lecture notes1 (4)

8/2/2019 Lecture Notes1 (4)

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NORTHERN CARIBBEAN UNIVERSITY

DEPARTMENT OF BIOLOGY and CHEMISTRY

LECTURE NOTES

CHEM122: GENERAL CHEMISTRY II

Assistant Professor: Dr. Nicole White

1. Electrochemistry

Before we can fully understand what happens in an electrochemical cell let us review redox

(reduction-oxidation) reactions. A redox reaction is one in which there is a transfer of electrons;

one species releases electrons and is oxidized while another picks up electrons and is reduced. The

oxidized species is known as the reducing agent and the reduced species is known as the oxidizing

agent. A simple way of remembering this relationship is the reducing agent is always oxidized and

the oxidizing agent is always reduced. The overall equation consists of 2 half reactions where 1

shows a reduction and the other an oxidation. It is the adding of these half reactions that gives the

final equation. It is important that the electrons cancel.

Recall

Oxidation : Loss of electrons (an increase in oxidation number)

Reduction: Gain of electrons (a decrease in oxidation number)

Eg. At anode Zno(s) Zn2+(aq) + 2e

- (oxidation)

At cathode Cu2+(aq) + 2e-

Cuo(s) (reduction)

Overall Reaction - Zno(s) + Cu2+

(aq) Zn2+(aq) + Cu

o(s)

There are instances, however, where the number of electrons in the half reactions are not equal and

so will not cancel out in the overall reaction. In these cases one must multiply the half reactions by

the appropriate factors to ensure that the final electron numbers are the same.


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Al(s) Al3+(aq) + 3e- and Br2(aq) + 2e- 2Br-(aq)

The above 2 equations differ in electron number and to rectify this first is multiplied by 2 and the

second by 3.

2Al(s) 2Al3+(aq) + 6e-

3Br2(aq) + 6e- 6Br-(aq)__________

Br2(aq) + 2Al(s) Br-(aq) +2Al3+(aq)

There are steps that are to be followed when balancing a redox reaction depending on whether the

reaction takes place in acidic or basic medium.

Acidic medium:

)()()()( 33

2

2

72 aqNOaqCraqHNOaqOCr+

++

1. Split the equation into its potential half reactions, grouping like elements together.

+

32

32

72

NOHNO

CrOCr

2. Balance all elements except hydrogen and oxygen.

+

32

32

72 2

NOHNO

CrOCr

3. Balance the oxygen atoms by adding H2O where oxygen is needed.

+

+

+

322

2

32

72 72

NOHNOOH

OHCrOCr

4. Balance the hydrogen atoms by adding H+ where hydrogen is needed.

+

++

++

++

HNOHNOOH

OHCrOCrH

3

7214

322

2

32

72

5. Balance the charge by adding electrons. The key is to add electrons to the side with the more

positive net oxidation number so that both sides will have the same oxidation number.

+

++

+++

+++

eHNOHNOOH

OHCrOCrHe

23

72146

322

2

32

72


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6. If the number of electrons is not equal in the half reactions then multiply by the appropriate

factors.

)23(3

72146

322

2

32

72

+

++

+++

+++

eHNOHNOOH

OHCrOCrHe

7. Add the two half reactions to get the overall equation, canceling like terms.

++

+

++

++++

+++

+++

32

32

722

322

2

32

72

34253

69333

72146

NOOHCrHOCrHNO

eHNOHNOOH

OHCrOCrHe

Basic medium:

For reactions done in basic media steps 1-6 are the same as those in acidic media, the differencecomes when the H+ ions must be replaced by OH- ions. Add the same numbers of OH- ions as there

are H+ ions to both sides of the reaction. This step can be done after addition of the half reactions

as well.

+

+

+

+

+++++

+++++

+++

+++

++

OHOHClOHeHClO

OHHCrOeOHOHOHCrstep

OHCleHClO

HCrOeOHOHCrstep

aqClaqCrOaqClOsOHCr

6336663

101026102)(2:7

33663

10262)(2:6

)()()()()(

2

2

423

2

2

423

2

43

The H+ and OH- ions together give the water molecule. Do the necessary cancellations and add the

resulting half equations to get the overall equation.

