Redox I

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Redox Reactions
• Redox is short for “oxidation-reduction.”
•Plating and combustion are redox reactions.
• Consider the following redox reaction:
Fe (s) + Ni(NO3)2 (aq) Fe(NO3)2 (aq) + Ni (s)
• Fe goes from an element
to an ion: Fe  Fe2+
Fe got oxidized
• Ni goes from an ion to an
element: Ni2+  Ni
Ni2+ got reduced
Oxidation
Is
Loss of electrons
Reduction
Is
Gain of electrons
Oxidation Numbers
• We determine whether a chemical species has
been oxidized or reduced by looking at the
change in its oxidation number.
Fe (s) + Ni(NO3)2 (aq) Fe(NO3)2 (aq) + Ni (s)
ox # 0
+2
+2
0
• The oxidation number of an element is 0.
• The oxidation number of a monatomic
ion is its charge.
• The ox # for Fe went up. It was oxidized.
• The ox # for Ni2+ went down. It was
reduced (reduced = down).
Oxidation
Is
Loss of electrons
Reduction
Is
Gain of electrons
Oxidation Numbers
You are responsible for knowing the
rules for assigning oxidation
numbers to elements in ions and
compounds.
Oxidation Numbers
1. The oxidation number of an element is 0.
2. The oxidation number of a monatomic ion is its
charge.
3. The oxidation number of oxygen in a compound is
usually -2 (exception is peroxide, O22-, where the
oxidation number is -1).
4. The oxidation number of hydrogen in a compound
is 1 when bonded to nonmetals and -1 when
bonded to metals.
5. The oxidation number of fluoride is always -1.
6. The sum of the oxidation numbers in a neutral
compound is zero.
7. The sum of the oxidation numbers in a polyatomic
ion is the charge of the ion.
Oxidation Numbers - Examples
Species
Fe
H2
O2
Al3+
Fe2+
S in SO42- :
Cr in Cr2O72S in SO32P in PO43-
oxidation
number
0
0
0
+3
+2
Species
Na+
H+
O2ClN3-
oxidation
number
+1
+1
-2
-1
-3
overall charge of ion is -2 = ox # of S + 4(ox # of O)
-2 = ox # of S + 4(-2)
-2 = ox # of S – 8
-2+8 = 6 = ox # of S
+6
+4
+5
N in NO3N in NO2C in CO32-
+5
+3
+4
Redox Reactions
To see if a redox reaction has occurred, check
to see if the oxidation numbers of any element
involved in the reaction have changed.
Mg (s) + Fe(NO3)2 (aq) Mg(NO3)2 (aq) + Fe (s)
ox #
0
+2 +5 -2
+2
+5 -2
0
•Mg goes from an element to an ion: Mg  Mg2+
Mg got oxidized. Fe2+ was the oxidizing agent.
•Fe goes from an ion to an element: Fe2+  Fe
Fe2+ got reduced. Mg was the reducing agent.
Oxidation
Is
Loss of electrons
This type of redox reaction, where the ion in solution
is replaced through the oxidation of an element, is
called a single replacement reaction.
Reduction
Is
Gain of electrons
Redox Reactions
• Single replacement reactions have the form:
A + BX (aq)  AX (aq) + B
• Single replacement reactions will occur if the
element A is more likely to get oxidized than the
ion B.
• The activity series gives this information. See
the text.
• Notice that H2 (g) is included in
this series. The series can tell us
whether a metal will dissolve in a
nonoxidizing acid.
Oxidation
Is
Loss of electrons
Reduction
Is
Gain of electrons
Will the following
pairs react?
Zn + Cu(NO3)2 (aq)
Cu + Zn(NO3)2 (aq)
Zn + NaNO3 (aq)
Au + AgNO3 (aq)
Ag + Au(NO3)3 (aq)
y
n
n
n
y
Will the following
metals dissolve in
acid?
Li
Al
Ni
Cu
Au
y
y
y
n
n
What gas will be
produced?
H2
Redox Reactions - Example
Molecular equation:
Mg (s) + Fe(NO3)2 (aq) Mg(NO3)2 (aq) + Fe (s)
Total Ionic equation:
Mg(s) + Fe2+(aq) + 2NO3-(aq) Mg2+(aq) + Fe(s) + 2NO3-(aq)
Net Ionic equation:
Mg (s) + Fe2+ (aq) Mg2+ (aq) + Fe (s)
Mg got oxidized. Fe2+ was the oxidizing agent.
Fe2+ got reduced. Mg was the reducing agent.
Redox Reactions - Example
When Zn dissolves in sulfuric acid, the molecular
equation is
Zn (s) + H2SO4 (aq) ZnSO4 (aq) + H2 (g),
the total ionic equation is
Zn(s) + 2H+(aq) + SO42-(aq) Zn2+(aq) + SO42-(aq) + H2 (g),
and the net ionic equation is:
Zn(s) + 2H+(aq) Zn2+(aq) + H2 (g)
Zn got oxidized. H+ was the oxidizing agent.
H+ got reduced. Zn was the reducing agent.
Balancing Redox Reactions
• A redox reaction consists of an oxidation
AND a reduction.
