Prentice Hall © 2003Chapter 20 For the SHE, we assign 2H + (aq, 1M) + 2e - H 2 (g, 1 atm) E red = 0

Prentice Hall © 2003 Chapter 20

• For the SHE, we assign

2H+(aq, 1M) + 2e- H2(g, 1 atm)

• Ered = 0.

anodecathode redredcell EEE


• For Zn:

Ecell = Ered(cathode) - Ered(anode)

0.76 V = 0 V - Ered(anode).

• Therefore, Ered(anode) = -0.76 V.

• Standard reduction potentials must be written as reduction reactions:

Zn2+(aq) + 2e- Zn(s), Ered = -0.76 V.


• Changing the stoichiometric coefficient does not affect Ered.

• Therefore,

2Zn2+(aq) + 4e- 2Zn(s), Ered = -0.76 V.


• Reactions with Ered < 0 are spontaneous oxidations relative to the SHE.

• The larger the difference between Ered values, the larger Ecell.


Oxidizing and Reducing Agents

• The more positive Ered the stronger the oxidizing agent on the left.

• The more negative Ered the stronger the reducing agent on the right.



• More generally, for any electrochemical process

processoxidation processreduction redredcell EEE


Example: For the following cell, what is the cellreaction and Eo

cell?

Al3+(aq) + 3e- → Al(s); EoAl = -1.66 V

Fe2+(aq) + 2e- → Fe(s); EoFe = -0.41 V

Al(s)|Al3+(aq)||Fe2+(aq)|Fe(s)


EMF and Free-Energy Change• We can show that

G is the change in free-energy, n is the number of moles of electrons transferred, F is Faraday’s constant, and E is the emf of the cell.

• We define

• Since n and F are positive, if G > 0 then E < 0.

Spontaneity of Redox Spontaneity of Redox ReactionsReactions

nFEG

J/V·mol 96,500C/mol 500,961 F


The Nernst Equation

QRTGG ln

QRTnFEnFE ln


• At a temperature of 298 K:

QnFRT

EE ln

Qn

EE log0592.0

This gives:


Cell EMF and Chemical Equilibrium• A system is at equilibrium when G = 0.

• E = 0 V and Q = Keq:

0592.0log

log0592.0

0

nEK

Kn

E

eq

eq


Al3+(aq) + 3e- → Al(s); EoAl = -1.66 V

Fe2+(aq) + 2e- → Fe(s); EoFe = -0.41 V

Al(s)|Al3+(aq)||Fe2+(aq)|Fe(s)

In the problem presented before, calculate ΔGo

and Keq


Example: For the reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

Calculate the Ecell given that Eocell = +1.10 V, [Cu2+] = 5.0 M

and [Zn2+] = 0.050 M


Exercise: Calculate the emf generated by the cell that employs the following cell reaction: 2Al(s) + 3I2(s) → 2Al3+(aq) + 6I-(aq)

Take Eocell to be +2.20 V; [Al3+] = 4.0 x 10-3M and [I-] = 0.010M

Ans: +2.36 V


• Battery- self-contained electrochemical power source with one or more voltaic cell.

• When the cells are connected in series, greater emfs can be achieved.

• Primary cell – nonrechargeable• Secondary cell-rechargeable

BatteriesBatteries


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How Hydrogen Fuels Cells Work

• Lead-acid battery: Pb and PbO2 are electrodes and are immersed in sulphuric acid. Electrodes are separated by glass fibres or wood.

• Alkaline battery: anode is Zn in contact with a concentrated solution of KOH. The cathode is a mixture of MnO2(s) and graphite separated from the anode by a porous fabric

http://inventors.about.com/od/sstartinventions/ss/Physics_Illustr.htm

http://z.about.com/d/inventors/1/0/R/5/fuelcell.gif


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• Nickel-cadmium, Ni-metal hydride and Li-ion batteries




Fuel cells

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Fuel cells differ from batteries in that they are not self-contained systems.They use conventional fuels e.g. H2 and CH4 to produce electricity.