++++

+++

+++

++++

++++

2

423

2

2

2

43

22

2

2

423

25334)(2

63633

82610)(2

633663

1026102)(2

CrOOHClClOOHOHCr

OHCleOHClO

OHCrOeOHOHCr

OHOHCleOHClO

OHCrOeOHOHOHCr


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Electrochemistry, defined as the study of the relationship between electron flow and oxidation-

reduction reactions, is concerned with the interconversion of chemical and electrical energies.

There are two types of electrochemical cells: galvanic (voltaic) and electrolytic. We shall first

discuss the galvanic cell in which spontaneous reactions take place. A spontaneous reaction in

electrochemistry is one in which the electrons tend to flow with no external stimulus electric

current. The general set up of a galvanic cell includes a salt bridge, electrodes, 2 half-cells and an

external circuit which measures the flow of electricity driven by the flow of electrons.

The salt bridge allows the flow of ions but prevents the mixing of the different solutions

that would allow direct reaction of the cell reactants.

The half-cell is the portion of an electrochemical cell in which a half reaction takes place.

The electrodes are electrical conductors, and are the contact points of the external circuit

with the electrolytes of both half cells. An electrolyte is a substance that ionizes or

dissociates in water to form an electrically conducting solution. There are 2 types of

electrodes. The electrode at which reduction takes place is the cathode - positive electrode.

This is so because cations (Mn+) accumulate around this electrode to gain electrons. Cations

from the salt-bridge also migrate to the cathode cell. The anode is the electrode at which


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oxidation takes place (negative electrode). This is so because cations are lost from this

electrode leaving behind a build-up of electrons which flow through the external circuit to

the cathode. Anions through the salt-bridge also migrate to the anode cell. When drawing

an electrochemical cell the anode is drawn on the left of the diagram with the cathode on

the right.

Ions & Electrons Flow: electrons flow in the external circuit from anode to cathode. Ions

flow from the salt-bridge to each half cell, thus completing the electrical circuit.

Can you determine in which half-cell reduction takes place?

A redox reaction is always associated with an electric potential called e.m.f. (electromotive force)

also known as the cell potential (E) or voltage (V). Under standard conditions, 25 oC, 1 atm

pressure, 1 M solutions and pure solids and liquids, the cell potential is termed the standard cell

potential (Eo). The half-cell reaction with the more positive emf value represents the reduction cell.

The half reactions are written as reductions and so when the cell that reduction occurs has been

determined it means the other half reaction now has to show an oxidation which is accompanied by

a change in the sign of the half cell potential.

Zn2+(aq) + 2e- Zno(s) Eo = -0.76 V

Cu2+(aq) + 2e- Cuo(s) Eo = 0.34 V

Cu2+(aq) + 2e- Cuo(s) Eo = 0.34 V

Zn2+ + 2e- Zn Eo = + 0.76 V

The standard cell potential is the sum of the standard half-cell potentials for oxidation an reduction.

Eocell = Eoox + E

ored

Some books have it as Eocell = Eored Eoox, both are correct and will lead to the same answer. It is

important to note that for the former equation the sign of the oxidation reaction is reversed while

for the latter equation the signs do not change.

Try it.


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How is reduction potential of a half-cell determined?

Standard Hydrogen Electrode (SHE)

This is done by placing the half-cell in question in an electrochemical cell with the standard

hydrogen electrode (S.H.E.) also known as the normal electrode.


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The S.H.E. has a reduction potential of zero (0 V) at all temperatures and so any voltage reading is

as a result of the other half-cell. The voltage obtained indicates whether oxidation or reduction is

preferred. Hydrogen electrode is based on the redox half cell:

2H+(aq) + 2e- H2(g)

The metal is Cu

Eo/V

At anode (oxidation) : H2(g) 2H+

(aq) + 2e-

0.00

At cathode (reduction) : Cu2+(aq) + 2e- Cuo(s) 0.34

Overall cell reaction : Cu2+(aq) + H2(g) Cuo(s) + 2H+(aq) 0.34

Sum = 0.34V, therefore the Cu2+(aq) /Cu(s) half cell = 0.34 V

Why platinum?

The choice of platinum for the hydrogen electrode is due to several factors:
http://en.wikipedia.org/wiki/Half_cellhttp://en.wikipedia.org/wiki/Half_cell


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inertness of platinum (it does not corrode)

the capability of platinum to catalyze the reaction of proton reduction a high intrinsic exchange current density for proton reduction on platinum excellent reproducibility of the potential (bias of less than 10 V when two well-made

hydrogen electrodes are compared with one another.

