Electrochemistry
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Readings for this section

Introduction 02/09/2006

Electrochemistry is the study of interchange of chemical and electrical energy.

Oxidation/Reduction involves the exchange of electrons from one chemical species to another. Normally, this is done when the two chemicals contact each other in the activated complex (when two species bump into each other in solution for example).

We are interested in separating the chemical species such that the electrons transfer via an external circuit. That way, we can measure the electrochemical effects.

To properly understand the connection between the redox reaction and the electricity, we should balance the overall redox reaction using a half-reaction method such as the one described in the previous section of these notes. We can set up the physical reaction vessel such that the chemicals from one half reaction are separated from those of the second half reaction. For reaction to occur, we still need to connect the solutions to complete the circuit. This is done by attaching wires between electrodes in the two half cells and by connecting the solutions of the two half cells via a salt bridge or by some other device such as a semi-permeable membrane.

In general, such a cell is called an electrochemical cell. These cells could be used in one of two types of situations:

1. The chemical reaction is spontaneous and produces electricity.
   This is called a voltaic cell or a galvanic cell.
2. The chemical reaction is non-spontaneous and is forced by electricity from an external source.
   This kind of cell is called an electrolysis cell.

We will look at the first situation first.

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Galvanic or Voltaic cells

Consider a piece of zinc foil placed in a beaker of copper sulphate. The copper sulphate solution is blue because of the presence of the Cu²⁺ ions. When the zinc is added, the solution changes to colourless and the zinc metal is dissolved and replaced by a reddish orange powder. The colourless solution no longer has Cu²⁺ in it and the reddish orange powder is Cu(s). The Zn(s) is dissolved and is now Zn²⁺. [I know this because of my vast knowledge of chemistry ;-)]

If we write an overall reaction for this process, we get:

Zn + Cu²⁺ ® Zn²⁺ + Cu. This doesn't help us in our quest for electrochemistry knowledge. Let's rewrite this in half-reaction form.

Zn ® Zn²⁺ + 2e^–

Cu²⁺ + 2e^– ® Cu

Now, we can set up two half cells, one with a zinc electrode in a Zn²⁺ solution (say ZnSO₄) and the other with a copper electrode in a Cu²⁺ solution (say CuSO₄) as follows.

The centre line in the diagram, recall, is either a semi-permeable membrane or a salt bridge.

Now, at the anode, we have the reaction

Zn(s) ® Zn²⁺(aq) + 2e^–(aq)

and at the cathode, we have

Cu²⁺(aq) + 2e^–(aq) ® Cu(s)

Electrons pass through the wires and SO₄^2– pass through the membrane to keep the solutions neutral.

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Cell Potentials

Electrons from solution pushed onto the anode, around the external circuit and onto the cathode where they are pulled out into the solution.

That's one way of thinking of the electrical circuit part of the electrochemical cell. The electrons pushed around the external circuit can do work (run a motor, illuminate a light bulb, etc). The amount of work possible is a function of both the voltage (potential) and of the current (number of electrons) in the circuit.

Pushing one Coulomb of charge around a circuit at a potential of 1 Volt does One Joule of work. OR, mathematically, 1J = 1V × 1C.

The cell potential (voltage of the cell) depends on the chemicals used. For example, the chemicals in dry-cells (batteries) are such that the potential is always about 1.5 V. This has become a standard and is now a limiting factor in deciding which chemicals can be used to create a battery.

The cell potential is given a symbol of E_cell. If all chemicals are at activity of 1 (conc. = 1 M, p = 1 bar) then the cell potential is the standard cell potential and is given as E°_cell.

Any redox reaction has the potential (pun) to be used in an electrochemical cell. We merely need to be able to divide the oxidizing and reducing agents into two half cells (half reactions).

Take for example, the reaction of zinc metal dissolving in hydrochloric acid. The reaction is:

Zn(s) + 2H⁺(aq) ® Zn²⁺(aq) + H₂(g)

We need to separate the zinc from the hydrogen. We can use zinc as an electrode but what about the hydrogen.

