LaurenHill Chemistry 534

LaurenHill Chemistry 534

107

20. Oxidation_Reduction and Oxidation Numbers

Oxidation-Reduction(Redox) Non Oxidation-Reduction(Redox)

These reactions involve a substance that

gains electrons from a substance, which of

course loses them.

An example is combustion. Can you think

of at least three other examples?

These reactions do not involve the gain and

loss of electrons.

Acid base neutralizations involve the loss

and acceptance of a proton.

Other examples?

Add calcium to a flask of water, and it will fizz vigorously. Light a match above the flask and you

will hear a pop suggesting the presence of hydrogen gas. This is an oxidation-reduction reaction, which

involves a give and take of electrons. Calcium gives up electrons; water takes them in, and that releases

hydrogen from its molecules.

Many reactions involve a gain or loss of electrons, including combustion, respiration,

photosynthesis and electrochemistry, the driving force behind batteries. In order to gain an understanding

of these reactions, we have to learn how to assign oxidation numbers.

Definition An oxidation number is a number assigned to an atom according to a set of rules. Its

purpose is to help you keep track of electrons as they move from one atom or molecule to the next.

They don't necessarily represent a true charge.

Rules:

1. The oxidation number of an atom of any free element is ZERO. The oxidation

number of any atom in a molecule of a single element is also ZERO.

Element Oxidation

Number of

each atom

Na

He

O2

S8

V. Electrochemistry

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2. a) If hydrogen is not exclusively bonded to a metal, the oxidation number of

hydrogen in H-containing compound is +1.

Compound Oxidation

number of each

H atom

Total

contribution

by H

H2O

HNO3

H2O2

b) in metallic hydrides the oxidation number of hydrogen is -1. A metal

hydride has hydrogen bonded to a metal.

Compound Oxidation

Number of

each H atom

Total

contribution

by H

CaH2

NaH

AlH3

A nice little mixture of what we have seen so far:

Compound Oxidation

Number of

each H atom

Total

contribution

by H

H2S

MgH2

H2


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3. The oxidation number of oxygen in compounds is usually -2.

Compound Oxidation

Number of

each O

atom

Total

contribution by

O

H2O

NO2

NO3-1

Exception: Since peroxides contain one more oxygen than expected, each oxygen is

assigned an oxidation number of -1.

normal oxide: H2O

peroxide: H2O2

4. The oxidation of a monoatomic ion is equal to its charge. (So you can apply this

rule to a compound containing monoatomic ions. For members of metallic

families, the periodic table can be consulted. Also, halides (formed from a metal +

a halogen) will be -1.

A basic review:

If a halogen (F, Cl, Br, etc) is only bonded to a metal, then the halogen’s

oxidation number will be -1. In fact F will be -1 in any compound because it is

the most electronegative compound.

All alkali metals are +1 when part of a compound

All alkaline earth metals are +2 when part of a compound

Al is +3 when part of a compound.

Compound Oxidation

Number

of each

atom

Total contribution

by ion

Os+4

V. Electrochemistry

110

Na2O

Al2S3

CaCl2

5. The sum of the oxidation numbers of all the atoms in a compound is ZERO.

The sum of all the oxidation numbers of all the atoms in a polyatomic ion is equal

to the charge of the polyatomic ion.

Example 1 H2O

Example 2 SO4-2

Example 3 NO2-1

Example 4 KMnO4 Example 7 Fe(NO3)2

Example 5 MnO2 Example 8 CuSO4

Example 6 C3H8


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Exercises

1. Use the five rules to assign oxidation numbers to each element in the following:

a. MnO4-1

b. C2H6

c. Xe

d. Fe3O4

e. OF2 (fluoride's charge takes precedence)

f. H2O2

g. CH4

h. Cr2O7-2

i. C2H6O

j. C2H2

k. XeOF4

l. Na2C2O4

m. Ca(NO3)2

n. UO2+2

o. NaBiO3

p. NH3

q. H3AsO4

r. Al(OH)3

s. N2

V. Electrochemistry

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21. Oxidations and Reductions

LEO the lion said “GERrrrr!”