The surface of platinum is platinized (i.e., covered with platinum black) because of:

necessity to employ electrode with large true surface area. The greater the electrode truearea, the faster electrode kinetics

necessity to use electrode material which can adsorb hydrogen at its interface. Platinization

improves electrode kinetics

When expressing an electrochemical cell without actually drawing the diagram it is useful to knowthe following shorthand notation. Using copper and zinc as the example,

Zn Zn2+ +Cu2 Cu

It is expressed as an oxidation followed by reduction similar to the flow of electrons where first we

have oxidation which releases electrons which flow to the cathode where reduction takes place.

The double bar, , represents the salt bridge and the single bar represents the different phases or

oxidation states in the half-cell.

The potentials used so far are determined under standard conditions but not every experiment is

carried out under those specifications and so a correction to the half cell potentials must be made

in the form of the Nernst equation.

[ ][ ]

[ ][ ]treacproduct

nEEor

treac

product

nF

RTEE oo

tanlog

0592.0

tanlog ==

R gas constant; F Faraday constant; T absolute temperature.
http://en.wikipedia.org/wiki/Platinum_blackhttp://en.wikipedia.org/wiki/Platinum_black


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Uses of galvanic cells

1. Lead storage batteries as used in automobiles. A battery consists of galvanic cells linked in

series. The voltage provided is the sum of the individual cell voltages.

Eg. Battery (Lithium)

At anode (oxidation) : Li(s) Li+ + e-

At Cathode (reduction): MnO2(s) + Li+ + e- LiMnO2(s)

Overall cell reaction : MnO2(s) + Li(s) + LiMnO2(s)

2. Dry cell batteries as used in smaller batteries such as AA or AAA used in the running of

smaller appliances such as your CD player. This is a galvanic cell with a pasty low-moisture e-

electrolyte. A wet cell on the other hand is a cell with a liquid electrolyte, such as the lead

batteries.

3. Fuel cells - A fuel cell is a galvanic cell in which one of the reactants is a traditional fuel such

as CH4(g) or H2(g). The reactants flow into the cell and the reaction products flow out of it while the

electrolyte remains within it. In this type of cell the reactants are consumed.

Fuel Cell (H2 O2)

At anode (oxidation) : 2H2(g) + 40H-(aq) 4H2O(i) + 4e-

At Cathode (reduction): O2(g) + 2H2O(l) + 4e- 40H-(aq)

Overall cell reaction : 2H2(g) + O2(g) 2H2O(l)

Burning of H2by O2 only produces heat + water i.e. environmentally most friendly fuel.

Electrolytic cell

In this electrochemical cell the reaction that takes place is non-spontaneous that is it needs an

external source of energy to drive it electric current. The basic set up for an electrolytic cell is as

shown below.


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Do you observe any differences between the electrolytic and galvanic cells?

The main difference is that in this type of cell the flow of electrons is the reverse of the galvanic

cell due to the addition of an electric current. As a result of the change in direction of electron flow

the role of the two electrodes are also reversed what was the cathode is a galvanic cell is now the

anode and vice versa. The sign of the reduction potential are also reversed compared to the

galvanic cell.

Cathode : Negative (-ve) electrode. Cations (Mn+) migrate to this electrode and become

reduced.

Anode : Positive (+ve) electrode. Anions (Xn-

) move to this electrode, give up electrons and

become oxidized.

Note that the cathode is connected to the negative terminal and the anode is connected to the

positive terminal of the external power supply. Also, both electrodes dip in the same electrolyte.

A term known as electrolysis is used to explain what is happening in electrolytic cells.


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Eletcrolysis the process of using an electric current to bring about a chemical change.

Uses of electrolytic cells

1. The electrolytic purification of copper (active electrode)

Purification of a metal by electrolysis is called ELECTROREFINING. Example, impure Cu

obtained from ores is converted to pure copper in an electrolytic cell.

Generally

Cu (impure) - anode

cu (pure) - cathode

Electrolyte - CuSO4 (aq)

At impure Cu anode : Cu is oxidized along with other metallic impurities eg. Zn and Fe. At the

pure Cu cathode, Cu2+ ions are reduced to Cu metal. Less easily reduced metal ions are Fe2+ and

Zn2+ remain in the electrolyte solution.