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The most common fuel cell is the one involving the reaction of H2(g) and O2(g) to produce H2O(l).This is based on that if H2O can be split by electricity, then combining H2 and O2 should produce water and electricity.




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Corrosion of Iron

• Since Ered(Fe2+) < Ered(O2) iron can be oxidized by oxygen.

• Cathode: O2(g) + 4H+(aq) + 4e- 2H2O(l).

• Anode: Fe(s) Fe2+(aq) + 2e-.• Dissolved oxygen in water usually causes the oxidation

of iron.• Fe2+ initially formed can be further oxidized to Fe3+

which forms rust, Fe2O3.xH2O(s).

CorrosionCorrosion


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Galvanised iron, i.e. Fe coated with Zn uses the principle of electrochemistry to protect the iron from corrosion even after the surface coat is broken.The standard red. Pot. for Fe and Zn are:

Fe2+(aq) + 2e- → Fe(s) Eored = -0.44 V

Zn2+(aq) + 2e- → Zn(s) Eored = -0.76 V


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Protecting a metal from corrosion by making it the cathode in an electrochemical cell is called cathodic protection.The metal that is oxidised while protecting the cathode is called the sacrificial anode.


Electrolysis of Aqueous Solutions• Nonspontaneous reactions require an external current in

order to force the reaction to proceed.• Electrolysis reactions are nonspontaneous.• In voltaic and electrolytic cells:

– reduction occurs at the cathode, and

– oxidation occurs at the anode.

– However, in electrolytic cells, electrons are forced to flow from the anode to cathode.


– In electrolytic cells the anode is positive and the cathode is negative. (In galvanic cells the anode is negative and the cathode is positive.)


• Example, decomposition of molten NaCl.• Cathode: 2Na+(l) + 2e- 2Na(l)

• Anode: 2Cl-(l) Cl2(g) + 2e-.

• Industrially, electrolysis is used to produce metals like Al.


Electroplating• Active electrodes: electrodes that take part in electrolysis.• Example: electrolytic plating.



Electroplating• Consider an active Ni electrode and another metallic

electrode placed in an aqueous solution of NiSO4:

• Anode: Ni(s) Ni2+(aq) + 2e-

• Cathode: Ni2+(aq) + 2e- Ni(s).• Ni plates on the inert electrode.• Electroplating is important in protecting objects from

corrosion.


Quantitative Aspects of Electrolysis• We want to know how much material we obtain with

electrolysis.• Consider the reduction of Cu2+ to Cu.

– Cu2+(aq) + 2e- Cu(s).

– 2 mol of electrons will plate 1 mol of Cu.

– The charge of 1 mol of electrons is 96,500 C (1 F).

– Since Q = It, the amount of Cu can be calculated from the current (I) and time (t) taken to plate.


Example: When an aqueous solution of CuSO4 is electrolysed, Cu metal is deposited:

Cu2+(aq) + 2e- → Cu(s)

If a constant current was passed for 5.00 h and 404 mg of Cu metal was deposited, what was the current?

Ans: 6.81 x 10-2 A


Example: The half reaction for formation of Mg metal upon electrolysis of molten MgCl2 is:

Mg2+(aq) + 2e- → Mg(s)

a) Calculate the mass of Mg formed upon passage of a current of 60.0A for a period of 4.00 x 103s

b) How many seconds would be required to produce 50.0g of Mg from MgCl2 if the current is 100.0A?

Ans: a) 30.2g, b) 3.97 x 103s


Electrical Work

∆G is a measure of maximum useful work, wmax, that can be extracted from the process: ∆ G = wmax = -nFE

For an electrolytic cell, an external source of energy is used to bring about a non-spontaneous electrochemical process. In this case, w = nFEext


Electrical work can be expressed in energy units of watts x time: 1 W = 1 J/s

Usually, the kWh is used

Example: Calculate the no. of kWh of electricity required to produce 1.00 kg of Mg from electrolysis of molten MgCl2 if the applied emf is 5.00 V

Ans: 11.0 kWh

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Prentice Hall © 2003Chapter 20 For the SHE, we assign 2H + (aq, 1M) + 2e - H 2 (g, 1 atm) E red = 0