L = Loss

E = of Electrons is

O = Oxidation.

G = Gain

E = of Electrons is

R = Reduction.

An oxidation-reduction reaction involves an exchange of electrons. One chemical

"beast" loses electrons; the other one gains them.

A. Reduction

The substance gaining electrons is reduced. This is accompanied by a reduction or

decrease in oxidation number. The number decreases from a gain of electrons,

which have a negative charge.

B. Oxidation

The substance losing electrons is oxidized. This is accompanied by an increase in

oxidation number. The number increases from a loss of electrons. Taking away

negatives is causing the oxidation number to increase.

Example: Given: CuCl2 + Fe FeCl2 + Cu

Figure out what is being oxidized and what is being reduced.

GERrrrr!


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C. Half-reactions

These reveal how atoms are being either oxidized

or reduced.

Example

Fe Fe+2 + 2e (electrons are products, so they

are being lost by iron)

Cu+2, meanwhile, gets reduced, so it gains

electrons:

Cu+2 + 2e Cu

In order for an atom to be reduced, it must gain

electrons. But these electrons must first be given

up by another atom.

So, oxidation and reduction always occur together.

One half reaction cannot occur on its own.

More Examples

1. In each of the following, identify what is being oxidized and what is being

reduced.

a. C + O2 CO2

b. Al + Cu+2 Al+3 + Cu (not balanced; this will be handled in the next section)

c. H+1 + NO3-1 + Fe+2 H2O + NO + Fe+3

2. Write half-reactions for the following:

a. Al + Cu+2 Al+3 + Cu

b. Ag+ + Zn Zn+2 + Ag

AFTER

BEFORE

V. Electrochemistry

114

Exercises

1. In each of the following, identify what is being oxidized and what is being

reduced.

a. Cu+2 + Fe Fe+2 + Cu

b. Co + Sn+2 Co+2 + Sn

c. H+1 + NO3-1 + Fe+2 H2O + NO + Fe+3

d. Cr + Sn+2 Cr+3 + Sn

e. S-2 + NO3-1 + H+1 SO2 + NO2 + H2O

f. CH4 + O2 CO2 + H2O

g. NH4+1 + NO3

-1 N2 + H2O + H+

h. Al + SO4-2

+ H+1 SO2 + Al+3 + H2O

i. Mg + CO2 MgO + C

j. Cr2O7-2 + SO3

-2 + H+1 Cr+3 + SO4-2 + H2O

2. Write half-reactions for exercises (a), (b) and (d)

3. Identify which of the following are redox reactions. Show work.

a. 2 NO2(g) N2O4(g)

b. 3 HCl(aq) + Al(OH)3(aq) AlCl3(aq) + 3 H2O(l)

c. PH3(g) + 2 Cl2(g) + 2 H2O(l) H3PO2 (aq) + 4 HCl(aq)

d. O + O2 O3


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22. Balancing Oxidation-Reduction Reactions

The basic idea is that, in each half reaction, you have to balance:

(1) the participating atoms (if necessary); so you have to figure out what is oxidized and reduced.

(2) the oxidation numbers by using electrons,

(3) the overall charge (if necessary) by using H+1 or OH-1. (The total charge on the left hand side of

the equation must equal the total right hand-side charge.)

(4) and the number of hydrogens with water (if necessary).

(5) Then you combine the two half-reactions algebraically to get the electrons to cancel.

Example 1 Balance MnO4-1 + Cl-1 Mn+2 + Cl2 using half-reactions.

Assume an acidic medium.

Steps 1 and 2

Figure out what is being oxidized and reduced. Make sure that those participating atoms are balanced and

write each half reaction.

