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Lecture #12
Redox Reactions
Chemistry 142 B
James B. Callis, Instructor
Autumn Quarter, 2004
This lecture covers two topics:
• Assignment of oxidation numbers
• Balancing of redox equations
Oxidation - A process by which a substance (reductant)
gives up electrons to another substance (oxidizing agent).
The reductant is oxidized.
Reduction - A process by which a substance (oxidant)
accept electrons from another substance (reducing
agent).
In a chemical reaction, the total number of electrons
are conserved; so also are the number of charges.
Thus, it is convenient to assign fictitious charges to
the atoms in a molecule and call them oxidation
numbers. Oxidation numbers are chosen so that (a)
conservation laws are obeyed, and (b) in ionic
compounds the sum of oxidation numbers on the
atoms coincides with the charge on the ion.
Rules for Assigning Oxidation
Numbers:
1. The oxidation numbers of the atoms in a neutral
molecule must add up to zero, and those in an ion
must add up to the charge on the ion.
2. Alkali metal atoms have oxidation number +1,
alkaline earth atoms +2, in their compounds.
3. Fluorine always has oxidation number -1 in its
compounds. The other halogens have oxidation
number -1 in their compounds except those with
oxygen and with other halogens, where they can
have positive oxidation number.
Oxidation Numbers (cont.):
4. Hydrogen is assigned oxidation number +1 in its compounds
except in metal hydrides such as LiH, where convention (2)
takes precedence and hydrogen has oxidation number -1.
5. Oxygen is assigned oxidation number -2 in compounds. There
are two exceptions: in compounds with fluorine, convention 3
takes precedence, and in compounds that contain O - O bonds,
conventions 2 and 4 take precedence. Thus the oxidation
number of oxygen in OF2 is +2: in peroxides (such as H2O2
and Na2O2) it is -1. In superoxides (such as KO2) oxygen's
oxidation number is -1/2.
Examples:
NH4NO3
N=
H=
N=
O=
B2H6
B=
H=
Examples (cont.):
• BaH2
Ba =
H =
• (S4O6)2S=
O=
Problem 12-1: Determining the Oxidation Number of an
Element in a Compound
Problem: Determine the oxidation number (ox. no.) of each element
in the following compounds.
a) iron(III)chloride b) nitrogen dioxide
c) sulfuric acid
Plan: We apply the rules in Table 4.3, always making sure that the
oxidation no. values in a neutral compound add up to zero, and in a
polyatomic ion, to the ion’s charge.
Solution:
a) FeCl3 This compound is composed of monoatomic ions. The
ox. no. of Cl- is -1, for a total of -3. Therefore the Fe must be +3.
b) NO2
c) H2SO4
Problem 12-2:Recognizing Oxidizing and Reducing Agents - I
Problem: Identify the oxidizing and reducing agent in each of the Rx:
a) Zn(s) + 2 HCl(aq)
ZnCl2 (aq) + H2 (g)
b) S8 (s) + 12 O2 (g)
8 SO3 (g)
c) NiO(s) + CO(g)
Ni(s) + CO2 (g)
Plan: First we assign an oxidation number (ox. no.) to each atom (or ion)
based on the rules in Table 4.3. A reactant is the reducing agent if it
contains an atom that is oxidized (ox. no. increased in the reaction). A
reactant is the oxidizing agent if it contains an atom that is reduced
( ox. no. decreased).
Solution:
a) Assigning oxidation numbers:
-1
+1
0
Zn(s) + 2 HCl(aq)
-1
0
+2
ZnCl2(aq) + H2 (g)
HCl is the oxidizing agent, and Zn is the reducing agent!
Problem 12-2:Recognizing Oxidizing and Reducing Agents - II
b) Assigning oxidation numbers:
S8 (s) + 12 O2 (g)
8 SO3 (g)
____ is the reducing agent and ____ is the oxidizing agent.
c) Assigning oxidation numbers:
NiO(s) + CO(g)
Ni(s) + CO2 (g)
___ is the reducing agent and ____ is the oxidizing agent.
Problem 12-3: Balancing Redox Equations - I
To proceed let us illustrate with the following problem:
x NH3(g)+ y CH4(g) -> z HCN(g) + w H2(g)
The objective is to find a set of 4 integer
coefficients {x, y, z, w} which are as small as
possible and also solve the set of atom
conservation equations. These later state that no
mass is created or destroyed and that the identity
of the atoms is preserved. Thus for the above
equation we have three atom balance equations,
one each for N, H, and C.
Problem 12-3: These equations are:
N:
H:
C:
We see that we have 3 equations in 4 unknowns. Clearly we have
one more unknown than equations. Thus, there are many possible
valid solutions to the problem as stated. Mathematically, we say that
the problem is underdetermined of ill-posed. We solve this problem
by requiring that the set of values of the solution, {x,y,z,w} be the
minimum set of integers. Now we have an additional equation which
in most cases leads to an unique solution. Unfortunately, such a set
of equations cannot be solved in one step. However for simple
equations we give a straight forward manual procedure.
