Unit 5 Electrochemistry S 2

VERSION: April 20, 1999

Electrochemistry

A. Introduction

1. Electrochemistry is _______________________________________________________

________________________________________________________________________

________________________________________________________________________

Electrochemical reactions involve the _________________________________________

________________________________________________________________________

(analogous to proton transfer in acids and bases).

2. Consider the reaction

2Al(s) + 3CuCl2(aq) → AlCl3(aq) + 3Cu(s)

the net ionic equation is

This reaction only involves the ______________________________________________

We can re-write this equation as two separate _________________________

OXIDATION HALF-REACTION

REDUCTION HALF-REACTION

__________ of electrons is ____________________

__________ of electrons is ____________________

2 Chemistry 12

When a substance becomes oxidized _________________________________________

________________________________________________________________________

________________________________________________________________________

Zn → Zn2+ + 2e-

U3+ → U4+ + e-

2Cl- → Cl2 + 2e-

When a substance becomes reduced __________________________________________

________________________________________________________________________

________________________________________________________________________

Cu2+ + 2e- → Cu

V3+ + e- → V2+

F2 + 2e- → 2F-

3. For every reduction reaction there ____________________________________________

________________________________________________________________________

________________________________________________________________________

Reactions involving the loss and gain of electrons are called REDUCTION-

OXIDATION reactions or _______________ reactions.

In the reaction, 2Al + 3Cu2+ → 2Al3+ + 3Cu, Al is referred to as the _______________

________________________________________________________________________

________________________________________________________________________

while Cu2+ is referred to as the ______________________________________________

________________________________________________________________________

________________________________________________________________________

Unit 5 Electrochemistry (S).doc 3

The ____________________ _______________ is _______________ during a reaction.

The ____________________ _______________ is _______________ during a reaction.

B. Oxidation Numbers

1. An oxidation number is ____________________________________________________

________________________________________________________________________

________________________________________________________________________

2. Rules for determining oxidation number:

a) __________________________________________________________________

b) __________________________________________________________________

__________________________________________________________________

i) Li+, Na+, K+, and all other group 1 ions have _______________________

____________________________________________________________

ii) Ca2+, Ba2+, Mg2+, and all other group 2 ions have ___________________

____________________________________________________________

iii) F-, Cl-, Br-, I- (halogens) are usually ______________________________

___________________________________________________________ .

c) The oxidation number of _____________________________________________

In peroxides, H2O2, it is 1-.

d) The oxidation number for ____________________________________________

except in metallic hydrides such as NaH or BaH2 where it is 1-.

e) All other oxidation numbers are _______________________________________

__________________________________________________________________

__________________________________________________________________

4 Chemistry 12

EXAMPLE 5.1 DETERMINING OXIDATION NUMBERS

Problem: Determine the oxidation numbers of each atom for the following:

Na, CCl4, N2O4, PO43-, PbSO4

Solution: Na = _____

CCl4 = 0 = C + 4(_____)

⇒ Cl = _____, C = _____

N2O4 = 0 = 2N + 4(_____)

⇒ O = _____, N = _____

PO43- = 3- = P + 4(_____)

⇒ O = _____, P = _____

PbSO4 = 0 = Pb2+ + SO42-

⇒ Pb = _____

SO42- = 2- = S + 4(_____)

⇒ O = _____, S = _____

3. An atom’s change in oxidation number indicates whether it is oxidized or reduced. When

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

ClO3- → ClO4

- (_____ → _____ ; _______________)

H2O2 → H2O (_____ → _____ ; _______________)

Cr3+ → CrO42- (_____ → _____ ; _______________)

NO2 → N2O3 (_____ → _____ ; _______________)


OXIDATION = __________________________________________________________

REDUCTION = _________________________________________________________

C. Predicting Spontaneous Reactions

1. An excerpt from the Table of Reduction Potentials is shown below.

F2 + 2e- →← 2F -

Ag+ + e- →← Ag

Cu2+ +2e- →← Cu

Zn2+ + 2e- →← Zn

Li + + e- →← Li

INCREASING TENDENCY TO REDUCE

INCREASING STRENGTH

AS AN OXIDIZING AGENT

INCREASING TENDENCY TO OXIDIZE

INCREASING STRENGTH

AS A REDUCING AGENT

The following observation can be made from the Table of Reduction Potentials:

i) In general, __________________________________________________

____________________________________________________________

____________________________________________________________

ii) In general, __________________________________________________

____________________________________________________________

____________________________________________________________

iii) Some metals such as Fe, Sn, Cr, Hg, and Cu ________________________

____________________________________________________________

As a result, __________________________________________________

____________________________________________________________

____________________________________________________________

Remember this point because it means that _________________________

6 Chemistry 12

____________________________________________________________

e.g. Cu+ + e- →← Cu(s)

Cu2+ + 2e- →← Cu(s)

Cu2+ +e- →← Cu+

The Cu+ ion is found on both left and right sides. Similarly, Sn2+ and

Fe2+ are also found on both sides of the table.

iv) ____________________________________________________________

____________________________________________________________

____________________________________________________________

OXIDIZING AGENTS ___________________________________________________

________________________________________________________________________

REDUCING AGENTS ___________________________________________________

________________________________________________________________________

2. The table of Standard Reduction Potentials is used in a similar way to the table of

Relative Strengths of Acids. The species at the _______________ __________ have a

tendency to go _______________ while the species at the _______________

__________ have a tendency to go _______________. Each reaction in the table can go

________________________________________________________________________

The half-reaction for Zn and Zn2+ is

If there is a piece of Zn(s) in a solution of Zn2+, you can write either:

Zn2+ + 2e- → Zn (written as a _______________)

or Zn → Zn2+ + 2e- (written as an _______________)


Note: When referring to an isolated half-reaction, use equilibrium arrows to show that

the reaction can go forward or backward.