MnO4-1 + 5e Mn+2 (5 electrons because Mn is being reduced from +7 to 2)

2Cl-1 Cl2 + 2e (2 electrons because each Cl is being oxidized from –1 to 0)

Step 3

Balance the total charge on each side of equation with H+1 if they specify acidic conditions and with OH-1 if

conditions are alkaline.

Acidic situation:

MnO4-1 + 5e + 8H+1

Mn+2

2Cl-1 Cl2 + 2e

Basic (alkaline) situation:

MnO4-1 + 5e Mn+2 + 8OH-1

2Cl-1 Cl2 + 2e-

Step 4 Balance the number of hydrogens on each side of the equation with water.

Acidic situation:

MnO4-1 + 5e + 8H+1

Mn+2 + 4H2O

2 Cl-1 Cl2 + 2e-

Basic (alkaline) situation:

MnO4-1 + 5 e + 4H2O Mn+2 + 8 OH-1

2 Cl-1 Cl2 + 2e-

V. Electrochemistry

116

Step 5

Manipulate the equations so that the sum of the two will get the electrons to cancel out.

Acidic situation:

2[MnO4-1 + 5e + 8H+1

Mn+2 + 4H2O]

= 2MnO4-1 + 10e + 16H+1

2Mn+2 + 8H2O

5[2Cl-1 Cl2 + 2e] = 10Cl-1 5Cl2 + 10e

Final answer (acidic solution): 2MnO4-1 + 16H+1 + 10Cl-1

5Cl2 +2Mn+2 + 8H2O

Example 2 Al + SO4-2

+ H+1 SO2 + Al+3 + H2O

Example 3 Cr2O7-2 + SO3

-2 + H+1 Cr+3 + SO4-2 + H2O


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Exercises

1. Balance each of the following half-reactions. (half the work!) Assume acidic

conditions.

a. VO+2 V+3

b. H3PO3 H3PO4

c. MnO2 Mn+2

2. Balance each of the following using two balanced half-reactions.(Assume acidic

conditions.)

a. C2O4-2 + MnO4

-1 Mn+2 + CO2

b. MnO2 + H+1 + NO2-1 NO3

-1 + Mn+2 + H2O

c. Sn+2 + Cr2O7-2 Sn+4 + Cr+3

d. I2 + NO3-1 IO3

-1 + NO

e. MnO4-1 + NO2

-1 MnO2 + NO3-1

f. NiO2 + S2O3-2 Ni+2 + SO3

-2

g. NO2-1 + Al NH3 + AlO2

-1

3. Balance each of the following using two balanced half-reactions.(Assume basic

conditions.)

In class example: C2O4-2 + MnO4

-1 Mn+2 + CO2

a. MnO2 + NO2-1 NO3

-1 + Mn+2

b. Sn+2 + Cr2O7-2 Sn+4 + Cr+3

c. I2 + NO3-1 IO3

-1 + NO

d. NiO2 + S2O3-2 Ni(OH)2 + SO3

-2

e. Cr + CrO4-2 Cr(OH)3

V. Electrochemistry

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23. Oxidizing and Reducing Agents

A. Definition: An oxidizing agent is a substance which takes electrons away

from a substance that is being oxidized. Oxidizing agents are electron

prowlers.

Oxidizing agents contain an atom with a relatively high oxidation

number.

The atom in an oxidizing agent is always reduced.

Examples of substances that often (not always) act as oxidizing agents:

Common oxidizing agent Element with relatively high oxidation

number

H2O2 ( peroxide)

MnO4-1

O2

Br2

NO3-1

Ag+1

B. Reducing Agent: a substance that gives electrons to a substance that is being

reduced.

Reducing agents contain an atom with a relatively low oxidation number.

