Download Balancing Redox Eqs-10

Oxidation-Reduction
Dr. Ron Rusay
Balancing Oxidation-Reduction
Reactions
© Copyright 2002-2010 R.J. Rusay
Oxidation-Reduction
� Oxidation
is the loss of electrons.
� Reduction is the gain of electrons.
� The reactions occur together. One does
not occur without the other.
� The terms are used relative to the
change in the oxidation state or
oxidation number of the reactant(s).
© Copyright 2002-2010 R.J. Rusay
Oxidation Reduction Reactions
QUESTION
ANSWER
B) NO2
Oxygen almost always has an oxidation state of
–2 when part of a compound. The exception is
when it is part of a peroxide. For example,
hydrogen peroxide H2O2. Then it has an
oxidation state of –1.
QUESTION
What is the oxidation number of chromium
in ammonium dichromate?
A) +3
B) +4
C) +5
D) +6
ANSWER
What is the oxidation number of chromium
in ammonium dichromate?
A) +3
B) +4
C) +5
(NH4)2Cr2O7
D) +6
Zinc
QUESTION
Select all redox reactions by looking for a change in oxidation
number as reactants are converted to products.
I) Ca + 2 H2O → Ca(OH)2 + H2
II) CaO + H2O → Ca(OH)2
III) Ca(OH)2 + H3PO4 → Ca3(PO4)2 + H2O
IV) Cl2 + 2 KBr → Br2 + 2 KCl
A) I and II
B) II and III
C) I and IV
D) III and IV
ANSWER
Select all redox reactions by looking for a change in oxidation number
as reactants are converted to products.
I) Ca + 2 H2O → Ca(OH)2 + H2
II) CaO + H2O → Ca(OH)2
III) Ca(OH)2 + H3PO4 → Ca3(PO4)2 + H2O
IV) Cl2 + 2 KBr → Br2 + 2 KCl
A) I and II
B) II and III
C) I and IV
D) III and IV
QUESTION
ANSWER
C) 2
If an element is found on the reactant’s side, this
is almost always a redox reaction, since an
element usually becomes part of a compound
during a chemical reaction.
Number of electrons gained must equal the number of electrons lost.
- 2 e-
+2 e-
Use oxidation numbers to determine what is oxidized
and what is reduced.
0
+2 e-
Cu 2+

0
Refer to Balancing
H2 (g)
Oxidation-Reduction Reactions
Cu (s)
- 2 e-

2H+
QUESTION
ANSWER
B) the oxidizing agent.
Metals lose electrons, so they are oxidized,
making the other reactant an oxidizing agent.
Balancing Redox Equations
in acidic solutions
1
1) Determine the oxidation numbers of atoms in both
reactants and products.
2) Identify and select out those which change
oxidation number (“redox” atoms) into separate
“half reactions”.
3) Balance the “redox” atoms and charges (electron
gain and loss must equal!).
4) In acidic reactions balance oxygen with water
then hydrogen from water with acid proton(s).
© Copyright 2002-2010 R.J. Rusay
Balancing Redox Equations
1
Fe+2(aq)+ Cr2O72-(aq) +H+(aq)---->
Fe3+(aq) + Cr3+(aq) + H2O(l)
?
Fe 2+(aq)+ Cr2O72-(aq) +H+(aq)---->
Fe 3+(aq) + Cr 3+(aq) + H2O(l)
x = ? Cr ; 2x+7(-2) = -2; x = +6
© Copyright 2002-2010 R.J. Rusay
Balancing Redox Equations
1
Fe 2+(aq)-e - ---> Fe 3+(aq)
Cr2O72-(aq) + 6 e - --> 2 Cr 3+(aq)
Cr = (6+)
6 (Fe 2+(aq) -e - ---> Fe3+(aq))
6 Fe 2+(aq) ---> 6 Fe3+(aq) + 6 e Cr2O72-(aq) + 6 e - --> 2 Cr3+(aq)
© Copyright 2002-2010 R.J. Rusay
Balancing Redox Equations
1
6 Fe2+(aq) ---> 6 Fe3+(aq) + 6 e Cr2O72-(aq) + 6 e - --> 2 Cr3+(aq)
6 Fe2+(aq)+ Cr2O72-(aq) + ? 2nd H+(aq) ---->
6 Fe3+(aq) + 2 Cr3+(aq)+ ? 1st Oxygen H2O(l)
Oxygen
=7
© Copyright 2002-2010 R.J. Rusay
2nd (Hydrogen)
= 14
Balancing Redox Equations
1
Completely Balanced Equation:
6 Fe2+(aq)+ Cr2O72-(aq) + 14 H+(aq) ---->
6 Fe3+(aq) + 2 Cr3+(aq)+ 7 H2O(l)
© Copyright 2002-2010 R.J. Rusay
QUESTION
Dichromate ion in acidic medium converts ethanol,
C2H5OH, to CO2 according to the unbalanced equation:
Cr2O72−(aq) + C2H5OH(aq) → Cr3+(aq) + CO2(g) +
H2O(l)
The coefficient for H+ in the balanced equation using
smallest integer coefficients is:
A) 8
B) 10
C) 13
D) 16
ANSWER
Dichromate ion in acidic medium converts ethanol, C2H5OH,
to CO2 according to the unbalanced equation:
Cr2O72−(aq) + C2H5OH(aq) → Cr3+(aq) + CO2(g) + H2O(l)
The coefficient for H+ in the balanced equation using smallest
integer coefficients is:
A) 8
B) 10
C) 13
D) 16
Balancing Redox Equations
in basic solutions
1
1) Determine oxidation numbers of atoms in
Reactants and Products
2) Identify and select out those which change
oxidation number into separate “half reactions”
3) Balance redox atoms and charges (electron gain
and loss must equal!)
4) In basic reactions balance the Oxygen with
hydroxide then Hydrogen from hydroxide with
water
© Copyright 2002-2010 R.J. Rusay
Balancing Redox Equations
in basic solutions
MnO2 (aq)+ ClO31-(aq) + OH 1-aq) ->
MnO41- (aq)+ Cl 1-(aq) + H2O(l)
Mn4+ (MnO2) ---> Mn7+ (MnO4 ) 1Cl+5 (ClO3 ) 1-+ 6 e- ---> Cl 1-
© Copyright 2002-2010 R.J. Rusay
Balancing Redox Equations
in basic solutions
Electronically Balanced Equation:
2 MnO2 (aq)+ ClO31-(aq) + 6 e - ---->
2 MnO4 1- + Cl 1- + 6 e-
© Copyright 2002-2010 R.J. Rusay
Balancing Redox Equations
in basic solutions
Completely Balanced Equation:
2 MnO2 (aq)+ ClO31-(aq) + 2 OH 1- (aq)---->
2 MnO4 (aq)1- + Cl 1- (aq)+ 1 H2O (l)
9 O in product
© Copyright 2002-2010 R.J. Rusay
QUESTION
Oxalate ion can be found in rhubarb and spinach (among other green
leafy plants). The following unbalanced equation carried out in a
basic solution, shows how MnO4– could be used to analyze samples
for oxalate.
MnO4– + C2O42–  MnO2 + CO32– (basic solution)
When properly balanced, how many OH– are present?
A.1
B.2
C.3
D.4
ANSWER
D). The redox equation could be balanced according to the steps for
an acid solution, but then OH– must be added to neutralize the H+
ions. When done properly 4 OH– ions will be present in the basic
equation.