Problem 12-3: Two Step Hand Solution
• Here, we start with a guess for one of the coefficients,
e.g. let x = 1, and then by back substitution we solve
for all of the other coefficients. Hopefully, this will
lead directly to the sought for solution in terms of the
set of integers of minimum value that satisfy all of the
atom balance equations.
• Often, our guess is not correct, and we are left with
solutions that are rational numbers instead of integers.
Now all we need to do is find the smallest common
multiplier which will convert the solution to one of the
set of integers of minimum values.
Problem 12-3: The Solution
Let x = 1,
Thus our solution is
We verify it by substituting the numerical
values of x, y, z and w into the atom balance
equations.
Balancing Chemical Equations Involving
Charged Species
• In this section we will show that balancing a chemical
equation involving charged species closely resembles
the previous approach and merely involves one
additional equation to ensure that charge is balanced in
addition to mass.
• One of the major difficulties of redox reactions is that it
is often necessary to add H2O and H3O+ to opposite
sides of the reaction to balance the equation. As
beginners we will assume that all pertinent species are
provided and only coefficients are to be determined.
Example: Problem 12-4
t CuS + u NO3- + v H3O+ -> w Cu2+ + x SO42- +y NO + z H2O
The objective is to find a set of integer
coefficients {t, u, v, w, x, y, z} which are as
small as possible and also solve the set of
atom conservation equations.
For the above equation we have five atom balance
equations, one each for Cu, S, N, O, H. These equations are:
Cu:
S:
N:
O:
H:
We see that we thus far have 5 equations in 7 unknowns, or
since we have the three equalities t =w, t = x and u = y, we
actually have 2 equations in 4 unknowns. Clearly we have
two more unknowns than equations. Mathematically, we
say that the problem is underdetermined.
In the case of charged species, the charges must
also balance. Here is the equation for charge
balance (CB):
CB:
Now that we have 6 equations in 7 unknowns, we
can add the constraint that the solution set
{x,y,z,t,u,v,w} be the minimum set of integers.
Problem 12-4: Hand Solution - Part I
t = 1 = = This leaves 4 equations:
Eliminate u
Eliminate v
Eliminate y
Thus z = 4.
Problem 12-4: Hand Solution - Part II
Now go backwards through the solutions solving
for each variable.
From z = 4 and y =2z/3
From v = y,
From u = y,
From v = u,
Problem 12-4: Hand Solution - Part III
The above are a set of integers and rational numbers. By multiplying by
3 we can convert to the minimum set of integers.
Thus the solution is
_CuS + _NO3- + _H3O+ -> _Cu2+ + _(SO4)2- +_NO + _H2O
Problem 12-5: Redox Titration- I
Volume (L) of KMnO4 Solution
a)
M (mol/L)
Moles of KMnO4
b)
Molar ratio
in redox rxn.
Moles of CaC2O4
c)
Chemical Formulas
Moles of Ca+2
Problem: Calcium Oxalate was
precipitated from 1.00 mL blood by
the addition of Sodium Oxalate so
the Ca2+ conc. in the blood could
be determined. This precipitate was
dissolved in a sulfuric acid solution,
which then required 2.05 mL of
4.88 x 10-4 M KMnO4 to reach the
endpoint via the rxn. of Fig. 4.14.
a) Calculate the moles of Ca2+.
b) Calculate the Ca2+ conc. in blood.
Plan: a) Calculate the moles of
Ca2+ in the H2SO4 solution (and
blood sample).
b) Convert the Ca2+ conc.into units
of mg Ca2+/ 100 mL blood.
Problem 12-5: Redox Titration - Calculation - II
Equation:
2 KMnO4 (aq) + 5 CaC2O4 (aq) + 8 H2SO4 (aq)
2 MnSO4 (aq) + K2SO4 (aq) + 5 CaSO4 (aq) + 10 CO2 (g) + 8 H2O(l)
a) Moles of KMnO4
b) Moles of CaC2O4
c) Moles of Ca+2
Problem 12-5: Redox Titration - III
Moles of Ca2+/ 1 mL of blood
a) Calc of mol Ca2+ per 100 mL
multiply by 100
Moles of Ca2+/ 100 mL blood
M (g/mol)
b) Calc of mass of Ca2+ per 100 mL
Mass (g) of Ca2+/ 100 mL blood
1g = 1000mg
Mass (mg) of Ca2+ / 100 mL blood
c) convert g to mg!
Problem 12-5: Redox Titration - IV
a) Mol Ca2+ per 100 mL Blood
b) mass (g) of Ca2+
c) mass (mg) of Ca2+
Answers to Problems in Lecture #12
1. (b) The ox. no. of oxygen is –2; the ox. no. of N is +4
(c) The ox. no. of H is +1, the ox. no. of each O is –2, the sulfur
atom is +6
2. (b) S8 is the reducing agent and O2 is the oxidizing agent
(c) CO is the reducing agent and NiO is the oxidizing agent
3. NH3 + CH4 -> HCN + 3H2
4. 3CuS + 8NO3- + 8H3O+ -> 3Cu2+ + 3(SO4)2- +8NO + 12H2O
5. (a) 2.50 x 10 -6 mol Ca2+ in H2SO4 soln; 2.50 x 10 -4 mol Ca2+ in
blood; (b) 10.0 mg Ca2+/100 mL blood