The atom in a reducing agent is always oxidized


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Examples of substances that often (not always)act as reducing agents:

Common reducing agent Element with relatively low oxidation

number

Li

Ca

Al

Zn

LiAlH4 (metal hydride)

H2

HSO3-1

Example 1 : Identify the oxidizing and reducing agents in the following:

a. Fe + Cu+2 Fe+2 + Cu

b. 2 Na + Cl2 2 NaCl

c. H2O2 + SO2 SO4-2 + H+1

d. Ga + I2 Ga+3 + I-1

V. Electrochemistry

120

Exercises

1. Identify the oxidizing and reducing agents in the following:

a. CH4 + O2 CO2 + H2O

b. C2O4-2 + MnO4

-1 Mn+2 + CO2

c. 4 Li + CO2 2 Li2O + C

d. 2 Na + Cl2 2 NaCl

e. MnO4-1 + H+1 + Cl-1 Cl2 + Mn+2 + H2O

2. a. Using the periodic table, find 3 good reducing agents.

b. Choose three excellent oxidizing agents.

3. In #1, (a), (b) and (e) are in desperate need of balancing. Balance them.

4. Balance MnO4-1 + Cl-1 Cl2 +Mn+2 in a basic medium.

5. Why is MnO4-1 a good oxidizing agent?

6. Why is H2O both a weak reducing agent and

a weak oxidizing agent?


121

24. The Electrochemical Cell

Observations:

1. In the above diagram, the voltmeter records a potential difference. If the voltmeter

is replaced with a galvanometer (very sensitive ammeter) it will record a small

current.

2. With time, the copper strip grows while the zinc strip gets smaller. Meanwhile the

blue colour of the copper solution fades.

3. For every mole of zinc that goes into solution, one mole of copper is deposited on

the original copper strip.

4. No current is produced without a salt bridge.

Explain each of the observations

V. Electrochemistry

122

Why a salt bridge?

Without the KNO3-salt bridge, excess positive charge(from Zn+2) would accumulate at

the Zn anode and too much negative charge would accumulate at the Cu cathode

(because Cu+2 is being consumed). Nitrate (NO3-1) from the salt bridge compensates for

the additional (+'s) formed by Zn, while K+1 moves into the Cu+2 beaker and replaces lost

(+'s).

Key Definitions:

anode An electrode (an electrical conducting rod or

strip) where oxidation occurs.

cathode An electrode where reduction occurs.

anion A negatively charged ion which is attracted to

the anode (where (+) charges are formed)

cation A positively charged ion which is attracted to

the cathode (where electrons are reducing the

(+) ions)

Example Label the anode1, cathode, anions and cations in the following diagram.

1 Note: the solution around the anode accumulates positive ions from the oxidation. For this reason, it

attracts negative ions (anions) from the salt bridge. But at the top of the anode, electrons are streaming out

(also form the oxidation). Consequently we assign the anode a negative charge.

Because reduction at the cathode consumes positive ions, in the solution around the cathode

negative ions accumulate. This is why cations (positively charged) are attracted there. But since the top of

the cathode is receiving electrons, the cathode is assigned a positive charge.

Cu(NO3)2(aq)

Ag I-1

K+

Cu

AgNO3(aq)

Cu(NO3)2(aq)


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Exercises

1. What is the difference between an anode and a cathode?

2. Why are cations attracted to the cathode?

3. a. Draw an electrochemical cell for:

Cr + Au+3 Cr+3 + Au

b. Label the anode and cathode.

c. Show the electron flow.

d. Write the half-reactions involved.

Examine the following voltaic cell. Use it to answer questions 4 and 5.

4. TRUE? Or FALSE?

a. The magnesium electrode is being oxidized.

b. The aluminum electrode is losing mass.

c. The aluminum electrode is the anode.

d. The anions migrate toward the Mg electrode.

5. a) What would happen to the voltage of the cell if the salt bridge was

removed?

b) What would happen to the current?

c) Would there be a reaction if you placed the Mg strip directly into the

beaker containing aluminum solution and solid aluminum? Explain.