Ag+ + e- →← Ag

If the half-reaction is made to undergo either reduction or oxidation as a result of

being part of a redox reaction, the use a one-way arrow.

Ag+ + e- → Ag

3. In a redox reaction, _______________________________________________________

________________________________________________________________________

________________________________________________________________________

Assume that you have Zn in a solution of Zn2+ and Cu in a solution of Cu2+, the two

possible half reactions are

Cu2+ + 2e- →← Cu

and Zn2+ + 2e- →← Zn

Of the two oxidizing agents, _____ and _____, the _____ is _______________ on the

left side of the table and has a greater tendency to become _______________. The

reduction reaction will be

Cu2+ + 2e- → Cu

8 Chemistry 12

Of the two reducing agents, _____ and _____, the _____ is _______________ on the

right side of the table and has a greater tendency to become _______________. The

oxidation reaction will be

Zn2+ + 2e- ← Zn

or re-written as

Zn → Zn2+ + 2e-

The overall redox reaction which will spontaneously occur is found by adding the two

half-reactions. Cu2+ + 2e- → Cu

Zn → Zn2+ + 2e-

When two half-reactions are joined, ________________________________________

________________________________________________________________________

________________________________________________________________________

4. If you are _____________________________________________, rather than the four

necessary for two complete half-reactions as above, you __________________________

________________________________________________________________________

a) If ________________________________________________________________

__________________________________________________________________

(i.e., both are oxidizing agents (left) or both reducing agents (right)) then

__________________________________________________________________


Assume the only reactants are Zn and Cu

Cu2+ + 2e- ← Cu

Zn2+ + 2e- ← Zn

Both Cu and Zn can only undergo

_______________, there is nothing present to

undergo _______________. A complete redox

reaction is not possible. ___________________

Assume the only reactants are Zn2+ and Cu2+

Cu2+ + 2e- → Cu

Zn2+ + 2e- → Zn

Both Cu2+ and Zn2+ can only undergo

_______________, there is nothing present to

undergo _______________. A complete redox

reaction is not possible. ___________________

b) If __________________________________________________ (oxidizing

agent) and __________________________________________________

(reducing agent) there are two possible cases.

i) The ____________________ _______________ is _______________ on

the table than the reducing agent.

Assume the reactants are Cu2+ and Zn. Cu2+ + 2e- → Cu

Zn2+ + 2e- ← Zn

The overall redox reaction is obtained by reversing the oxidation reaction

and adding the two half-reactions. Cu2+ + 2e- → Cu

Zn → Zn2+ + 2e-

This is a ____________________ _______________

10 Chemistry 12

ii) The ____________________ _______________ is _______________ on

the table than the reducing agent.

Assume the reactants are Zn2+ and Cu. Cu2+ + 2e- → Cu

Zn2+ + 2e- ← Zn

In this case, _____ ____________________ occurs.

Notice that this is the reverse reaction to (i). Since the reaction

Cu2+ + Zn → Cu + Zn2+

is ____________________ , the reverse reaction is

_________________________

A reaction will be _________________________________________________________

________________________________________________________________________

(reactant to be reduced – left) which is _________________________________________

______________________________________________ (reactant to be oxidized – right).


EXAMPLE 5.2 PREDICTING SPONTANEOUS REACTIONS

Problem: Predict whether or not a reaction will occur when the following are mixed.

a) Cl2 and Br-

b) Sn and Mn

c) Ni2+ and Pb

Solution: a) Cl2 and Br-

Reaction is _________________________and the overall reaction is

b) Sn and Mn

Both reactants are ____________________________________________

___________________________________________________________

c) Ni2+ and Pb

___________________________________________________________

___________________________________________________________

12 Chemistry 12

5. In _____________________________________________________________________

________________________________________________________________________

When H+ is present, _______________________________________________________

For example, if you asked whether the reaction of SO42- in a solution of Na2SO4 will

reduce:

SO42- + 4H+ + 2e- → H2SO3 + 2H2O …… +0.17

your answer should be _____________________________________________________

________________________________________________________________________

________________________________________________________________________

Note: H+ is necessary in many reduction half-reactions; however, there is only ONE

reaction is which H+ is the only substance involved in the reduction half-reaction:

2H+ + 2e- → H2(g) …… 0.00


D. Balancing Half Reactions

1. A half-reaction ___________________________________________________________

When balancing half-reactions ______________________________________________

________________________________________________________________________

The steps to balancing a half-reaction are as follows:

Typically you will be starting with a “skeleton equation” containing the major atoms

involved.

a) Balance ___________________________________________________________

__________________________________________________________________

b) Balance ___________________________________________________________

__________________________________________________________________

c) Assume that solutions are acidic and ____________________________________

__________________________________________________________________

d) Balance ___________________________________________________________

__________________________________________________________________

14 Chemistry 12

EXAMPLE 5.3 BALANCING HALF-REACTIONS IN ACIDIC SOLUTION

Problem: Balance the following half-reaction:

Cr2O72- → Cr3+

Solution: Step 1: Balance ATOMS.

Cr2O72- → ____Cr3+

Step 2: Balance OXYGEN.

Cr2O72- → 2Cr3+ + ________

Step 3: Balance HYDROGEN.

________ + Cr2O72- → 2Cr3+ + 7H2O

Step 4: Balance CHARGE.

________ + 14H+ + Cr2O72- → 2Cr3+ + 7H2O

Note: Electrons are negatively charged, ___________________________

______________________________________________________

2. Sometimes it will be necessary to balance half-reactions in BASIC solutions. For these

problems, _______________________________________________________________

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________


EXAMPLE 5.3 BALANCING HALF-REACTIONS IN BASIC SOLUTION


MnO4- → MnO2

Solution: Step 1: Balance ATOMS.