• Although an oxidation cannot occur without a
reduction, it is useful to be able to look at each
separately.
• A redox reaction can be considered to be the
sum of an oxidation half reaction and a
reduction half reaction.
Balancing Redox Reactions
The Method of Half-Reactions:
Step 1. Use oxidation numbers to identify which
species is oxidized and which is reduced.
Cr2O72- (aq) + Fe2+(aq)  Cr3+ (aq) + Fe3+(aq)
+6 -2
+2
+3
+3
Cr2O72- is reduced. Fe2+ is oxidized.
The Method of Half-Reactions
Step 2. Separate the reaction into the two half-reactions.
oxidation half-reaction:
Fe2+(aq)  Fe3+(aq)
reduction half-reaction:
Cr2O72- (aq)  Cr3+ (aq)
The Method of Half-Reactions
Step 3. Balance each half-reaction.
Step 3a. Balance the elements other than H and O
Fe2+(aq)  Fe3+(aq) already balanced
Cr2O72- (aq)  Cr3+ (aq)
Cr2O72- (aq)  2Cr3+ (aq)
Step 3b. Balance O by adding H2O:
Fe2+(aq)  Fe3+(aq) already balanced
Cr2O72- (aq)  2Cr3+ (aq)
Cr2O72- (aq)  2Cr3+ (aq) + 7H2O(l)
The Method of Half-Reactions
Step 3c. Balance H by adding H+:
Fe2+(aq)  Fe3+(aq) already balanced
Cr2O72- (aq)  2Cr3+ (aq) + 7H2O(l)
14H+ (aq) + Cr2O72- (aq)  2Cr3+ (aq) + 7H2O(l)
Step 3d. Balance the charge by adding electrons (e-) to
the side that is relatively more positive:
Fe2+(aq)  Fe3+(aq) + e6e- + 14H+(aq) + Cr2O72- (aq)  2Cr3+ (aq) + 7H2O(l)
The Method of Half-Reactions
Step 4. Multiply each half-reaction by an integer so that the
electrons (e-) in each equation balance:
6( Fe2+(aq)  Fe3+(aq) + e- )
6e- + 14H+ (aq) + Cr2O72- (aq)  2Cr3+ (aq) + 7H2O(l)
Step 5. Add the two half-reactions, canceling species where
possible:
6Fe2+(aq)+6e- +14H+(aq)+Cr2O72-(aq) 6Fe3+(aq)+6e- +2Cr3+(aq)+7H2O(l)
6Fe2+(aq) + 14H+(aq)+ Cr2O72-(aq)  6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)
this is the balanced equation
The Method of Half-Reactions
Step 6. CHECK THE EQUATION to make sure it is
balanced both for mass and for charge.
6Fe2+(aq) + 14H+(aq)+ Cr2O72-(aq)  6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)
The Method of Half-Reactions
Reactions in Basic solution
If the redox reaction takes place in BASIC solution, use steps
1-6 (as before) to balance the equation as if it took place in
acidic solution. Then perform one more step:
Step 7. (ONLY for redox reactions taking place in basic
solution!) Add OH- to BOTH sides of the equation to cancel all
of the H+, then make sure H2O appears only on one side of the
equation. If, in our example, the reaction between Fe3+ and
Cr2O72- took place in basic solution:
6Fe2+(aq) + 14H+(aq) + 14OH- (aq) + Cr2O72-(aq)
6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l) + 14OH- (aq)
The Method of Half-Reactions
6Fe2+(aq) + 14H+(aq) + 14OH- (aq) + Cr2O72-(aq)
6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l) + 14OH- (aq)
Remember that
H+(aq) + OH –(aq)  H2O(l):
14H+(aq) + 14OH- (aq)  14H2O(l)
6Fe2+(aq) + 14H2O(l) + Cr2O72-(aq)
6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l) + 14OH- (aq)
The Method of Half-Reactions
Reactions in Basic solution
Step 7. Add OH- to BOTH sides of the equation to cancel all of
the H+, then make sure H2O appears only on one side of the
equation.
7H2O
6Fe2+(aq) + 14H2O(l) + Cr2O72-(aq)
6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l) + 14OH- (aq)
Balanced equation for the reaction in basic solution:
6Fe2+(aq) + 7H2O(l) + Cr2O72-(aq)
6Fe3+(aq) + 2Cr3+(aq) + 14OH- (aq)
Stoichiometry of Redox Reactions
Once you have a balanced redox equation, you may use it to
perform stoichiometric calculations.
Example: How many grams of potassium dichromate are
needed to oxidize 10.0 g of iron(II) nitrate according to the
following balanced equation?
6Fe2+(aq) + 7H2O(l) + Cr2O72-(aq) 6Fe3+(aq) + 2Cr3+(aq) + 14OH- (aq)
10.0 g Fe(NO3)2 x
1 mol Fe(NO3)2 x 1 mol Fe2+
x 1 mol Cr2O72179.86 g Fe(NO3)2 1 mol Fe(NO3)2
6 mol Fe2+
x 1 mol K2Cr2O7 x
1 mol Cr2O72-
294.18 g K2Cr2O7
1 mol K2Cr2O7
= 2.73 g K2Cr2O7

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