6. What would happen to the voltage if the concentration of the salt bridge was

increased?

salt bridge

Al Mg

1.0 mol/L 1.0 mol/L

Al(NO 3 ) 3(aq) Mg(NO 3 ) 2(aq)

volts

Electrons

V. Electrochemistry

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25. Competition for Electrons and Standard Reduction Potentials

A. Competition Some substances are better than others at stealing electrons. This

makes them better oxidizing agents. Those that are better at losing electrons make

better reducing agents. To see who is "better", compare two or more reactions and

see which atom is getting its way by examining the direction of the reaction.

Example 1

So far we've seen the following reaction:

Zn + Cu+2 Zn+2 + Cu . (1)

An electrochemical cell can also be set up using

Cu + 2Ag+1 2 Ag + Cu+2 . (2)

Imagine each of the above as a competition to gain electrons. In other words, see

it as a competition between prowlers or oxidizing agents.

a. Which is the better oxidizing agent?

b. Which is the better reducing agent?

Example 2 Consider two more reactions which proceed as written:

Zn + 2 H+1 Zn+2 + H2 (3)

H2 + Cu+2 Cu + 2 H+1 (4)

Based on (3) and (4) extend our list of oxidizing agents from strongest to

weakest.

_______________________ in terms of oxidizing ability, and

_______________________ in terms of reducing ability.

B. Standard Reduction Potentials (E values) In order to make quantitative

predictions about whether equilibrium favours reactants or products, we use the

hydrogen half-cell as a reference:

2H+1 + 2e = H2 E = 0.00 V *2

*2 under standard conditions of 1.0 M solution for each electrode at a temperature of 25 oC and 101.3 kPa


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When we set the above reaction at 0V, then we measure the following of other

reactions we mentioned:

Ag+1 + 1e Ag E = 0.80 V

Cu+2 + 2e Cu E = 0.34 V

2H+1 + 2e = H2 E = 0.00 V

Zn+2 + 2e Zn E = -0.76 V

Note that the higher E value, the better the oxidizing agent. When we combine

two half-reactions, in order to have a real reaction, only one beast can do the stealing.

The other atom has to give up the electrons. So in order for the reaction to proceed

spontaneously, it must have an overall (+) E value (standard reduction potential).

Note: For a reverse reaction, E 's sign changes. However since the values represent

relative potential differences and not amounts of energy released, the value of E does not

change if we multiply the whole equation by a coefficient.

Example 1 A student sets up an electrochemical cell using Zn and Ag. Use standard

reduction potentials to predict what the anode will be.

Example 2 Repeat for Fe in a FeCl2 and Cr in Cr(NO3)2

V. Electrochemistry

126

Another Competition Example

In the laboratory, you are given different metallic strips and different solutions containing

metal ions.

You place the strips in the solutions and you observe the following:

V 3+

Q

V 3+

M

2. Reaction

X 2+

V

3. Reaction

Q 3+

X

4. No reaction 1.No reaction

Which of the above metals is the strongest reducing agent? The strongest oxidizing

agent?

Answer


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Exercises

1. Given:

Au+3 + 3 Fe+2 Au + 3 Fe+3

Fe+3 + Cu Fe+2 + Cu+1

a. Is Cu a stronger reducing agent than Fe+2?

b. Is Fe+2 a stronger reducing agent than Au?

c. What would be favoured, then, in the following equilibrium?

Au + 3 Cu+1 3 Cu + Au+3

d. What is the strongest oxidizing agent in (c)?

2. Given: 3 Cs + Al+3 3 Cs+1 + Al

Li + Cs+1 Li+1 + Cs

a. Of the above which metal has the strongest tendency to lose electrons?

b. Between Cs and Li, which is the stronger reducing agent?

c. Write an equation to represent the reaction between Li and Al+3.

d. Explain (c) in terms of competition for electrons.

3. The following data was obtained from an experiment in which five solid metallic

elements were placed into aqueous solutions of different metallic ions.

A(s) + D2+

(aq) reaction

C(s) + A2+

(aq) reaction

D(s) + B2+

(aq) no reaction

B(s) + E2+

(aq) no reaction

E(s) + C2+

(aq) no reaction

Arrange the metallic ions D2+, B2,+ E2+, and C2+ in decreasing order of their

tendency to be reduced.