MnO4- → MnO2

Atoms already balanced.

Step 2: Balance OXYGEN.

MnO4- → MnO2 + ________

Step 3: Balance HYDROGEN.

________ + MnO4- → MnO2 + 2H2O

Step 4: Balance CHARGE.

________+ 4H+ + MnO4- → MnO2 + 2H2O

Step 5: Add OH- and cancel out H2O.

________+ 3e- + 4H+ + MnO4- → MnO2 + 2H2O + ________

3e- + ________ + MnO4- → MnO2 + 2H2O + ________

3e- + ________ + MnO4- → MnO2 + ________

16 Chemistry 12

E. Balancing Redox Equations Using Half-Reactions

1. An overall redox equation can be obtained by __________________________________

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

EXAMPLE 5.4 BALANCING A REDOX REACTION BY THE HALF-REACTION METHOD


ClO4- + I2 → Cl

- + IO3

-

Solution: Step 1: Break equation into two half-reactions.

________________________

________________________

Step 2: Balance each half-reaction.

________________________________________

________________________________________

Step 3: Lowest common multiple between electrons.

In this case LCM is ________, multiply reduction by ____ and oxidation by ____.

5 x (8e- + 8H+ + ClO4- → Cl

- + 4H2O)

4 x (6H2O + I2 → 2IO3- + 12H+ + 10e-)

____e- + ____H+ + ____ClO4- → ____Cl

- + ____H2O

____H2O + ____I2 → ____IO3- + ____H+ + ____e-


Step 4: Add two half-reactions, cancelling out electrons and common particles.

40e- + 40H+ + 5ClO4- → 5Cl

- + 20H2O

24H2O + 4I2 → 8IO3- + 48H+ + 40e-

______________________________________

Balanced redox equations ___________________________________________________

________________________________________________________________________

________________________________________________________________________

2. Once again, in basic solutions the final equation can be converted by ________________

________________________________________________________________________

________________________________________________________________________

5ClO4- + 4I2 + 4H2O → 5Cl- + 8IO3

- + 8H+

For basic solution, add 8OH- to both sides of equation

____ + 5ClO4- + 4I2 + 4H2O → 5Cl- + 8IO3

- + 8H+ + ____

____ + 5ClO4- + 4I2 + 4H2O → 5Cl- + 8IO3

- + ____

Cancel out water

8OH- + 5ClO4- + 4I2 + 4H2O → 5Cl- + 8IO3

- + 8H2O

8OH- + 5ClO4- + 4I2 → 5Cl- + 8IO3

- + ________

18 Chemistry 12

3. In some redox reactions, ___________________________________________________

________________________________________________________________________

________________________________________________________________________

Such a reaction is called a ________________________

ClO2- → ClO3

- + Cl-

The two half-reactions are

ClO2- → ClO3

-

ClO2- → Cl-

Balanced redox equation is

3ClO2- → 2ClO3

- + Cl-

F. Balancing Redox Equations Using Oxidation Numbers

1. This method involves ______________________________________________________

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

Remember that an increase in oxidation number is oxidation and a decrease in oxidation

number is reduction. In this method __________________________________________

________________________________________________________________________

________________________________________________________________________


EXAMPLE 5.5 BALANCING A REDOX REACTION BY OXIDATION NUMBERS


ClO4- + I2 → Cl

- + IO3

-

Solution: Step 1: Balance atoms.

ClO4- + I2 → Cl

- + 2IO3

-

Step 2: Assign oxidation numbers to all atoms involved in a change in oxidation number (∆ON = “change in oxidation number”).

ClO4- + I 2 ⎯⎯→ Cl

- + 2IO 3

-

Step 3: Balance change in oxidation number

In this case LCM is ____, multiply decrease by ____ and increase by ____.

____ClO4- + ____I2 ⎯⎯→ ____Cl

- + ____IO3

-

Step 4: Balance OXYGENS by adding H2O and HYDROGENS by adding H+. DO NOT ADD ELECTRONS.

________ + 5ClO4- + 4I2 ⎯⎯→ 5Cl

- + 8IO3

- + ________

Remember, ______________________________________________________________

________________________________________________________________________

________________________________________________________________________

20 Chemistry 12

SAMPLE

PROBLEM 5.1 BALANCING A REDOX REACTION BY OXIDATION NUMBERS

Problem: Balance the following redox reactions by the oxidation number method:

a) MnO4- + Fe2+ → Mn2+ + Fe3+ (acidic)

b) S2- + ClO3- → Cl- + S (basic)

c) CN- + IO3- → I- + CNO- (basic)

d) H2O2 + Cr3+ → CrO42- + H2O (acidic)

Solution:

MnO4- + Fe2+ → Mn2+ + Fe3+

S2- + ClO3- → Cl - + S

CN - + IO3- → I - + CNO -

H2O2 + Cr3+ → CrO42- + H2O


G. Redox Titrations

1. Acid-base titrations are useful for determining the concentration of an unknown sample

of acid or base. In a similar manner, __________________________________________

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

2. OXIDIZING AGENTS

One of the most common oxidizing agents used in redox titrations is ________________

The half-reaction:

MnO4- + 8H+ + 5e- → Mn2+ + 4H2O ……… E° = 1.49 V

has such a strong tendency to reduce that it will be able to oxidize a large number of

other substances. (The K+ in KMnO4 is left out because it is a spectator ion.)

In acid-base titrations, an ____________________ which changes colour is added to help

identify the ____________________ _______________ of the titration. An indicator is

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

If MnO4- (oxidizing agent) in a burette is added to a solution containing Fe2+ (reducing

agent), the MnO4- is converted to Mn2+ (colourless). ____________________________

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

22 Chemistry 12

EXAMPLE 5.6 REDOX TITRATION INVOLVING THE PERMANGANATE ION

Problem: A 100.0 mL sample containing FeCl2 is titrated with 0.100 M KMnO4 solution. If 27.45 mL of KMnO4 solution were required to reach the end-point, what was the [FeCl2]?

