2 K(s) → 2 KCl(s)
2 × (+1) = +2
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Ready for more
complex reactions?
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YES!
Acid / Base
Redox Reactions
Some redox reactions have equations
that must be balanced by special
techniques.
If reactions occur in the “presence of”
acid or base
MnO4- + 5 Fe2+ + 8 H+
---> Mn2+ + 5 Fe3+ + 4 H2O
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Permanganate and iron(II) ions are
reacted in an acidic solution
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MnO4- + Fe2+ ---> Mn2+ + Fe3+
Step 1: Divide into half-reactions
Ox Fe2+ ---> Fe3+
Red
MnO4- ---> Mn2+
Need to balance MASS
Step 2:
Balance each for mass.
Already done for Iron
Fe2+ ---> Fe3+
Need to have “O” on both sided: Add
water
MnO4- ---> Mn2+ + H2O
Never add O2, O atoms, or O2- to balance
oxygen.
MnO4- ---> Mn2+ + 4H2O
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Step 2:
Balance each for mass.
MnO4- ---> Mn2+ + 4H2O
Need to have “H” on both sided.
Told in an acidic solution: need H+
8 H+ + MnO4- ---> Mn2+ + 4H2O
Never add H2 or H atoms to balance
hydrogen.
Step 3:
Balance each half-reaction for
charge by adding electrons.
8 H+ + 5 e- +MnO4- ---> Mn2+ + 4H2O
5Fe2+ ---> 5Fe3+ + 5eMultiply each half-reaction by a
factor to have the electrons lost
equal to number gained
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Acid / Base
Redox Reactions
Step 5: Add to obtain the overall
equation
MnO4- + 5 Fe2+ + 8 H+
---> Mn2+ + 5 Fe3+ + 4 H2O
Check by adding charges on both sides
and by counting atoms
The equation is now balanced for
BOTH charge and mass.
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Balancing Equations
• Never add O2, O atoms, or O2- to
balance oxygen.
• Never add H2 or H atoms to
balance hydrogen.
• Be sure to write the correct
charges on all the ions.
• Check your work at the end to
make sure mass and charge are
balanced.
• PRACTICE!
Reduction of VO2+ with Zn
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Balancing Equations
Balance the following in acid solution—
VO2+ + Zn ---> VO2+ + Zn2+
Step 1:
Write the half-reactions
Ox
Zn ---> Zn2+
Red
VO2+ ---> VO2+
Balance each for mass.
Step 2:
2
H+
+ ---> VO2+ + H O
VO
+
2
2
Add H2O on O-deficient side and add H+
on other side for H-balance.
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Balancing Equations
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Step 3: Add electrons to half reaction.
Zn ---> Zn2+ + 2ee- + 2 H+ + VO2+ ---> VO2+ + H2O
Step 4:
Multiply by an appropriate factor.
2e- + 4 H+ + 2 VO2+ ---> 2 VO2+ + 2 H2O
Zn ---> Zn2+ + 2e-
Balancing Equations
Step 5:
Add balanced half-reactions
Zn + 4 H+ + 2 VO2+
---> Zn2+ + 2 VO2+ + 2 H2O
Check by adding charges on both
sides and by counting atoms
20.3 Describing Redox
Equations
>
Balancing
Redox
Equations
by
Half-Reactions
Balance this redox equation
using the half-reaction
method.
KMnO4(aq) + HCl(l) → MnCl2(aq) + Cl2(g) + KCl(aq)
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20.3 Describing Redox Equations >
Sample Problem 20.7
Step 1:Write the equation in ionic form.
K+(aq) + MnO4–(aq) + H+(aq) + Cl–(aq) →
Mn2+(a) + 2Cl–(a) + Cl2(g) + H2O + K+(a) + Cl–(a)
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20.3 Describing Redox Equations >
Step 2: Write half-reactions. Determine the
oxidation and reduction process.
Oxidation half-reaction:
–1
0
Cl– → Cl2
Reduction half-reaction:
+7
+2
MnO4– → Mn2+
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20.3 Describing
Redox Equations
Step
3: Balance
the>atoms in each half-reaction.
• The solution is acidic, so use H2O and H+
to balance the oxygen and hydrogen.
Oxidation:
2Cl–(aq) → Cl2(g) (atoms balanced)
Reduction:
MnO4–(aq) + 8H+(aq) →
Mn2+(aq) + 4H2O(l) (atoms balanced)
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20.3 Describing Redox Equations >
Step 4: Balance the charges by adding
electrons.
Oxidation:
2Cl–(aq) → Cl2(g) + 2e– (charges balanced)
Reduction:
MnO4–(aq) + 8H+(aq) + 5e– →
Mn2+(aq) + 4H2O(l) (charges balanced)
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20.3 Describing Redox Equations >
Step 5: Make the numbers of electrons equal.
• Multiply the oxidation half-reaction by 5
and the reduction half-reaction by 2.
Oxidation:
10Cl–(aq) → 5Cl2(g) + 10e–
Reduction:
2MnO4–(aq) + 16H+(aq) + 10e– →
2Mn2+(aq) + 8H2O(l)
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20.3 Describing Redox Equations >
Step 6: Add the half-reactions. Then, subtract
the terms that appear on both sides.
10Cl–(aq) + 2MnO4–(aq) + 16H+(aq) + 10e– →
5Cl2(g) + 10e– + 2Mn2+(aq) + 8H2O(l)
Step 7: Add the spectator ions, making sure the
charges and atoms are balanced.
10Cl– + 2MnO4– + 2K+ + 16H+ + 6Cl– →
5Cl2 + 2Mn2+ + 4Cl– + 8H2O + 2K+ + 2Cl–
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Combine the spectator and nonspectator Cl– on
each side.
20.3 Describing Redox Equations >
16Cl–(a) + 2MnO4–(a) + 2K+(a) + 16H+(a) →
5Cl2(g) + 2Mn2+(a) + 6Cl–(a) + 8H2O(l) + 2K+(a)
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20.3 Describing Redox Equations >
Show the balanced equation for the substances
given in the question (rather than for ions).
2KMnO4(aq) + 16HCl(aq) →
2MnCl2(aq) + 5Cl2(g) + 8H2O(l) + 2KCl(aq)
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20.3 Describing Redox Equations >
Use the half-reaction method to
balance the following redox equation.
FeCl3 + H2S → FeCl2 + HCl + S
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20.3 Describing Redox Equations >
Glossary Terms
• oxidation-number-change method: a method
of balancing a redox equation by comparing the
increases and decreases in oxidation numbers
• half-reaction: an equation showing either the
oxidation or the reduction that takes place in a
redox reaction
• half-reaction method: a method of balancing a
redox equation by balancing the oxidation and
reduction half-reactions separately before
combining them into a balanced redox equation
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20.3 Describing Redox Equations >
BIG IDEA
Reactions
Redox equations can be balanced by two
methods, the oxidation-number-change
method and balancing the oxidation and
reduction half-reactions.
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