Transcript Balancing Redox Equations in acidic solutions

Oxidation-Reduction
Dr. Ron Rusay
Balancing Oxidation-Reduction
Reactions
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).
Aqueous Reactions:
Oxidation - Reduction
 In
the following reaction, identify what is being
oxidized and what is being reduced. What is
the total number of electrons involved in the
process?
Oxidation Reduction Reactions
QUESTION
In a redox reaction, oxidation and reduction must both occur.
Which statement provides an accurate premise of redox
chemistry?
A.The substance that is oxidized must be the oxidizing agent.
B.The substance that is oxidized must gain electrons.
C.The substance that is oxidized must have a higher
oxidation number afterwards.
D.The substance that is oxidized must combine with oxygen.
ANSWER
In a redox reaction, oxidation and reduction must both occur.
Which statement provides an accurate premise of redox
chemistry?
A.The substance that is oxidized must be the oxidizing agent.
B.The substance that is oxidized must gain electrons.
C.The substance that is oxidized must have a higher
oxidation number afterwards.
D.The substance that is oxidized must combine with oxygen.
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
Reactivity Tables
(usually reducing)
show relative
reactivities:
In the examples from
the previous slide, the
acid solution (H+) will
react with anything
below it in the Table
but
not above.
Nickel
and Zinc,
….but not Copper.
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
How many of the following are oxidationreduction reactions?
NaOH + HCl ® NaCl + H2O
Cu + 2AgNO3 ® 2Ag + Cu(NO3)2
Mg(OH)2 ® MgO + H2O
N2 + 3H2 ® 2NH3
ANSWER
C) 2
Cu + 2AgNO3  2Ag + Cu(NO3)2
N2 + 3H2  2NH3
If an element is a reactant, this
usually 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).
Balancing Redox Equations
in acidic solutions
1
Fe+2(aq)+ Cr2O72-(aq) +H+(aq) 
Fe3+(aq) + Cr3+(aq) + H2O(l)
? Cr oxidation number?
Fe 2+(aq)+ Cr2O72-(aq) +H+(aq) 
Fe 3+(aq) + Cr 3+(aq) + H2O(l)
x = ? Cr ; 2x+7(-2) = -2; x = +6
Balancing Redox Equations
in acidic solutions
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)
Balancing Redox Equations
in acidic solutions
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
2nd (Hydrogen)
= 14
Balancing Redox Equations
in acidic solutions
1
Completely Balanced Equation:
6 Fe2+(aq)+ Cr2O72-(aq) + 14 H+(aq) 
6 Fe3+(aq) + 2 Cr3+(aq)+ 7 H2O(l)
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
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-
Balancing Redox Equations
in basic solutions
Electronically Balanced Equation:
2 MnO2 (aq)+ ClO31-(aq) + 6 e - 
2 MnO4 1- + Cl 1- + 6 e-
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
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.
2(MnO4– + 3 e–  MnO2)
3(C2O42–  2 CO32– + 2 e–)
+ 4 OH–  + 2 H2O