V. Electrochemistry

128

4. In the laboratory, you are given different metallic strips and different solutions

containing metal ions.

You place the strips in the solutions and you observe the following.

Classify the 4 metals, Cr,

Mg, Al and Ni, in

descending order of

tendency to give up

electrons.

5. A scientist must store a 1 mol/L solution of Cr(NO3)3(aq) at room conditions.

Can the scientist use a copper container to store this solution? Explain.

6. During an oxidation reduction experiment, a student records the following

information:

e1 + Ag _+

(aq) Ag(s)

+ 0.80 V

e2 + Cu_+2

(aq) Cu(s) + 0.34 V

e2 + Pb_+2

(aq) Pb(s) - 0.13 V

e2 + Ni_+2

(aq) Ni(s) - 0.25 V

e2 + Zn_+2

(aq) Zn (s) - 0.76 V

Which pair of electrodes offers the greatest electric potential?

A)

Cu - Pb

C)

Ag - Cu

B)

Pb - Zn

D)

Ni - Zn

Al 3+

Cr

Al 3+

Mg

Reaction

Ni 2+

Al

Reaction

Cr 3+

Ni

No reactionNo reaction


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7. A chromium, Cr, rod is placed into a beaker of 1.0 mol/L chromium nitrate,

Cr(NO3)2, and a lead, Pb, rod is placed into a beaker of 1.0 mol/L lead nitrate,

Pb(NO3)2.

An electrochemical cell is then constructed and is illustrated below.

Use the table of reduction potentials (see next page) to answer the following:

1. What is the half-cell reaction at the anode?

2. Which electrode increases in mass?

3. What is the oxidizing agent of the cell?

4. What is the complete cell reaction?

5. What is the cell voltage?

V. Electrochemistry

130

Standard Reduction Potentials

ION CONCENTRATION of 1 mol/L at 25°C and 101.3 kPa.

Reduction Half-reaction Reduction Potential (V)

F2(g) + 2e 2F (aq) E° = + 2.87

Au3+(aq) + 3e Au(s) E° = + 1.50

Cl2(g) + 2e 2Cl (aq) E° = + 1.36

Br2(aq) + 2e 2Br-(aq) E° = + 1.09

Br2(l) + 2e 2Br-(aq) E° = + 1.07

Ag+(aq) + e Ag(s) E° = + 0.80

Hg 2+(aq) + 2e Hg(l) E° = + 0.78

Fe3+(aq) + e Fe2+(aq) E° = + 0.77

I2(s) + 2e 2I (aq) E° = + 0.53

Cu+(aq) + e Cu(s) E° = + 0.52

Cu2+(aq) + 2e Cu(s) E° = + 0.34

2H+(aq) + 2e H2(g) E° = + 0.00

Pb2+(aq) + 2e Pb(s) E° = - 0.13

Sn2+(aq) + 2e Sn(s) E° = - 0.14

Ni2+(aq) + 2e Ni(s) E° = - 0.26

Co2+(aq) + 2e Co(s) E° = - 0.28

Fe2+(aq) + 2e Fe(s) E° = - 0.44

Cr3+(aq) + 3e Cr(s) E° = - 0.74

Zn2+(aq) + 2e Zn(s) E° = - 0.76

Cr2+(aq) + 2e Cr(s) E° = - 0.91

Mn2+(aq) + 2e Mn(s) E° = - 1.18

Al3+(aq) + 3e Al(s) E° = - 1.66

Be2+(aq) + 2e Be(s) E° = - 1.85

Mg2+(aq) + 2e Mg(s) E° = - 2.37

Na+(aq) + e Na(s) E° = - 2.71

Ca2+(aq) + 2e Ca(s) E° = - 2.87

Sr2+(aq) + 2e Sr(s) E° = - 2.89

Ba2+(aq) + 2e Ba(s) E° = - 2.91

Cs+(aq) + e Cs(s) E° = - 2.92

K+(aq) + e K(s) E° = - 2.93

Rb+(aq) + e Rb(s) E° = - 2.98

Li+(aq) + e Li(s) E° = - 3.04


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26. Applications of Redox: Batteries and Corrosion Prevention

A. How an Alkaline Battery Works

Like the electrochemical cells we

studied, an alkaline battery has an

anode. At the anode, electrons are

lost, and these are the electrons

that leave the battery and that

power battery-operated gadgets.