At anode (oxidation) : M(s) M2+(aq) + 2e- ( M = Cu, Zn, Fe)

At Cathode (reduction): Cu2+(aq) + 2e- Cuo(s)

Overall cell reaction : M(s) + Cu2+(aq) Cuo(s) + M2+(aq)

During electrolysis, the anode becomes smaller while the cathode gets larger. The net cell reaction

is simply a transfer of Cu metal form the impure anode to the pure cathode. The process takes

about 4 weeks, which gives ~ 99.5% pure copper.

2. Electroplating : The use of one metal to coat another metal. An electric current is used to

reduce cations of the desired metal from a solution and coat a conductive object with a thin layer.


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2. Thermochemistry

This is study of heat changes that is the quantity of heat absorbed or given off, during chemical

reactions. Energy is defined as the ability to do work or the capacity to move matter. Energy

cannot be created or destroyed but converted from one form to another LAW OF

CONSERVATION OF ENERGY. The internal energy of a substance is the sum of its kinetic and

potential energies (KE + PE). Kinetic energy is the energy a substance has due to motion =2

21 mv

whereas potential energy is the energy a substance has because of its position can also be termed

as stored energy.

E = q + w in Joules

The SI unit of energy is thejoule (J):1 J = 1 kg m2 s-2

Where q is heat and w is the work done.

w = -PV

Hence, E = q + (-PV)

At constant pressure

qp = E + PV = H (Joules)

Heat is the energy that flows into or out of a system because of a difference in temperature

between the system and its surroundings. Heat flows from a region of higher temperature to one of

lower temperature stopping only when the temperature of the two sides is equal.

Temperature, pressure, energy, and volume are state functions a function or property whose

value depends only on the present state of the system and not on the path used to arrive at the

condition. Path-independence means that no matter what path (sequence of changes in chemical

composition and other state variables) a system takes in going from state 1 to state 2, the change in


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value of the state function is the same. The overall change in a state function is zero when the

system returns to its original condition.

Work and heat are not state functions.

Ho the change in enthalpy can be used to determine the heat absorbed or evolved in a chemical

reaction. The sign on the enthalpy indicates the type of reaction, whether exothermic (heat is given

off) or endothermic (heat is absorbed).

2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g); H = -367 kJmol-1

What type of reaction is shown above?

The superscript o indicates that the measurement was done under standard conditions.

Heat capacity (C) and specific heat capacity (c)

The heat capacity of a sample of substance is the quantity of heat needed to raise the temperature

of the sample of substance 1 oC/ K.

=

qC


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The specific heat capacity is the quantity of heat required to raise the temperature of 1 g of a

substance by 1 oC/ K at constant pressure.

=

m

qc

Can you differentiate between the two?

The heat of a reaction can be measured by using a calorimeter which is used to measure the heat

change during a chemical and physical change.

Heats of reaction, Ho

Enthalpy of a reaction or energy change of a reaction H, is the amount of energy or heat involved

in a reaction. Reverse the signs for the reactants if this equation is used.

Ho = (Hproducts) - (Hreactants)

Or you can follow Hess law the overall enthalpy change for a reaction is equal to the sum of the

enthalpy changes for the individual steps in the reaction.

For example,

C2H5OH + 3 O2 2 CO2 + 3 H2O

Individual steps:

C2H5OH(l) 2C(graphite) + 3H2(l) + 21 O2(g) H = 228 kJ/mol

2[C(graphite) + 2O2(g) CO2(g)] H = -394 2 kJ/mol

3[H2(g) + 21 O2(g) H2O(l)] H= -286 3 kJ/ mol

Adding all three equations and energies leads to the following

C2H5OH(l) + 3 O2(g) 2 CO2(g) + 3 H2O (l) Ho = - 1 4 1 8 k J / m o l


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Or in the case where one or more of the individual steps must be reversed the sign for the enthalpy

change must also be reversed.

