Chemistry - A Molecular Science

Example 11.2

Identify the oxidants and reductants in the following redox reactions.

a) C

H 12

O 22

(s) + 12O 11

(g) 2

→

12CO

(g) + 11H 2

O(l) 2

O^2

is elemental and is the oxidant as the

oxidation state of O changes from 0 in O

to -2 2

on the right. C

H 12

O 22

is the reducing agent. Assigning oxidation states of +1 to H and -2 11

to O allows us to determine that C is zero in C

H 12

O 22

, while it is +4 in CO 11

. Each of the 2

24 O atoms in 12O

undergoes a two-electron oxidation (0 to -2), so it is a 48-electron 2

reduction. Similarly, each of t

he 12 C atoms undergoes a four-ele

ctron oxidation (0 to +4),

so it is a 48-electron oxidat

ion.* Thus, n = 48 electrons.

* This is a redox reaction that d

oes not involve an electron transfer

because the transfer of 48 electrons requires the formation of C

4+ and

2-O

ions, while all of the product and reactant bonds are covalent, and

all of the atoms have zero formal charge. The bonds become more polar but not ionic, so electrons

move

away from carbon atoms and

toward oxygen atoms, but

they are not transferred.

b) 3Cu(s) + 2NO

1- 3

(aq) + 8H

1+(aq)

→

3Cu

2+(aq) + 2NO(g) + 4H

O(l) 2

Elemental copper undergoes a two-electron oxi

dation from 0 to +2, so it is the reducing

agent. Each of the three Cu atoms undergoes a tw

o-electron oxidation, so this is a 3 x 2 =

6 electron oxidation. NO

1- is then identified as the oxi 3

dant. The oxidation state of N

changes from +5 in NO

1- 3

to +2 in NO. Each N undergoes a three-electron reduction so

this is a 2 x 3 = 6 electron r

eduction. Consequently, n = 6.

Example 11.3

How many electrons are required for the reduction of 0.15 mol of Cr

O 2

2- to Cr 7

3+?

First, balance the Cr atoms in the electron transfer: Cr

O 2

2- 7

→

2Cr

3+^

The oxidation state of chromium is +6 in Cr

O 2

2- and +3 in Cr 7

3+, so 3 mol electrons are

required for

each

mole of Cr. Use the factor-label meth

od to determine moles of electrons.

27

0.15 mol Cr O

2-

2 mol Cr

×

27

1 mol Cr O

1-

2-

3 mol e×1 mol Cr

1-

= 0.90 mol e

11.2

HALF-REACTIONS A redox reaction can be written as the sum of

two half-reactions: oxi

dation and reduction.

The two half-reactions for the reaction of iron and copper (II) are:

Oxidation half-reaction:

Fe

→

Fe

2+ + 2e

1-^

Reduction half-reaction:

Cu

2+ + 2e

1-^

→

Cu

Net redox reaction:

Cu

2+ + Fe

→

Cu + Fe

2+^

The two half-reactions indicate the changes that

occur in each of the two redox couples. In

this reaction, two electrons are given up in

the oxidation half-reaction, and two electrons

Chapter 11 Electron Transfer and Electrochemistry