Redox Reactions

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Redox Reactions
Lecture 12

• Chemistry 142 B
• James B. Callis Instructor
• Autumn Quarter 2004

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This lecture covers two topics

• Assignment of oxidation numbers
• Balancing of redox equations

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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.
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Rules for Assigning Oxidation Numbers

• 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.
• Alkali metal atoms have oxidation number 1
  alkaline earth atoms 2 in their compounds.
• 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.

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Oxidation Numbers (cont.)

• 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.
• 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) oxygens oxidation
  number is -1/2.

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Examples

• NH4NO3N H N O
• B2H6B H

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Examples (cont.)

• BaH2Ba H
• (S4O6)2-S O

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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 ions 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
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Problem 12-2Recognizing 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
1
0
0
2
Zn(s) 2 HCl(aq) ZnCl2(aq)
H2 (g)
HCl is the oxidizing agent and Zn is the
reducing agent!
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Problem 12-2Recognizing 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.
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To proceed let us illustrate with the following
problem
Problem 12-3 Balancing Redox Equations - I
x NH3(g) y CH4(g) -gt 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.
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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 xyzw 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.
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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.

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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.
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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.

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Example Problem 12-4 t CuS u NO3- v H3O
-gt 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.
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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.
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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 xyztuvw be the minimum set
of integers.
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Problem 12-4 Hand Solution - Part I t 1
This leaves 4 equations Eliminate
u Eliminate v Eliminate y Thus z 4.
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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
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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 -gt _Cu2 _(SO4)2- _NO _H2O
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Problem 12-5 Redox Titration- I
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.
Volume (L) of KMnO4 Solution
a)
M (mol/L)
Moles of KMnO4
Molar ratio in redox rxn.
b)
Moles of CaC2O4
Chemical Formulas
c)
Moles of Ca2
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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 Ca2
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Problem 12-5 Redox Titration - III
Moles of Ca2/ 1 mL of blood
multiply by 100 a) Calc of mol Ca2 per
100 mL
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 c) convert g to mg!
Mass (mg) of Ca2 / 100 mL blood
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Problem 12-5 Redox Titration - IV
a) Mol Ca2 per 100 mL Blood
b) mass (g) of Ca2
c) mass (mg) of Ca2
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Answers to Problems in Lecture 12

• (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
• (b) S8 is the reducing agent and O2 is the
  oxidizing agent
• (c) CO is the reducing agent and NiO is the
  oxidizing agent
• NH3 CH4 -gt HCN 3H2
• 3CuS 8NO3- 8H3O -gt 3Cu2 3(SO4)2- 8NO
  12H2O
• (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