2/1/2012 1 ELECTROCHEMISTRY Review: Balancing REDOX reactions • Balance the following redox reactions: 1. MnO 4 - (aq) + Cl - (aq) ⇆ Mn 2+ (aq) + Cl 2(aq) acidic solution 2. H 2 O 2(aq) + ClO 2(aq) ⇆ ClO 2 - (aq) + O 2(g) basic solution 3. NO 2 - (aq) + Cr 2 O 7 2- (aq) ⇆ Cr 3+ (aq) + NO 3 - (aq) acidic solution Electrochemistry • Oxidation-Reduction reactions. – LEORA (loose electron oxidation reducing agent) – GEROA (gain electron reduction oxidizing agent) Cu 2+ + Zn Zn 2+ + Cu – Reduction: Cu 2+ + 2e - Cu – Oxidation: Zn Zn 2+ + 2e - (Half-reactions of the redox process) • The two parts of the reaction can be physically separated. – The oxidation reaction occurs in one cell . – The reduction reaction occurs in the other cell. – A “cell” is a compartment for the half-reaction • The cell must contain all physical forms of the species involved – Reduction half-reaction cell contains aqueous Cu 2+ and solid Cu – Oxidation half-reaction cell contains aqueous Zn 2+ and solid Zn • The combination of two cells (reduction and oxidation cell) is called an electrochemical cell Electrochemistry There are two kinds electrochemical cells. 1. Electrochemical cells containing in nonspontaneous chemical reactions are called electrolytic cells. 2. Electrochemical cells containing spontaneous chemical reactions are called voltaic or galvanic cells.
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The motion of the ions through the solution = electric current.
Electrolytic conduction
o Positive ions migrate toward the negative electrode (cathode).
o Negative ions migrate toward positive electrode (anode).
Electrolytic Cell: Electrolysis of NaCl
• Requires molten (melted) NaCl since ions conduct electricity– Consists of Na+ and Cl- ions.
• If direct current (greater than 3.852 V) is applied (by way of two inert electrodes) through the cell containing the molten NaCl, we observe the following:– Pale green gas (Cl2) is formed in one electrode.
– Molten, silvery Na forms at the other electrode. The Na particles floats on top of the molten NaCl.
• Liquid Na is produced at the cathode (-).
2{Na+ + e� Na(l)*}
• Gaseous Cl2 is produced at the anode (+).
2Cl- � Cl2 + 2e-
---------------------------------
2Na++2Cl-�2Na(l)+Cl2(g)
*Na(l) � Na(s)
• This is non-spontaneous except at very high T (>801°C)
• Direct current (dc) source supplies energy to force the reaction forward.
AnodeCathode
• Electrons are used in the cathode half-reaction (reduction) and produced in the anode half-reaction (oxidation).
• Travel of e- : from ANODE(+) to CATHODE(-).
• The dc source forces the e- to flow non-spontaneously from the positive electrode to the negative electrode.
• Na and Cl2 must NOT be allowed to come into contact with each other because they react spontaneously, rapidly and explosively to form NaCl.
• In Downs’ Cell, the liquid Na is drained off, cooled and cast into blocks, then stored in inert mineral oil to prevent reaction with atmosphere (O2) or H2O.
Downs’ Cell – Industrial
production of Na(s)
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Electrolytic Cells: Electrolysis of Aqueous KI
K+ (aq) + e- ���� K(s) -2.924 V
2H2O + 2e- ���� 2H2 (g) + OH- -0.828 V
I2(s) + 2e- ���� 2I- (aq) +0.535 V
O2 (g) + 4H+ +4e- ���� 2H2O +1.229 V
Calculations of different redox combinations and the minimum applied voltages required for electrolysis:
Calculations of different redox combinations and the minimum applied voltages required for electrolysis:
E1 = E(Na+/Na) – E(Cl-/Cl2)
= (-2.714 V) – (+1.360 V) = -4.074 V
E2 = E(H2O/H2,OH) – E(Cl2/Cl-)
= (-0.828V) – (1.360 V) = -2.188 V
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Counting Electrons: Coulometry and Faraday’s
Law of Electrolysis
• Faraday’s Law –
mole substance oxidized or reduced ∝ mole e-
• A faraday is the amount of electricity that reduces one equivalent of a species at the cathode and oxidizes one equivalent of a species at the anode.
• A coulomb is the amount of charge that passes a given point when a current of one ampere (A) flows for one second.
• 1 amp = 1 coulomb/second
Counting Electrons: Coulometry and
Faraday’s Law of Electrolysis
• Faraday’s Law states that during electrolysis, one faraday of electricity (96,487 coulombs) reduces and oxidizes, respectively, one equivalent of the oxidizing agent and the reducing agent.
– This corresponds to the passage of one mole of electrons
through the electrolytic cell.
Counting Electrons: Coulometry and
Faraday’s Law of Electrolysis
• Calculate the mass of palladium produced by the reduction of palladium (II) ions during the passage of 3.20 amperes of current through a solution of palladium (II) sulfate for 30.0 minutes.
Counting Electrons: Coulometry and
Faraday’s Law of Electrolysis
• Calculate the volume of oxygen (measured at STP)
produced by the oxidation of water.
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Commercial Applications of Electrolytic Cells
Electrolytic Refining and Electroplating of Metals
• Impure metallic copper can be purified
electrolytically to ≈ 100% pure Cu.
– The impurities commonly include some active metals plus
less active metals such as: Ag, Au, and Pt.
• The electrolytic solution is CuSO4 and H2SO4
• The impure Cu dissolves to form Cu2+.
• The Cu2+ ions are reduced to Cu at the cathode.
Commercial Applications of Electrolytic Cells
• Any active metal impurities are oxidized to cations that are more difficult to reduce than Cu2+.
– This effectively removes them from the Cu metal.
Effect of Concentrations (or Partial
Pressures) on Electrode Potentials
The Nernst Equation
• The Nernst equation describes the electrode potentials at nonstandard conditions.
The Nernst EquationThe Nernst Equation
• Substitution of the values of the constants into the
Nernst equation at 25o C gives:
The Nernst Equation
• For this half-reaction:
• The corresponding Nernst equation is:
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The Nernst Equation
• Substituting E0 into the above expression gives:
• If [Cu2+] and [Cu+] are both 1.0 M, i.e. at standard conditions, then E = E0 because the concentration term equals zero.
The Nernst Equation
• Calculate the potential for the Cu2+/Cu+ electrode at
250C when the Cu+ ion concentration is 1/3 of the Cu2+
ion concentration.
The Nernst Equation
• Calculate the initial potential of a cell that consists of
an Fe3+/Fe2+ electrode in which [Fe3+]=1.0 x 10-2 M
and [Fe2+]=0.1 M connected to a Sn4+/Sn2+ electrode
in which [Sn4+]=1.0 M and [Sn2+]=0.10 M . A wire
and salt bridge complete the circuit.
• Calculate the E0 cell by the usual procedure.
• Substitute the ion concentrations into Q to calculate