Solution: First we need the balanced redox equation.

Fe2+

+ MnO4- → Fe

3+ + Mn

2+

___ Fe2+

+ MnO4- + _______ → ___Fe

3+ + Mn

2+ + _______

The remainder of the calculation is similar to acid-base titration calculations.

Calculate MOLES of MnO4-

0.100 M MnO4- x

27.45 mL1000 mL = _______________ mol MnO4

-

Use MOLE RATIO from balance equation.

0.002745 mol MnO4- x

5 mol Fe2+

1 mol MnO4- = _______________ mol Fe

2+

Determine CONCENTRATION.

0.0137 mol Fe2+

0.100 L = _______________M Fe2+

and since Fe2+ and FeCl2 are in a 1:1 ratio, [FeCl2] = _______________.


3. REDUCING AGENTS

A commonly used reducing agent is __________ or __________. A large number of

substances can oxidize I- to I2 according to the half-reaction

2I- → I2 + 2e-

Titrations involving I- generally involve two consecutive steps:

a) first, _____________________________________________________________

__________________________________________________________________

b) second, ___________________________________________________________

__________________________________________________________________

__________________________________________________________________

An example of a reaction involving I- is the reduction of laundry bleach, NaOCl. The

reaction between I- and OCl- proceeds according to:

2I- → I2 + 2e-

2e- + 2H+ + OCl- → Cl- + H2O ___________________________________

No attempt is made to add exactly enough I- to react with the OCl-. Rather, __________

________________________________________________________________________

The excess I- does not affect the results; however, _______________________________

________________________________________________________________________

The above reaction between OCl- and I- is the “initial” reaction because the actual redox

titration involves a second reaction between the I2 produced and the reducing agent

sodium thiosulphate, Na2S2O3.

I2 + 2e- → 2I-

2S2O32- → S4O6

2- + 2e- ___________________________

24 Chemistry 12

When the addition of S2O32- has reacted most of the I2 present, the brown colour of I2

almost disappears (pale yellow). Some starch solution is then added to the titration,

producing a dark blue colour (from the reaction of starch with the remaining I2 in

solution). After adding the starch, ___________________________________________

________________________________________________________________________

________________________________________________________________________


EXAMPLE 5.7 REDOX TITRATION INVOLVING THE IODIDE ION

Problem: A 25.00 mL sample of bleach is reacted with excess KI according to the equation:

2H+ + OCl- + 2I- → Cl- + H2O + I2

The I2 produced requires exactly 46.84 mL of 0.7500 M Na2S2O3 to bring the titration to the endpoint according to the equation:

2S2O32- + I2 → S4O6

2- + 2I-

using starch solution as an indicator. What is the [OCl-] in the bleach?

Note: The TWO reactions are linked together. The I2 produced in the first reaction is used up in the second

2H+ + OCl- + 2I- → Cl- + H2O + I2

2S2O32- + I2 → S4O6

2- + 2I-

Solution: First, calculate the moles of S2O32-

moles S2O32- = __________ M x

46.84 mL1000 mL/L = _______________ mol S2O3

2-

Next, calculate the moles of I2

moles I2 = __________ mol S2O32- x

I22S2O3

2- = _______________ mol I2

Now moles of OCl - can be calculated

moles OCl - = __________ mol I2 x OCl -

I2 = __________ mol OCl -

Finally, the [OCl -] can be calculated

[OCl -] = 0.01757 mol0.02500 L = _______________

26 Chemistry 12

H. Electrochemical Cells

1. An _________________________ __________ (galvanic or voltaic cell) is a chemical

system in which __________________________________________________________

________________________________________________________________________

Consider the reaction

Cu2+ + Zn → Cu + Zn2+

A spontaneous reaction will occur when zinc metal is placed into a solution of CuSO4;

however, ________________________________________________________________

________________________________________________________________________

This same reaction can used to produce electricity if _____________________________

________________________________________________________________________

2. An electrochemical cell:

a) In the electrochemical cell above, we have Zn(s), Zn(NO3)2(aq), Cu(s), and

CuSO4(aq). The half-reactions are:

Cu2+ + 2e- →← Cu …… + 0.34

Zn2+ + 2e- →← Zn …… - 0.76


Since Cu2+ is a ______________________________________ than Zn2+, Cu2+

will become ____________________ and since Zn is a _____________________

than Cu, Zn will become ____________________. The reduction and oxidation

half-reactions that occur will be →

→ __________________________

Cu2+ + Zn → Cu + Zn2+

28 Chemistry 12

b) In the copper cell, ___________________________________________________

__________________________________________________________________

__________________________________________________________________

__________________________________________________________________

__________________________________________________________________

At the copper electrode, copper ions, Cu2+, ______________________________

__________________________________________________________________

The copper electrode is ______________________________________________

__________________________________________________________________

__________________________________________________________________

Cu2+ + 2e- → Cu

In the zinc cell, _____________________________________________________

___________________________________. At the zinc electrode, zinc metal ___

The zinc electrode is ________________________________________________

__________________________________________________________________

Zn → Zn2+ + 2e-


_______________ is the electrode where ____________________ occurs.

_______________ is the electrode where ____________________ occurs.

c) The two electrodes are connected by a __________________________________

_________________________. Electrons produced at the zinc electrode flow

through the wire to the copper electrode where they reduce copper ions. The

electrons will ______________________________________________________

__________________________________________________________________

__________________________________________________________________

__________________________________________________________________

A __________ _______________ composed of a concentrated solution of a

strong electrolyte, usually KNO3 or KCl, connects the two cells to complete the

circuit. As the two half-reactions occur, _________________________________

__________________________________________________________________

__________________________________________________________________

Water and certain ions, Zn2+ and SO42-, will pass through the salt bridge from

one cell to the other. The positive Zn2+ ions are attracted to the CATHODE so

they are called _______________ while the negative SO42- ions are attracted to

the ANODE so they are called _______________.