In this case, we need to set up a special electrode, which allows H₂ gas molecules to interact directly with H⁺ dissolved in water and with the electrons from the external circuit simultaneously. Such a system is pictured below.

A blow-up of the surface of the platinum electrode is shown below so the location of reaction can be better understood.

The other half-cell would look much like that pictured in a previous diagram. The whole cell diagram, of course would include the external part of the circuit and the salt bridge or membrane to complete the circuit. We can abbreviate this diagram as follows:

Zn(s)/ZnSO₄(aq)//H₂SO₄(aq)/H₂(g),Pt(s)

  [Anode             //            Cathode]

Where the single slash mark / represents the boundary between solution and electrode and the double slash // represents the salt bridge or semi-permeable membrane. The external circuit, of course, joins the two electrodes (solid) and is not explicitly shown here.

The overall cell voltage can be summed from the half-cell potentials of the oxidation and of the reduction reactions.

E_cell = E_Red + E_ox.

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Standard Reduction Potentials

As before, we wish to tabulate thermodynamic data in some useable way. We define a standard set of conditions under which all half-reactions can theoretically operate and we tabulate the half-cell potentials for the reactions under standard conditions. Since the potential of an oxidation reaction (loose electron) is merely the negative of the potential for a reduction reaction reaction (gain electron), we choose to tabulate only the reduction half-reactions. Any oxidation half reactions are merely the reverse of the tabulated reduction reaction and the oxidation potential is the negative of the tabulate reduction potential.

Table D.4 from Petrucci lists many reduction half-reactions that you will need for problems, etc in the text.  Herein, I have duplicated only a few in the following table.

Let's look at this small reduction potential table and see if we can use it to explain certain experimentally observable properties of matter.  First off, I should point out that the more positive reduction potentials refer to more spontaneous reduction reactions.

The first reaction in this mini-table shows the reduction of oxygen. This is higher in the table than many other things. That makes sense since we know that oxygen is a good oxidizing agent (it makes other things oxidize while it itself is reduced).

We can see that water can be both oxidized (-1) and reduced (2). It is oxidized to oxygen and reduced to hydrogen. These two reactions, in fact, serve as a set of boundaries for aqueous solutions. Any materials harder to oxidize than water will not oxidize since the water will.  Similarly, any materials harder to reduce than water will never reduce (in water). For example, sodium ions will never reduce to sodium metal in the presence of water since the water will reduce first.  In fact, sodium metal in water will spontaneously (explosively) oxidize to Na⁺ ions and the water will reduce to hydrogen gas and leave a basic solution (equation 6). The overall reaction for this process is the sum  (#6) – 2×(#7).

2Na + 2 H₂O ® 2Na⁺ + H₂ + 2OH^–

Thus, the overall standard cell potential for this reaction is

Eº_cell = Eº_red + Eº_ox = –0.83 + 2.71 (reverse reaction => change sign) = 1.88 V.

Positive cell potential means spontaneous reaction.

When pairing up metals, we know that zinc will dissolve in a copper (II) solution. Here we see that the copper (II) (3) is easier to reduce than the zinc (II) (5).

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Gibbs' Free Energy and Cell potential

Let's now relate the electrochemistry ideas we've explored with the thermodynamic parameter called Gibbs' Free Energy.

We saw,  

w = –QE                       (work = – charge times potential), 

where Q is the charge and can be defined as Q = nF where F is the Faraday constant (96485 C/mol), named after Michael Faraday.

In our case, we're interested in the maximum work since this can be related to the thermodynamic parameter DG.

Thus, for the case where the work is done infinitely slowly (chemical system is always at equilibrium with the electrodes, electrical resistance is zero since current is essentially zero, etc…) we have
w_max = – QE_max where E_max for standard conditions is simply E° as calculated from the tabulated half-cell potentials.

