thermochemistry (4 lectures)...slide 25-1 highlights of last lecture introduction to...

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Highlights of last lectureIntroduction to electrochemistry..

CONCEPTS

Revision of oxidation No., and redox;

Half-reactions

Galvanic cell

Cell potential (voltage)

CALCULATIONS

none from last lecture

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Reduction potential tableHalf-reaction Half-cell

potential (V)Au3+(aq) + 3e

Au(s) +1.50Cl2

(g) + 2e

2Cl

(aq) +1.36O2

(g) + 4H+(aq) + 4e

2H2

O(l) +1.23Ag+(aq) + e

Ag(s) +0.80Cu2+(aq) + 2e

Cu(s) +0.342H+(aq) + 2e

H2

(g) 0.00*Sn2+(aq) + 2e

Sn(s) -0.14Fe2+(aq) + 2e

Fe(s) -0.44Zn2+(aq) + 2e

Zn(s) -0.762H2

O(l) + 2e

H2

(g) + 2OH(aq) -0.83Mg2+(aq) + 2e

Mg(s) -2.37* by definition

Strong oxidising

agent

Weak oxidising

agent

Weak reducing

agent

Strong reducing

agent

3

Active and inactive electrodesWe have seen voltaic (galvanic) cells where the electrodes themselves

are part of the chemical reaction (remember the Zn electrode in the Cu2+

solution?)These are called “active electrodes”.

There are many species for which there is no appropriate electrode material (remember the H2

electrode?). In these cases we use other inert materials, such as C(graphite), or Pt as the electrode.These are called “inactive electrodes”.The inactive electrodes conduct electrons in and/or out of the half-

cell, but do not take part in the reaction.

Electrodes (both active and inactive), are placed at the outside

of the shorthand notation for a voltaic cell. (Check back to last lecture at the shorthand involving Cu and Zn electrodes.)

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Active and inactive electrodesFor example:

Half-reactions:

2I-

I2

+ 2e-

MnO4-

+ 8H+

+ 5e-

Mn2+

+ 4H2

O(l)

10I-(aq) + 2MnO4-(aq) + 16H+(aq)

5I2

(s) + 2Mn2+(aq) + 8H2

O(l)

graphite |I2

(s)| I-(aq) || H+, MnO4-, Mn2+| graphite

Cell notation (note phase boundaries and electrodes):

Overall reaction*:

* Check that you can balance redox

reactions (see homework at end of lecture)

5

Standard Electrodes

For routine measurements, it is nice to have one half-reaction that we use all the time, and just change the other half-cell.

Several choices:

Normal calomel electrode, E 0

= 0.28 V

Saturated calomel electrode (SCE), E0

= 0.242 V

Silver/Silver chloride, E 0

= 0.22 V

Calomel: Hg2

Cl2

(s) + 2e

2Hg(l) + 2Cl(aq);

E0=0.24V

Zn

Zn2+

+ 2e- ; E0=0.76V

Q: The standard reduction potential of Zn2+/Zn is -0.76 V. What would be the observed cell potential for the Zn/Zn2+

couple when measured using the SCE as a reference?

Ans: The Zn will be oxidised (lower reduction potential), soE0(cell) = 0.76 + 0.24 = 1.00 V

So to get E 0

using the SCE as cathode, subtract 0.24 V from the reading.

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Let’s explore what happens when the concentration of the cell is changed…

[I-] (M) [I3-] (M) Eobs

(V) E0

(I3-,I-) (V)

0.2 0.050.2 0.005

0.05 0.005

graphite|I3-(aq), I(aq)||calomel

The effect of concentration

Calomel: Hg2

Cl2

(s) + 2e

2Hg(l) + 2Cl(aq); E0

= 0.24 VI3

-,I-: I3-(aq)

+ 2e-

3 I-(aq);

E 0

= +0.53 V

2Hg(l) + 2Cl(aq) + I3-(aq)

Hg2

Cl2

(s) + 3I-(aq); E0

= +0.29V

Add 0.24V to E(SCE) as anode get E0

7

Standard concentration

What happened in this experiment when I diluted the solutions?

Dilution Change in potential (V)

constant I,I3

diluted

0.05M

0.005 Mconstant I3

,I

diluted

0.2M

0.05 M

Conclusion: The cell potential depends on concentration.

Implication: We need to stipulate concentration if referring to cell potential.

Corollary: It is useful to define a standard concentration, which is 1 M

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Do these results make sense in terms of what you have learned about equilibrium?

The effect of concentration

I3-(aq)+ 2e-

3 I-(aq) ;

E 0

= +0.53 V

1.

Reduced [I-], reaction shifts to right, cell potential increases.

2.

Reduced [I3-], reaction shifts to left, cell

potential decreases.