__________________________________________________________________

__________________________________________________________________

30 Chemistry 12

Note: There are NO electrons flowing in the solution, only ions. ____________

____________________________________________________________

The salt bridge allow __________________________________________

____________________________________________________________

Since oxidation occurs at the anode, ______________________________

and since reduction occurs at the cathode, the _______________________

__________ as the electrochemical cell operates.


EXAMPLE 5.8 DRAWING AN ELECTROCHEMICAL CELL

Problem: Assume two half-cells consisting of Pb(s) in a Pb(NO3)2 solution and Zn(s) in a ZnCl2 solution are connected to make an electrochemical cell. Draw and label the parts of the cell, write the equations for the individual half-reactions and overall reaction, and indicate the directions in which the ions and electrons move.

Solution: The two half-reactions are:

Pb2+ + 2e- →← Pb …… - 0.13

Zn2+ + 2e- →← Zn …… - 0.76

Since Pb2+ is the stronger oxidizing reagent, it will become reduced and Zn(s) will become oxidized.

The two half-reactions are

Pb2+ + 2e- → Pb

Zn → Zn2+ + 2e- __________________________

Pb2+

+ Zn → Pb + Zn2+

32 Chemistry 12

I. Standard Reduction Potentials

1. The tendency of electrons to flow in an electrochemical cell is called the

____________________, or ____________________ ____________________to do

work.

________________________________________________________________________

________________________________________________________________________

Since electrons cannot flow in an isolated half-cell, ______________________________

________________________________________________________________________

However, the ____________________________________________________________

________________________________________________________________________

A ______________________________________________________________________

________________________________________________________________________

The voltage for the HYDROGEN HALF-CELL is defined to be

2H+ + 2e- → H2 …… E° = __________

Where E° is the _____________________________________________, in volts and the

“ ° ” in E° means _________________________ ____________________

An electrochemical cell is said to be at STANDARD STATE if

• it is at __________, and

• all gases are at _________________________, and

• all elements are in their standard states (normal phases at 25 °C), and

• all solutions involved in the cell have a ______________________________


2. All voltages listed in the table of STANDARD REDUCTION POTENTIALS are

determined at standard state and are _________________________________________

________________________________________________________________________

Cu2+ + 2e- →← Cu …… E° = + 0.34

This half-cell has a voltage which is 0.34 V ____________________ than that of the

hydrogen half-cell.

Zn2+ + 2e- →← Zn …… E° = - 0.76

This half-cell has a voltage which is 0.76 V __________ than that of the hydrogen half-

cell.

Since the voltage is a measure of the work done, reversing a reduction reaction such as

Zn2+ + 2e- → Zn …… E° = - 0.76

produces ________________________________________________________________

Zn → Zn2+ + 2e- …… E° = + 0.76

________________________________________________________________________

________________________________________________________________________

3. When two half-reactions are combined, _______________________________________

________________________________________________________________________

________________________________________________________________________

Hg2+ + 2e- →← Hg …… E° = + 0.85 V

Cu2+ + 2e- →← Cu …… E° = + 0.34 V

34 Chemistry 12

The voltage for the overall reaction is

→ …… E° = __________ V

→ …… E° = __________ V _______________________________________________

→ …… E°CELL = __________ V

The potential of an electrochemical cell is ____________________________________

________________________________________________________________________

________________________________________________________________________

EXAMPLE 5.9 CALCULATING CELL POTENTIALS

Problem: Calculate the potential of the cell: Ni2+ + Fe → Ni + Fe2+.

Solution: The half-reactions involved are:

Ni 2+ + 2e- →← Ni …… E° = - 0.26 V

Fe2+ + 2e- →← Fe …… E° = - 0.45 V


→ …… E° = __________V

→ …… E° = __________ V _________________________________________ → …… E°CELL = __________

Note: This reaction is SPONTANEOUS since Ni2+ is higher than Fe and the reaction has a POSITIVE VOLTAGE.



Problem: Calculate the potential of the cell: Ni + Fe2+ → Ni 2++ Fe.


Ni 2+ + 2e- →← Ni …… E° = - 0.26 V

Fe2+ + 2e- →← Fe …… E° = - 0.45 V


→ …… E° = __________ V

Fe2+ + 2e- → Fe …… E° = __________ V _______________________________________________

→ …… E°CELL = __________

Note: This reaction is NOT SPONTANEOUS since Fe2+ is lower than Ni and the reaction has a NEGATIVE VOLTAGE.


Problem: Calculate the potential of the cell: 3Ag+ + Al → 3Ag + Al3+.


Ag+ + e- →← Ag …… E° = + 0.80 V

Al3+ + 3e- →← Al …… E° = - 1.66 V


→ …… E° = __________ V

→ …… E° = __________ V _______________________________________________

→ …… E°CELL = __________

Note: Although the Ag+/Ag half-reaction is multiplied by 3 to balance the electrons, the voltage is NOT multiplied by 3.

36 Chemistry 12

If E° is _______________ for a redox reaction, the reaction is expected to be

____________________

If E° is ____________________ for a redox reaction, the reaction is

_________________________

Note: Although E° can be used to predict if a reaction is spontaneous, it ___________

__________________________________________________________________

__________________________________________________________________

Remember the ____________________ _______________ of a reaction

determines the rate.