Take for example, a cell with a maximum cell potential of 2.50 V. If 1.33 mol of e^– passes through the cell at an average potential E = 2.10 V. What is the efficiency?

w = – Q E = – nFE. = 1.33 mol × 96485 C/mol × 2.10 V              (V = J/C)

w = – 2.69 × 10⁵ J = – 269 kJ

w_max = – nFE_max = 1.33 mol × 96485 C/mol × 2.50 V

        = – 321 kJ

efficiency = w/w_max × 100% = –269/–321 × 100% = 83.8 %

Of course, since w_max is only achievable if the work is done reversibly (infinitely slowly), we can never reach 100% efficiency in any system in the real world.

We already have seen that DG is a measure of the maximum work obtainable from a system. Thus,

DG = w_max

DG = – Q E

DG = – nFE. In this case, the potential is the cell potential E_cell.

DG° = – RT lnK = – nFE°_cell.

Thus, we now have a link between free-energy, equilibrium and electrochemical thermodynamic parameters.

Example:

Is Fe²⁺ spontaneously oxidized by the oxygen of the air in acidic solution? Calculate DG and K.

Two half-reactions can be determined by looking in the table of standard reduction potentials.

Oxidation of iron:	4[Fe2+ ® Fe3+ + e– ]	–E°ox =  –0.77 V
Reduction of O2	O2 + 4H+ + 4e– ® 2 H2O	E°red =  1.23 V
Overall:	4Fe2+ + O2 + 4H+ ® 4Fe3+ + 2H2O n = (4 mol electrons / mol equation)	E°cell = 0.46 V (Positive means spontaneous®)

DG = –nFE°_cell = – 4 mol × 96485 C/mol × 0.46 V = – 1.8×10⁵ J = –180 kJ (per mole of reaction)

     K  =  e^72.7  =  3.7×10³¹.

SUMMARY

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Non-Standard Conditions: Nernst Equation

Consider the following set of half-reactions

Reduction	Fe2+ + 2e– Fe	E°red = –0.44 V
Oxidation	Zn Zn2+ +  2e–	E°ox = +0.76 V
Overall	Zn + Fe2+ Zn2+ + Fe	E°cell = 0.32 V

This is valid only if all species have concentrations of 1 M.

 What if not all the species are at Standard Conditions?

 Le Châtelier's principle gives us some ideas. Look at the overall reaction. Consider the reaction to be at equilibrium (E_cell = 0 V). If we add more Fe²⁺ to the reaction mixture, The equilibrium will shift to the right to use up the excess Fe²⁺. This will, of course cause a positive cell potential to be measurable.

Conversely, if we removed Fe²⁺ from the reaction mixture, the equilibrium would shift to the left to try to replace some of the missing Fe²⁺. This would result in a measurable negative potential (reaction goes left).

If the reaction were not at equilibrium then these preceding two changes would me additive to the measured voltage (say standard voltage). So, if we add Fe²⁺ to a mixture at Standard Conditions, then the voltage would be more positive and if we removed some Fe²⁺ from a mixture at Standard Conditions the measured voltage would be lower (less positive).

The reverse results would be found for addition or removal of Zn²⁺ (a product). More Zn²⁺ would give a lower positive potential and vice versa.

Now we try to quantify the previous hand-waving discussion.

Recall the relation: DG = DG° + RT ln Q                          (for non-standard conditions)

Substitute the new definition of DG and we get

                         –nFE = –nFE° + RT ln Q

                                                     (This is called the Nernst Equation).

Thus, we can calculate the cell potential of any cell, given only the initial conditions (to calculate Q) and the reduction potentials (to calculate E°).

Now, looking back at the reaction cell discussed at the beginning of this section we can calculate the voltage for the situation where the concentrations are not standard. For example, consider [Fe²⁺] = 0.1 M and [Zn²⁺] = 1.9 M.

For this reaction, we also know that there are two electrons transferred for each equation (n = 2).