Can we quantify this?…

9

The equation that fits the experimental measurements is:

][][log

3

30

IIAEEobs

nFRTA 303.2

where R = 8.314 J K-1

mol-1T = temperature (K)n = number of electrons transferred per mole of reagentF = charge on 1 mol of electrons = 96485 C = “Faraday constant”

][][log303.2

3

30

II

nFRTEEcell

The effect of concentrationI3

-(aq) + 2e-

3 I-(aq);

E 0

= +0.53 V

10

Nernst Equation

The ratio looks like something we have seen before: ][][

3

3

IIQ

)log(303.20 QnFRTEEcell “Nernst equation”

I3-(aq) + 2e-

3 I-(aq);

E 0

= +0.53 V

][][log303.2

3

30

II

nFRTEEcell

11

Nernst Equation

)log(303.20 QnFRTEEcell “Nernst equation”

At 25ºC,

)log(0592.00 Qn

EEcell V 0592.0303.2

F

RT

Walther Nernst derived this equation in 1889 at the age of 25. He also derived the 3rd

Law of Thermodynamics and explained the principle of the solubility product. He was the awarded the Nobel Prize in Chemistry in 1920.

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Use the Nernst Equation to calculate the expected cell potential for two similar experiments to the ones in the last lecture, i.e.

i) [Cu2+] = 1.0 M; [Zn2+] = 10-5M

ii) [Cu2+] = 10-5M; [Zn2+] = 1.0 M

Example calculation (1)

)log(0592.00 Qn

EEcell

Zn(s) + Cu2+(aq)

Zn2+(aq) + Cu(s)

Cu2+

+ 2e

Cu

Zn

Zn2+

+ 2e

Firstly, work out the value of n :n = 2

2 mol e-

transferred per mole of reaction

Zn(s) + Cu2+(aq)

Zn2+(aq) + Cu(s) E0

= 1.1 V

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[Zn2+] (M)

[Cu2+] (M)

Q log(Q) Ecell

(V)

1.0 1.010-5 1.01.0 10-5

Zn(s) + Cu2+(aq)

Zn2+(aq) + Cu(s)

)log(0592.00 Qn

EEcell

Example calculation (1)

(n=2)

1.0

10-5

105

0.0

-5.0

5.0

1.10

1.25

0.95

][][

2

2

CuZnQ

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Nernst Equation

)log(0592.00 Qn

EEcell where n is the number of moles of electrons transferred in the reaction (2 in our case above for Zn/Cu).

15

E 0 and K

Q: What do you think happens as the reaction proceeds until equilibrium is reached?

A: The reaction stops, therefore the voltage, or electrical potential, is zero (the battery is flat!)

In mathematical terms:

0)log(0592.00 Qn

EEobs

So the equilibrium constant determines the cell potential!

Large K

products favoured

large cell potential, E

)log(0592.00 Kn

E

At eq’m

Q=K

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Let’s examine the relationship between E 0 and K…

E 0 and K

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Example question (2)

Co(s) + Ni2+(aq)

Co2+(aq) + Ni(s) E 0

= 0.03 V

Q: A voltaic cell consisting of a Ni/Ni2+

half-cell and Co/Co2+

half-cell is constructed with the following

initial concentrations: [Ni2+] = 0.80M; [Co2+]=0.2M.

a) What is the initial Ecell

?b)

What is the [Ni2+] when the voltage reaches 0.025V?

c)

What are the equilibrium concentrations of the ions?Given: E 0

(Ni2+|Ni) = -0.25 V; E 0

(Co2+|Co) = -0.28 V

Ni2+

+ 2e-

Ni

E 0

= -0.25V

Co

Co2+

+ 2e-

E 0

= +0.28V

18


Co(s) + Ni2+(aq)

Co2+(aq) + Ni(s) E 0

= 0.03 V

25.08.02.0

][][

2

2

NiCoQ

)log(0592.00 Qn

EEcell

)25.0log(2

0592.003.0

)602.0(0296.003.0

V 048.0

a) What is the initial Ecell

?

19


)log(0592.00 Qn

EEcell )log(0296.003.0025.0 Q

169.00296.0005.0)log( Q

Q = 1.47

)8.0()2.0(

][][47.1 2

2

xx

NiCoQ

Co(s) + Ni2+(aq)

Co2+(aq) + Ni(s)0.80-x

0.20+x

)2.0()8.0(47.1 xx

xx 2.047.1176.1976.047.2 x

x = 0.40

So when Ecell

= 0.025 V

[Co2+] = 0.60 M

[Ni2+] = 0.40 M

b)

What is the [Ni2+] when the voltage reaches 0.025V?

20


)log(0592.00 Kn

E )log(0296.003.0 K

014.10296.0

03.0)log( K

K = 10.24Co(s) + Ni2+(aq)

Co2+(aq) + Ni(s)0.80-x

0.20+x

)8.0()2.0(

][][24.10 2

2

xx

NiCoK

)2.0()8.0(24.10 xx

xx 2.024.10192.8986.724.11 x

x = 0.71

So at equilibrium,

[Co2+] = 0.91 M

[Ni2+] = 0.09 M

c) What are the equilibrium concentrations of the ions?