In some texts, the reduction of neutral water is given as

2H+ (10-7 M) + 2e- →← H2(g) …… E = - 0.41 V

Other texts show this same reaction as

2H2O + 2e- →← H2(g) + 2OH- (10-7 M) …… E = - 0.41 V

If a redox reaction occurs in acidic water, the reduction of H+ (at 0.00 V) may

occur and if a redox reaction occurs in a neutral solution, the reduction of neutral

water (at - 0.41 V) may be possible.

4. The SURFACE AREA of an electrode _______________________________________

________________________________________________________________________

This is because ___________________________________________________________

________________________________________________________________________

Therefore, increasing the surface area of a solid electrode has no effect on the

concentration of the solid. As a result there is no equilibrium shift and no change in half-

cell potential.


Increasing the surface are of an electrode does however, __________________________

____________________. The number of electrons transferred per second increases.

This increases the ____________________ of the cell but not the voltage. Increasing the

surface area of the electrode, also _____________________________________________

________________________________________________________________________

5. When a half-reaction is NOT at standard state, it is ______________________________

________________________________________________________________________

Consider the following half-reaction

Cu2+ + 2e- →← Cu …… E° = + 0.34 V

If the [Cu2+] is _______________ than 1.0 M the equilibrium shifts __________ and the

reduction potential ____________________.

Cu2+ + 2e- →← Cu …… E > + 0.34 V

If the [Cu2+] is __________ than 1.0 M the equilibrium shifts __________ and the

reduction potential ____________________.

Cu2+ + 2e- →← Cu …… E < + 0.34 V

Notice that the “ ° ” is omitted from the E° symbol since it is not at standard state.

6. Operating electrochemical cells ______________________________________________

Consider the cell

2Ag+ + Cu → 2Ag+ Cu2+

THE REDUCTION REACTION: 2Ag+ + 2e- → 2Ag

As the [Ag+] decreases, ____________________________________________________

________________________________________________________________________

As a result, the ___________________________________________________________

________________________________________________________________________

38 Chemistry 12

THE OXIDATION REACTION: Cu2+ + 2e- ← Cu

As the [Cu2+] increases, ____________________________________________________

________________________________________________________________________

(the reduction potential of Cu2+ + 2e- → Cu increases) Therefore, the ______________

________________________________________________________________________

________________________________________________________________________

Overall, the following occurs as the cell goes to equilibrium.

2Ag+ + 2e- → 2Ag

Cu2+ + 2e- ← Cu

Voltage 2Ag+ + 2e- → 2Ag

Cu2+ + 2e- ← Cu

INITIALLY AS TIME PASSES

FINALLY

E° = + 0.46 V

E = 0.00 V

CELL

CELL

________________________________________________________________________

________________________________________________________________________


J. Selecting Preferred Reactions

1. When a cell contains a mixture of substances, __________________________________

________________________________________________________________________

Consider the following cell:

Zn AgCu

Zn

NO

AgCu

3

2++

2+-

NO3-

The possible half-reactions are:

_____________________________________________

_____________________________________________

_____________________________________________

When several different reduction half-reactions can occur, _____________________

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

Therefore, the __________ will reduce and form __________

40 Chemistry 12

When several different oxidation half-reactions can occur, _____________________

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

Therefore, the __________ will oxidize to form __________

2. Determining the preferred half-reactions:

• First, _____________________________________________________________

__________________________________________________________________

• Next, starting at the _________________________________________________

__________________________________________________________________

__________________________________________________________________

• Then, starting at the _________________________________________________

__________________________________________________________________

__________________________________________________________________


EXAMPLE 5.12 SELECTING PREFERRED REACTIONS

Problem: An iron strip is placed in a mixture of Br2(aq) and I2(aq). What is the preferred reaction which occurs?

Solution: The species present and the possible reactions are

Br2 + 2e- →← 2Br- …… E° = + 1.09 V

I2 + 2e- →← 2I- …… E° = + 0.54 V

Fe2+ + 2e- →← Fe …… E° = - 0.45 V

The preferred reduction involves __________ and the only oxidation possible involves __________. The overall reaction is

→

→

___________________________

→

Note: Any ion capable of being reduced ______________________________________

__________________________________________________________________

__________________________________________________________________

__________________________________________________________________

Similarily, any ion capable of being ____________________________________

__________________________________________________________________

__________________________________________________________________

__________________________________________________________________

The following ions are generally considered to be spectator ions:

Na+, K+, Ca2+, Mg2+, SO42- (in neutral solution), and Cl-

42 Chemistry 12

K. Corrosion of Metals

1. The term CORROSION simply _____________________________________________

The most common form of corrosion _________________________________________

When a drop of water rests on an iron surface, a spontaneous reaction occurs.

Fe

H O

cathode region (oxygen-rich)

Precipitate of Fe(OH)

anode region (oxygen-poor)

2

2

At the __________________________________________________________________

________________________________________________________________________

Fe(s) → Fe2+ + 2e-

After being formed, the Fe2+ ions ____________________________________________

________________________________________________________________________

________________________________________________________________________

As the Fe2+ ions move away, they expose fresh Fe(s) for further oxidation. In the

meantime, the reaction 12 O2 + H2O + 2e- → 2OH-

is proceeding at the oxygen-rich outer surface of the drop. When the Fe2+ meets the OH-

, it _____________________________________________________________________

The Fe(OH)2 is eventually __________________________________________________

________________________________________________________________________


“Rust” is just Fe2O3•XH2O, where “X” is variable. The variability of “X” explains the

different colours rust can have (red, brown, yellow, black) since differing numbers of

water molecules attached to the Fe2O3 will change the colour of the component.