Although the reaction is 90% complete (from Standard Conditions initial) the cell potential has only dropped by a small amount (0.04 V). This is normal. For example, the batteries in your flashlight will put out almost full voltage until the last traces of chemical are almost used up at which point the voltage will drop off rather sharply. This is an especially useful characteristic of cells powering electronic equipment, (like a calculator or CD player, etc.) which require a certain minimum (and dependable) voltage to operate successfully.

We can also use the Nernst equation to calculate such things as equilibrium constants.

At equilibrium, Q = K and E_cell = 0 so the Nernst equation becomes

or

Example:

What is the equilibrium constant for the reaction of copper metal with bromine to form copper(II) ions and bromide ions in aqueous solution at 25°C?

Cathode	Br2 + 2e– 2 Br–	E°red = 1.09 V
Anode	Cu Cu2+ + 2e–	E°ox = –E°red = –0.34 V
Cell	Cu + Br2 Cu2+ + 2 Br–	E°cell = 0.75

Example:

What is the value of the solubility product constant for AgCl?

We could calculate this by breaking the overall K_sp reaction into a series of redox reactions as follows:

Anode	Ag(s) ® Ag+ + e–	E°ox = –0.80 V
Cathode	AgCl + e– ® Ag(s) + Cl–	E°red = 0.22 V
Overall  (º Ksp reaction)	AgCl Ag+ + Cl–	E°cell = –0.58 V (not very spontaneous)

K = 6.3 × 10^-9

Ion Concentrations can also be calculated...

Consider the Platinum-Hydrogen electrode coupled with a copper/copper(II) electrode.

The overall reaction is Cu²⁺ + H₂ ® Cu + 2H⁺.

The Nernst equation is

If [Cu²⁺] = 1 M and P(H₂) = 1 bar (both their activities are 1) then

E  = E° – 0.0257 V ln [H⁺]               (Note:  ln Y = 2.303 × log Y and .-257×2.303=.05917)

E  = E° + 0.05917 V (–log [H⁺])

 E  = E° + 0.05917 V × pH.

We see here that the cell potential is a function of pH. The probes in pH meters are set up this way. A complete electrochemical cell is contained within the probe casing. All chemicals are at standard conditions and a porous glass membrane allows only H⁺ ions to pass through. The diagram of the set-up is given here.

Concentration cells

A concentration cell is one in which the half-reactions in the two half cells are identical but the concentration of the ions is different.  In this case, the reduction and the oxidation half reaction are identical.  Take for example a concentration cell where both half cells have an iron electrode in a Fe²⁺ solution.  We would have the following:

Consider the following set of half-reactions

Reduction (half-cell 1)	Fe2+ + 2e– Fe	E°red = –0.44 V
Oxidation (half-cell 2)	Fe Fe2+ +  2e–	E°ox = +0.44 V
Overall	Fe + Fe2+ Fe2+ + Fe	E°cell = 0.00 V

if all concentrations are identical (or if they are all standard) then obviously the cell voltage is zero. If, however there is a different concentration on one half cell than on the other one, we could use the nernst equation to calculate, for example the cell potential.

Half-cell calculations

The Nernst equation can also be used to calculate half-cell potentials. In this case, Q is simply the ratio of ions of the half-cell equation rather than the full-cell and the value for E° is that of the half-cell. No other changes to the equation are made.

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Batteries

There are three classes of batteries:

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Electrolysis and Electrolytic Cells

We've seen how spontaneous reactions can be harnessed to produce electricity. In electrolytic cells, electricity is used to force non-spontaneous reactions to proceed. Electrolysis is used to prepare certain compounds and metals. Products of electrolysis are found in every day life: Chrome plated bumpers, Silver and gold plating on some inexpensive jewellery, among other things.

In setting up an electrolysis cell, we must take into account all possible reactions and make sure the conditions are set so that the desired reaction occurs. We will do this at first by using only inert electrodes and then by using pure electrolyte liquids rather than aqueous (or other) solutions.

