Redox Titration

Only available on StudyMode
  • Download(s) : 749
  • Published : May 14, 2011
Open Document
Text Preview
CHEMISTRY 205
REDOX TITRATIONA. Purpose:
• To learn some technique in volumetric
analysis: Redox titration.
• To review the stoichiometry of an oxidation-
reduction reaction.
• To determine the concentration of an
unknown sodium oxalate (Na2C2O4) solution
by titrating it against standardized potassium
permanganate solution (KMnO4).
• To determine the percent by mass of Fe(II) in
the form of ferrous ammonium sulfate
Fe(NH4)2(SO4)2.6H2O in a mixture by redox
titration. B. Theory:
(electron transfer reactions)
2Mg (s) + O2 (g) 2MgO (s)
2Mg 2Mg2+ + 4e-
O2 + 4e-
2O2-
Oxidation half-reaction (lose e-
)
Reduction half-reaction (gain e-
)
2Mg + O2 + 4e-
2Mg2+ + 2O2-
+ 4e-
2Mg + O2 2MgO
* Copyright: The McGraw-Hill Companies, Inc.
Oxidation-Reduction Reactions*Zn (s) + CuSO4 (aq) ZnSO4 (aq) + Cu (s)
Zn is oxidized, Zn Zn2+ + 2e-
Cu2+ is reduced, Cu2+ + 2e-
Cu
Zn is the reducing agent
Cu2+ is the oxidizing agent
Copper wire reacts with silver nitrate to form silver metal. What is the oxidizing agent in the reaction?
Cu (s) + 2AgNO3 (aq) Cu(NO3)2
(aq) + 2Ag (s)
Cu Cu2+ + 2e-
Ag+ + 1e-
Ag Ag+ is reduced, Ag+ is the oxidizing agentOxidation number*: The charge the atom would have in a molecule (or an
ionic compound) if electrons were completely transferred.
1. Free elements (uncombined state) have an
oxidation number of zero.
Na, Be, K, Pb, H2, O2, P4 = 0
2. In monatomic ions, the oxidation number is equal
to the charge on the ion.
Li
+, Li = +1; Fe3+, Fe = +3; O2-
, O = -2
3. The oxidation number of oxygen is usually –2. In
H2O2
and O2
2-
it is –1.
* Copyright: The McGraw-Hill Companies, Inc.4. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds.
In these cases, its oxidation number is –1.
6. The sum of the oxidation numbers of all the atoms
in a molecule or ion is equal to the charge on the
molecule or ion.
5. Group IA metals are +1, IIA metals are +2 and
fluorine is always –1.
HCO3
-
O = -2 H = +1
3x(-2) + 1 + ? = -1
C = +4
Oxidation numbers of
all the elements in
HCO3
-
?The oxidation numbers of elements in their compoundsNaIO3
Na = +1 O = -2
3x(-2) + 1 + ? = 0
I = +5
IF7
F = -1
7x(-1) + ? = 0
I = +7
K2Cr2O7
O = -2 K = +1
7x(-2) + 2x(+1) + 2x(?) = 0
Cr = +6
Oxidation numbers of
all the elements in the
following ?
Examples*:
* Copyright: The McGraw-Hill Companies, Inc.Balancing Redox Equations*: 1. Write the unbalanced equation for the reaction ion ionic form.
The oxidation of Fe2+ to Fe3+ by Cr2O7
2-
in acid solution?
Fe2+ + Cr2O7
2-
Fe3+ + Cr3+
2. Separate the equation into two half-reactions.
Oxidation:
Cr2O7
2-
Cr3+
+6 +3
Reduction:
Fe2+ Fe3+
+2 +3
3. Balance the atoms other than O and H in each half-
reaction.
Cr2O7
2-
2Cr3+
* Copyright: The McGraw-Hill Companies, Inc.4. For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms.
Cr2O7
2-
2Cr3+ + 7H2O
14H+
+ Cr2O7
2-
2Cr3+ + 7H2O
5. Add electrons to one side of each half-reaction to
balance the charges on the half-reaction.
Fe2+ Fe3+ + 1e-
6e-
+ 14H+
+ Cr2O7
2-
2Cr3+ + 7H2O
6. If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by
appropriate coefficients.
6Fe2+ 6Fe3+ + 6e-
6e-
+ 14H+
+ Cr2O7
2-
2Cr3+ + 7H2O7. Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both
sides must cancel.
6e-
+ 14H+
+ Cr2O7
2-
2Cr3+ + 7H2O
6Fe2+ 6Fe3+ + 6e-
Oxidation:
Reduction:
14H+ + Cr2O7
2-
+ 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O
8. Verify that the number of atoms and the charges are
balanced.
14x1 – 2 + 6x2 = 24 = 6x3 + 2x3
9. For reactions in basic solutions, add OH-
to both sides of
the equation for every H+ that appears in the final
equation.Reaction of KMnO4 with...
tracking img