A metal _________________________________________________________________

________________________________________________________________________

________________________________________________________________________

For example, if iron touches some copper wire and the place where they touch gets wet,

then:

Fe(s) → Fe2+ + 2e- (Fe has a greater tendency to oxidize than Cu)

The copper ______________________________________________________________

________________________________________________________________________

________________________________________________________________________ 12 O2 + H2O + 2e- → 2OH-

2. PREVENTING CORROSION

A. Isolating The Metal From The Environment

i) Apply ______________________________________________________

____________________________________________________________

If oxygen and water cannot reach the metal, it will not corrode.

ii) Apply ______________________________________________________

____________________________________________________________

Tin can be applied to the surface of steel cans. The tin is quickly oxidized

to produce a thin layer of tin oxide which protects the underlying metal

from further corrosion.

44 Chemistry 12

B. Electrochemical Methods

i) Cathodic Protection

Recall that __________________________________________________

____________________________________________________________

____________________________________________________________

____________________________________________________________

This implies that we can prevent the corrosion of iron ________________

____________________________________________________________

____________________________________________________________

The term CATHODIC PROTECTION is applied to the process of ____

____________________________________________________________

____________________________________________________________

____________________________________________________________

For example, both ____________________ and __________ are stronger

reducing agents than iron, if pieces of magnesium or zinc are attached to

the surface of iron, the magnesium and zinc will ____________________

____________________________________________________________

Since the substance with the greater tendency to oxidize, Mg, acts as the

_______________, the Fe acts as a _______________ and remains in its

reduced form.

e.g. Zinc blocks are bolted to iron-hulled ships to prevent them from rusting.

Some modern ships pass a low voltage electric current into the hull from

an electrical generator. This forces electrons into the metal and prevents it

from becoming oxidized.

Galvanized iron involves coating zinc onto the surface of iron sheets.


Some buried gas and oil tanks made from steel have thick braided wire

connected to them. The wire is attached to posts made of magnesium or

zinc.

ii) Change the Condition of the Surroundings

When iron is placed in contact with water containing oxygen, the half-

reaction 12 O2 + H2O + 2e- → 2OH-

oxidizes the iron. If _______________ is removed from the solution, the

reduction tendency drops drastically.

In addition to removing the oxygen, ______________________________

____________________________________________________________

____________________________________________________________

L. Electrolysis

1. ELECTROLYSIS is ______________________________________________________

________________________________________________________________________

________________________________________________________________________

An ____________________ __________ or ELECTROLYSIS CELL is an apparatus

in which electrolysis can occur.

____________________ __________ involve ____________________ redox reactions

that require energy to occur ____________________

____________________ __________ involve ____________________ redox reactions

that release energy ____________________

46 Chemistry 12

2. ELECTROLYSIS OF A MOLTEN BINARY SALT

A _______________ __________ is made up of only two different elements, e.g. NaCl,

KBr, MgI2, AlF3, etc. When a salt such as NaCl is melted, the ions are free to move

(Note: molten NaCl is NaCl(l), there are only Na+ and Cl- ions present. Do not confuse

with NaCl(aq) which is a solution and contains Na+, Cl-, and H2O).

– +

Na Cl+

–

direct current (DC) power supply or battery.

(the symbol is sometimes used)

There is ________________________________________________________________

________________________________________________________________________

The only reactants present are Na+ and Cl-

Cl2 + 2e- → 2Cl-

Na+ + e- → Na

The _______________ reaction is 2Cl- → Cl2 + 2e- E° = - 1.36 V

The _______________ reaction is Na+ + e- → Na E° = - 2.71 V

The overall reaction is 2Na+ + 2Cl- → 2Na + Cl2 E°cell = - 4.07 V

Note: This is a _________________________ reaction since the oxidizing agent is

below the reducing agent and since the voltage is negative.

To make this cell operate, ___________________________________________________


Since the half-cells are not at standard state, the reduction potentials will be different

from those listed on the table.

3. ELECTROLYSIS OF AN AQUEOUS SOLUTION

Consider the electrolysis of aqueous sodium iodide, NaI(aq). This __________________

________________________________________________________________________

Once again, _______________ electrodes are used and the cell is set up as follows,

– +

D.C.

Note: During the electrolysis of aqueous solutions, _____________________________

__________________________________________________________________

There are two possible oxidations: 12 O2(g) + 2H+ (10-7 M) + 2e- ←

I2 + 2e- ←

E� = + 0.82 V

E� = + 0.54 V

The oxidation of __________________________________________________________

________________________________________________________________________

48 Chemistry 12

There are two possible reductions:

+ 2e- → H2(g) + 2OH- (10-7 M)

+ e- → Na

E� = - 0.41 V

E� = - 2.71 V

The reduction of __________________________________________________________

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

The half-reactions and the overall reaction involving the electrolysis of NaI(aq) are

2H2O + 2e- → H2(g) + 2OH- (10-7 M) E� = - 0.41 V

2I- → I2 + 2e- E� = - 0.54 V

E� = - 0.95 V

so at least + 0.95 V must be applied to the cell.


Note: The concentrations of the materials in cells will not be of concern; as long as

there is sufficient material in the cell you can assume the reactions proceed as

predicted.

In NEUTRAL aqueous solutions, there are two “water equations” which must be

considered.

OXIDATION: 12 O2(g) + 2H+ (10-7 M) + 2e- ← H2O E� = + 0.82 V

REDUCTION: 2H2O + 2e- → H2(g) + 2OH- (10-7 M) E� = - 0.41 V

In ACIDIC solutions there are also two “water equations” which must be

considered.