We will use a non-aqueous system such as molten NaCl as our medium, and non reactive electrodes like platinum. We can successfully separate out pure Na. Consider the following electrolytic cell.

This is a very important method of producing sodium metal. At the anode, Cl₂ is evolved in the reaction
2Cl^– ® Cl₂(g)+ 2e^–.

At the Cathode, the molten sodium ions are converted to sodium metal (liquid at these temperatures) in the reaction
Na⁺ + e^– ® Na(l).

Alkali earth metals which are also highly reactive can be prepared using electrolysis of molten chloride salts as in the reaction MgCl(l) ® Mg(l) + Cl₂(g)

Other metals which are not quite as reactive as the above two families but are sufficiently reactive that they cannot be produced in an aqueous cell can be produced this way. Aluminum is one such metal.

The Hall process invented before the turn of the century allowed for the production of aluminum on an industrial scale.

A molten mixture of aluminum ore (Al₂O₃) and cryolite (Na⁺)₃(AlF₆^–)(l). The cryolite is the solvent and it is used because it has a lower melting point than pure Al₂O₃. A carbon electrode acts as the anode and aluminum forms the cathode such that the following reactions occur:

Al³⁺ + 3e^– ® Al(l)

C(s) + 2O^2– ® CO₂(g) + 4e^–

This process needs lots of electricity per kg of aluminum so you will most commonly find aluminum production plants very close to electric power stations to reduce the cost of electricity.

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Quantitative Aspects of Electrolysis:

Consider the reaction in the molten NaCl electrolysis cell.

2Cl–  ®   Cl2(g) + 2e–.	2 moles of electrons per mole of Cl2(g)
Na+ + e– ® Na(l).	1 mole of electrons per mole of Na metal

Recall that the charge on the electron is Q = nF where F is the Faraday constant (96485 C/mol), named after Michael Faraday.

Thus, to produce one mole (23 g) of sodium, we need Q = 1×96485 C = 96500 C.

To produce one mole (70.9 g) of chlorine gas, we need 2×96500 = 1,930,000 C.

Current is defined as the amount of charge passing a point in a circuit in one second.

I = Q/s                   The units are            1 Ampere = 1 Coulomb/second (1A = 1C/s).

Now, if we have a current of 50.0 A passing through an NaCl(l) electrolysis cell in 1 hour, how much sodium and chlorine will we produce.?

50.0 A × 3600 s = 180,000 C                                 (NOTE: 1As = 1C)

n = Q/F = 180,000/96,485 = 1.87 mol e^–

It takes lots of current to produce very little sodium and chlorine.

We can do other types of calculations similar to this.

What mass of aluminum will be produced in 1.00 h by electrolysis of Molten AlCl₃ using a current of 10.0 A?

Q = I×t = 10.0 A × 3600 s = 3.60×10⁴ C.

It takes three moles of electrons for every mole of Aluminum. So...

n = Q/F = 3.60×10⁴ C / 96485C/mol e^– ×1/3 × 27.0 g/mol = 3.36 g Al.

What volume of Cl₂(g) at STP (0°C, 1 atm) will be produced by a current of 20.0 A in 2.00 h in the same cell as used in the previous example?

Note that in this problem, we used
R=0.08206 Latm/mol K.

This was to facilitate the conversion to volume. You could have also used the more standard value of R=8.314 J/mol K but conversion to volume would have involved one more step of calculations.

If we had tried to electrolyse sodium Chloride in water, we would need to consider not only the Na⁺ and Cl^– as possible reactants but also the water.  We can never produce sodium in aqueous solution, because it will spontaneously react with the water to produce Na⁺(aq). Whether we tried to produce sodium metal either chemically or electrolytically, the chance of success would be equally bad. Thus, at the cathode of an aqueous NaCl cell, we would get the reduction of water happening

2H₂O + 2e^– ® H₂ + 2OH^–                       E°_red = –0.83 V

rather than the reduction of the sodium

Na⁺ + e^– ® Na                                        E°_red = –2.71 V

Notice that the reduction potential for the water reduction is a lot more positive than that for the sodium. This is a way we can look up reduction potentials and tell what will happen in our solution before hand.