The anode is made of zinc:

Zn + 2 OH-1 ZnO + H2O + 2e-

The electrons eventually return to

the battery, where they are

accepted at the cathode by a black

powder known as MnO2.

2 MnO2 +H2O +2e- Mn2O3 + 2 OH-

The anode and the cathode are kept apart by a plastic membrane that still allows hydroxide

ion and water to flow back and forth. The overall reaction has a positive potential; the reaction

is spontaneous, and it provides the driving force for the work that batteries do.

Example Given that dry cells use ammonium paste as a reducing agent and produce

hydrogen gas as an unwanted side-product (one that has to be oxidized), give

two main reasons why alkaline batteries last longer than dry cells.

B. How Corrosion Happens

Corrosion occurs when a water droplet facilitates the oxidation of iron:

Fe(s) Fe+2 (aq) + 2e At the anode a pit or hole develops because the solid iron

turns into aqueous iron. This is typical of all anodes: they lose mass.

At the cathode oxygen, with the help of water, collects the electrons and turns into

hydroxide:

Example: Write a half reaction for the conversion of oxygen to hydroxide.

Thanks to

battery-power

I can zoom

right out of

this class!!!

V. Electrochemistry

132

Still at the cathode the Fe+2 "swims" over from the anode, and eventually becomes Fe+3.

This ion combines with hydroxide and becomes Fe(OH)3 and eventually Fe2O3 = rust.

Example 1 Draw what has just been explained

Salt on the road accelerates corrosion because it acts as a salt bridge for the

electrochemical cell created by rusting, but also because FeCl3 (from NaCl) that's formed

is more soluble than Fe(OH)3. The consumption of Fe+3 drives the equilibrium (overall

reaction of two mentioned half-reactions) to the right, and forms more corroded iron.

Example 2 To prevent corrosion of your car:

What can you do to prevent....

1. ...dirt from trapping moisture, which triggers both half-reactions?

2. ...oxygen and water from reaching the metal?

3. ... metallic stress? When metal is stressed it loses electrons more easily.

4. ...electrolytes from accumulating on the surface in winter?

Also Make sure the manufacturer used Zn in the alloy. Zn sacrifices itself in the sense

that it is better than iron at losing electrons, so it will reverse the initial reaction of

corrosion:

Zn + Fe+2 Zn+2 + Fe


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Exercise

1. In the alkaline battery shown below, identify the following components:

a) reducing agent

b) oxidizing agent

c) anode

d) cathode

2. What in the battery

is being recycled?

3. Why don't electrons

go directly from the Zn to

MnO2 and not leave the

battery?

4. Write the half

reaction for zinc.

5. Look up the electrode reduction potential of zinc, and calculate the Eo for the

overall reaction, given:

2 MnO2 +H2O +2e Mn2O3 + 2 OH- E = 1.2 V

6. a) What is the overall reaction for the following:

2 MnO2(s) +H2O(l) +2e Mn2O3(s) + 2 OH - (aq)

Zn(s) + 2 OH-1(aq) ZnO(s) + H2O(l) + 2e

b) Use Le Chatelier's principle to predict the following effects on the above

overall equilibrium.

(1) adding MnO2

(2) adding base

(3) increasing pressure

7. Specifically what role do oxygen and water play in rusting?

8. Is it important to fix car dents immediately? Why?

9. Why doesn’t the rust accumulate at the site of oxidation?

LaurenHill Chemistry 534

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