OXIDATION: 12 O2(g) + 2H+ + 2e- ← H2O E� = + 1.23 V

REDUCTION: 2H+ + 2e- → H2(g) E� = 0.00 V

4. In practice, it is found that a higher potential than calculated must be applied to cause

electrolysis. The causes are _________________________________________________

________________________________________________________________________

The difference between actual potentials required for electrolysis and the calculated

potential is termed ___________________________________. Overpotentials vary for

each half-cell but the difference in potentials is _________________________________

________________________________________________________________________

As a result of the overpotential effect, when dilute neutral solutions (< 1.0 M) containing

_____ or _____ are electrolyzed

50 Chemistry 12

OXIDATION:

Cl2 + 2e- ← 2Cl - E� = + 1.36 V

12 O2(g) + 2H+ (10-7 M) + 2e- ← H2O E� = + 0.82 V

Br2 + 2e- ← 2Br - E� = + 1.09 VStrength of Reducing

Agent

we would expect that _____ will be produced. In practice however, we find that _____

or _____ are actually produced.

In addition when dilute solutions of ____________________ are electrolyzed

REDUCTION: 2H2O + 2e- → H2(g) + 2OH- (10-7 M) E� = - 0.41 V

Fe2+ + 2e- → Fe E� = - 0.45 V

Cr3+ + 3e- → Cr E� = - 0.74 V

Zn2+ + 2e- → Zn E� = - 0.76 V

Strongest Oxidizing

Agent

we would expect that _____ is produced but in practice, ____________________ will be

produced.

Electrolysis of aqueous solutions containing Cl- or Br- will produce Cl2 or Br2 at the

anode.

Electrolysis of aqueous solutions containing Fe2+, Cr3+, or Zn2+ will produce Fe, Cr,

or Zn at the cathode.


EXAMPLE 5.13 ELECTROLYSIS REACTIONS

Problem: What products are formed at the anode and cathode wand what is the overall reaction when a solution containing NiSO4(aq) is electrolyzed using inert electrodes? Determine the minimum voltage required.

Solution: First, the species present ______________________________ and the possible reactions are

REDUCTION

+ 4H+ + 4e- → H2SO3 + H2O E� = +0.17 V

+ 2e- → Ni E� = - 0.26 V

+ 2e- → H2 + 2OH- (10-7 M) E� = -0.41 V

OXIDATION

12 O2(g) + 2H+ (10-7 M) + 2e- ← E� = +0.82 V

______________________________________________________________

______________________________________________________________

so the half -reactions that occur are:

Anode: H2O → 12 O2(g) + 2H+ (10-7 M) + 2e- E� =

Cathode: Ni2+ + 2e- → Ni E� = - 0.26 V

H2O + Ni2+ → 12 O2(g) + 2H+ (10-7 M) + Ni E�CELL =

_____ is produced at the anode, _____ is produced at the cathode and a

voltage of at least __________ must be applied for the reaction to occur.

52 Chemistry 12

EXAMPLE 5.14 ELECTROLYSIS REACTIONS

Problem: What is the overall reaction which occurs when a 1 M solution of HCl(aq) is electrolyzed using carbon electrodes?

Solution: First, the species present H+, Cl -, H2O (acidic) and the possible reactions are

REDUCTION

+ 2e- → H2 E� = +0.00 V

+ 2e- → H2 + 2OH- (10-7 M) E� = -0.41 V

OXIDATION

Cl2(g) + 2e ← E� = +1.36 V

12 O2(g) + 2H+ (10-7 M) + 2e- ← E� = +0.82 V

-

Due to the high overpotential of H2O, the Cl- is oxidized so the two half reactions are

H+ + 2e- → H2 E� = + 0.00 V

2Cl- → Cl2(g) + 2e- E� =

H+ + 2Cl- → H2 + Cl2 E�CELL =

Anode

Cathode

_____ is produced at the anode, _____ is produced at the cathode and a

voltage of at least __________ must be applied for the reaction to occur.


5. ELECTROPLATING

ELECTROPLATING is __________________________________________________

________________________________________________________________________

________________________________________________________________________

• The ____________________is made out of the material which will receive the metal

plating.

• The _________________________ ____________________ contains ions of the

metal which is to be “plated” onto the cathode.

• The _______________ may be made of the same metal which is to be “plated out”

onto the cathode. (This is normal but an inert electrode can also be used.)

54 Chemistry 12

EXAMPLE 5.15 ELECTROPLATING CELLS

Problem: Design a cell to electroplate a copper medallion with nickel metal. Include in the design: the ions present in the solution, the direction of ion flow, the substance used for the anode and cathode, and the direction of electron flow when the cell is connected to a DC power source.

Solution: Since nickel metal must be produce, the reduction reaction is:

The cell is setup as follows:

DC Power Source

• The medallion must have the Ni(s) plated onto it, so it must be the _______________.

• Make the _______________ out of Ni(s) so that the oxidation reaction

Ni(s) → Ni2+ + 2e-

will ensure a _________________________________________________

___________________________________________________________

• Since electrons always flow from ______________________________, connect DC power in such a way as to supply electrons to the cathode.

• The ions needed in solution will Ni2+ and some anion that does NOT form a precipitate with Ni2+. The NO3

- is always a good choice since all nitrates are soluble.

• As with spontaneous reactions, __________________________________

___________________________________________________________

___________________________________________________________


6. ELECTROREFINING

ELECTROREFINING is _________________________________________________

________________________________________________________________________

________________________________________________________________________

DC Power Source

At the anode, the small amounts of Zn or Pb present is preferentially oxidized as it is

exposed at the surface. When any exposed Pb/Zn atoms have oxidized and gone into

solution as ions, only the Cu atoms are available to be oxidized. Any Au, Ag, or Pt

atoms present cannot be oxidized because the anode is mostly copper which is oxidized

in preference to Au, Ag, or Pt which simply drop off and accumulate on the bottom of the

cell. This “anode sludge” can be purified to obtain the valuable metals.

Unit 5 Electrochemistry S 2

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Transcript of Unit 5 Electrochemistry S 2