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Redox Reactions

Lecture 12

- Chemistry 142 B
- James B. Callis, Instructor
- Autumn Quarter, 2004

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.

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.

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) oxygen's oxidation

number is -1/2.

Examples

- NH4NO3N H N O
- B2H6B H

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

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!

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.

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.

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

-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.

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 -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

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

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

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

- (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

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