For the SHE, we assign 2H + ( aq , 1 M ) + 2e -  H 2 ( g , 1 atm) E  red = 0.

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# For the SHE, we assign 2H + ( aq , 1 M ) + 2e -  H 2 ( g , 1 atm) E  red = 0. - PowerPoint PPT Presentation

For the SHE, we assign 2H + ( aq , 1 M ) + 2e -  H 2 ( g , 1 atm) E  red = 0. For Zn: E  cell = E  red (cathode) - E  red (anode) 0.76 V = 0 V - E  red (anode). Therefore, E  red (anode) = -0.76 V. Standard reduction potentials must be written as reduction reactions:

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Presentation Transcript
For the SHE, we assign
• 2H+(aq, 1M) + 2e- H2(g, 1 atm)
• Ered = 0.

Chapter 20

For Zn:
• Ecell = Ered(cathode) - Ered(anode)
• 0.76 V = 0 V - Ered(anode).
• Therefore, Ered(anode) = -0.76 V.
• Standard reduction potentials must be written as reduction reactions:
• Zn2+(aq) + 2e- Zn(s), Ered = -0.76 V.

Chapter 20

• Therefore,
• 2Zn2+(aq) + 4e- 2Zn(s), Ered = -0.76 V.

Chapter 20

Reactions with Ered < 0 are spontaneous oxidations relative to the SHE.
• The larger the difference between Ered values, the larger Ecell.

Chapter 20

Oxidizing and Reducing Agents
• The more positive Ered the stronger the oxidizing agent on the left.
• The more negative Ered the stronger the reducing agent on the right.

Chapter 20

Example: For the following cell, what is the cell

reaction and Eocell?

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

Al3+(aq) + 3e-→ Al(s); EoAl = -1.66 V

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

Chapter 20

Spontaneity of Redox Reactions
• 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.

Chapter 20

This gives:
• At a temperature of 298 K:

Chapter 20

Cell EMF and Chemical Equilibrium
• A system is at equilibrium when G = 0.
• E = 0 V and Q = Keq:

Chapter 20

In the problem presented before, calculate ΔGo and Keq

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

Al3+(aq) + 3e-→ Al(s); EoAl = -1.66 V

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

Chapter 20

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

and [Zn2+] = 0.050 M

Chapter 20

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

Chapter 20

Batteries
• 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

Chapter 20

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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

Chapter 20

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How Hydrogen Fuels Cells Work

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

Chapter 20

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How Hydrogen Fuels Cells Work

Fuel cells

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.

Chapter 20

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How Hydrogen Fuels Cells Work

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.

Chapter 20

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How Hydrogen Fuels Cells Work

Chapter 20

Corrosion
• Corrosion of Iron
• Since Ered(Fe2+) < Ered(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).

Chapter 20

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How Hydrogen Fuels Cells Work

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

Chapter 20

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How Hydrogen Fuels Cells Work

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.

Chapter 20

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.

Chapter 20

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

Chapter 20

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.

Chapter 20

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

Chapter 20

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.

Chapter 20

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.

Chapter 20

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

Chapter 20

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

Chapter 20

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

Chapter 20

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

Chapter 20