2C(s) + 3H2(g) C2H6(g)Data Equations:

1 C(s) + O2(g) CO2(g) H = -394kJ mol-1

2 H2(g) + O2(g) H2O(l) H = -286kJ mol-1

3 C2H6(g) + (7/2)O2(g) 2CO2(g) + 3H2O(l) H = -1542kJ mol-1

2x1 2C(s) + 2O2(g) 2CO2(g) H = 2(-394kJ mol-1)

3x2 3H2(g) (3/2)O2(g) 3H2O(l) H = 3(-286kJ mol-1

)Reverse 3 2CO2(g) + 3H2O(l) C2H6(g) (7/2)O2(g) H = 1542kJ mol

-1

Add: 2C(s) + 3H2(g) C2H6(g) Ho = -104kJ mol-1

Hof= -104 kJ mol-1

Standard enthalpy of formation, Hof

This is the enthalpy change when 1 mole of a compound is formed from its constituent elements in

their standard states (state at room temperature, 25C and at a pressure of 1 atmosphere). Thismeans a substance, written as the product of a chemical equation, is formed DIRECTLY from the

elements involved. The substance in question is always written with a coefficient of one.

The Hfo for any element in its standard state is zero (0 kJ/mol). It is important that you write the

state symbol for the reaction because different phases have varying enthalpy changes,

Here are some examples:

C (s) + O2 (g) CO2 (g)

C (s) + 21 O2 (g) CO (g)

H2 (g) + O2 (g) H2O2 (l)


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H2 (g) + CO2 (g) H2O (l)

C (s) + 2 H2 (g) + 21 O2 (g) CH3OH (l)

An example,

CH4(g) + 4Cl2(g) CCl4(l) + 4HCl(g); = ?

C(s) + 2H2(g) CH4 (g); = 74.9 kJ mol-1C(s) + 2Cl2(g) CCl4(l) ); = 139 kJ mol

-1

21 H2(g) + 21 Cl2(g) HCl(g) ; = 92.3 kJ mol

-1

Applying Hess law: CH4 (g) C(s) + 2H2(g); = 74.9

kJ mol

-1

C(s) + 2Cl2(g) CCl4(l) ); = 139 kJ mol-1

4[ 21 H2(g) + 21 Cl2(g) HCl(g)] ; = 4 92.3 kJ mol-11

424 433);(4)()(4)(

=++ kJmolgHCllCClgClgCH o

These steps can also be placed on an enthalpy diagram. An enthalpy diagram shows H for theinitial and final states of a process.

Steps:

1. Draw a line at level 0 to represent the elements.

2. Look up the values ofHof for each compound and enter these on the enthalpy diagrams, taking

account of the signs negative values are below zero, positive values above.

3. Find the difference in levels between the two compounds which represents the difference intheir enthalpies.

4. o is the difference in levels taking account of the direction of change. Up is positive anddown is negative.

For example,


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Bond enthalpy/ Bond dissociation energies

The Bond Enthalpy is the energy required to break a chemical bond. The standard molar enthalpy

change of bond dissociation refers to a specific bond in a specific molecule. Bond breaking is anendothermic process, and the bond enthalpy involved is given a +ve sign. Bond making is an

exothermic process.

Ho = Hbroken + Hformed

Add them and note the sign of the final enthalpy whether it is endothermic or exothermic. The

bond energies for all types of bonds is not known and so what is used is the average bond

dissociation energy and the resultant bond enthalpy is just an approximation.

For example,

CH4(g) + 2O2(g)CO2(g) + 2H2O(l)

Bonds Broken Bonds Formed

4(C-H)=4(+414)= + 1656 2(C=O)=2(-724)= - 14482(O=O)=2(+497)= + 994 4(O=H)=4(-458)= - 1832

= + 2650 = - 3280

The enthalpy of combustion of methane is -630kJ mol-1.


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As seen the sign on the value is representative of the reaction whether bond breaking or formation.

If you want you may use this equation in which the values do not adopt a negative sign and you

subtract the bonds formed from the bonds broken.

Ho = Hbroken - Hformed

Uses of thermochemistry

Thermchemistry reactions are used in everyday life the most common involves lighting a matchand driving a car. What happens is that we are burning fuel yielding heat as a product. The amount

of energy released on burning a substance is called its heat of combustion, Hoc, and is the

standard enthalpy change for the reaction of 1 mole of the substance with oxygen.

H2(g) + 21 O2(g) H2O(l); Ho

c = -285.8 kJ/mol

CH4(g) + 2O2(g) CO2(g) + 2H2O(l); Hoc = -890.3 kJ/mol

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