21

Summary

CONCEPTS

Half-reaction

Table of standard reduction potential

Effect of concentration

Link between E and Q

CALCULATIONS

Work out cell potential from reduction potentials;

Work out cell potential for any concentration (Nernst equation)

22

The lecture should finish about here.

After this are some slides on balancing redox

equations for home

revision.

23

Balancing Redox

Reactions

Step 1: Determine the half-reactions.

Cr2

O72

Cr3+

I

I2

Step 2: Balance the atoms and charges in each half reaction:

Balance atoms other than O and H

Balance O by adding H2

O

Balance H by adding H+

Balance charge by adding e

Cr2

O72(aq) + I(aq)

Cr3+(aq) + I2

(s)

24


Balancing Redox

Reactions

Balance atoms other than O and HCr2

O72

2Cr3+

Cr2

O72

Cr3+

Reduction

Balance O by adding H2

OCr2

O72

2Cr3+

+ 7H2

O

Balance H by adding H+

14H+

+ Cr2

O72

2Cr3+

+ 7H2

O


6e

+ 14H+

+ Cr2

O72

2Cr3+

+ 7H2

O

25

Balancing Redox Reactions

I

I2


Balance atoms other than O and H

2I

I2


2I

I2

+ 2e Oxidation

26

Step 3: Multiply each half reaction by an integer so that the number of e

in each reaction are the

same.6e

+ 14H+

+ Cr2

O72

2Cr3+

+ 7H2

O(2I

I2

+ 2e)

3

Step 4: Add the balanced half reactions and include the states of matter.

6e

+ 14H+

+ Cr2

O72

2Cr3+

+ 7H2

O6I

3I2

+ 6e

____________________________________________________________________________________________________

6I(aq)

+ 14H+(aq)

+ Cr2

O72

(aq)

3I2(s)

+ 7H2

O(l)

+ 2Cr3+(aq)

Balancing Redox

Reactions

• Step 5: Check the atoms and charges are balanced.Reactants (6I, 14H, 2Cr, 7O; 6+)

Products (6I, 14H, 2Cr, 7O; 6+)

27

Balancing Redox

Reactions in Basic Solutions

Balance the following reaction under basic conditions:

MnO4(aq)

C2

O42

(aq)

MnO2(s)

+ CO32

(aq)

Initially balance the reaction as an acidic reactionC2

O42

CO3

2

Identify ½

EqnC2

O42

2CO3

2

Balance CC2

O42

+ 2H2

O 2CO32

Balance O

C2

O42

+ 2H2

O 2CO32

+ 4H+

Balance H

C2

O42

+ 2H2

O 2CO32

+ 4H+

+ 2e

Balance e

Loss of electrons = OXIDATION

MnO4

MnO2

Identify ½

EqnMnO4

MnO2

+ 2H2

O

Balance OMnO4

+ 4H+

MnO2

+ 2H2

O

Balance HMnO4

+ 4H+

+ 3e

MnO2

+ 2H2

O

Balance e

Gain of electrons = REDUCTION

STEPS

1 & 2

28

Balancing Redox


STEP 3

C2

O42

+ 2H2

O 2CO32

+ 4H+

+ 2e

3

MnO4

+ 4H+

+ 3e

MnO2

+ 2H2

O 2

3C2

O42

+ 6H2

O 6CO32

+ 12H+

+ 6e

2MnO4

+ 8H+

+ 6e

2MnO2

+ 4H2

O

4H+2H2

OSTEP 4

3C2

O42

+ 6H2

O 6CO32

+ 12H+

+ 6e

2MnO4

+ 8H+

+ 6e

2MnO2

+ 4H2

O

2MnO4

+ 2H2

O + 3C2

O42

2MnO2

+ 6CO32

+ 4H+

29

One additional step is necessary at Step 4.The acidic result would be:

2MnO4

+ 2H2

O + 3C2

O42

2MnO2

+ 6CO32

+ 4H+

Add OH

to each side to cancel the H+, therefore add 4OH

2MnO4

+ 2H2

O + 3C2

O42

+ 4OH

2MnO2

+ 6CO32

+ (4H+

+ 4OH)

2MnO4

+ 2H2

O + 3C2

O42

+ 4OH

2MnO2

+ 6CO32

+ 4H2

O

2MnO4(aq)

+ 3C2

O42

(aq)

+ 4OH(aq)

2MnO2(s)

+ 6CO32

(aq)

+ 2H2

O(l)

Balancing Redox


2H2 O

thermochemistry (4 lectures)...slide 25-1 highlights of last lecture introduction to...

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