Looking at the two possible reactions at the anode of the NaCl (aq) cell, we see that the Cl^– and the H₂O are both candidates for oxidation. The two possible half-reactions are:

2H₂O ® O₂ + 4H⁺ + 4e^–                         E°_ox = –E°_red = –1.23

2Cl^– ® Cl₂ + 2e^–                                     E°_ox = –E°_red = –1.36

Since the water has a more positive oxidation potential than the chlorine reaction, it should be oxidized more readily. There is a complication, however. The water oxidation requires a considerable over-voltage (extra voltage) to make the reaction rate appreciable. Since the two possible reactions are quite close in potentials the extra voltage can quickly be sufficient to cause the chlorine oxidation to occur. Since this latter reaction is quite rapid in comparison to the water ® oxygen one, it will dominate at only slightly elevated voltages.

If we had used Na₂SO₄ rather than NaCl as the electrolyte, then the anode reaction would have been the water one since SO₄^2– is very difficult to oxidize (very negative oxidation potential in comparison to water).

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Electrolytic Purification of Metals

Consider the electrolysis of Copper(II)sulfate solution.

We find O₂ produced at the anode and copper solid deposited at the cathode. This means, as we saw before that the water is more easily oxidized (to oxygen) than are the sulphate ions while the copper(II) ions are more easily reduced (to copper metal) than is water (to hydrogen). Hence, it is quite easy to set up an electrolytic purification process for copper metal. Unlike aluminum, where we needed molten baths of aluminum ore, we need only dissolve the copper in a water solution and then electrolyse it out as copper metal. If we do a crude separation chemically of the copper from the ore, we can use this impure copper as the anode and use pure copper as the cathode. Then, we will electrolyse copper ions off of the impure copper anode and then reduce them onto the pure cathode in the same solution. Impurities in the anode would merely settle to the bottom of the cell to be disposed of. We need to take care to adjust the voltage so that only the copper is oxidized from the anode and not some of the impurities. This done, we can successfully purify copper using electrolysis. The slower we take the reaction (less over voltage) the more pure will be the copper.

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Products of Electrolysis

We've seen that the particular product of electrolysis we see at the electrodes depends on the ease of production of that particular species. There are three factors that determine product that we will find in any given electrolysis cell: the electrode material, the solvent and the electrolyte. Often, we use inert electrodes and/or pure liquids to help alleviate some of the complications from competing chemistry, other times, we must simply carefully adjust conditions (concentrations and voltages) to ensure that the desired product is achieved.

Complication: Cl^– we've seen to be more easily oxidized than water even though it has an oxidation potential (negative of the reduction potential for Cl₂) which is more negative than that for water. This is due to kinetics. The water reaction is quite slow and requires a large voltage to force it to proceed with any speed. The Cl^– oxidizes to Cl₂ much more rapidly and the over voltage we apply in an attempt to force the water to oxidize is sufficient to produce Cl₂ which occurs faster.

Summary:

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Corrosion of metals

We've seen that the oxygen from the air can be reduced to water and that the reduction potential for this half-reaction is very positive. More positive, in fact than that for most metal reduction process. This means that any metal in contact with oxygen from the air and in the presence of an electrolyte solution will oxidize. Since mining and production of metal is a fairly energy-intensive process, it is very useful to be able to find ways to reduce or eliminate the oxidation of these metals if at all possible. There are several ways to reduce oxidation of metals

Eliminate the oxygen/metal contact

This is achieved by coating the metal with some material that is impervious to either the oxygen or to any aqueous salts. This prevents the electrical circuit from being complete in the 'electrochemical cell' that is spontaneously set up at the point of rusting.

To keep the O₂ or the salt solution, we can use several methods
1)Insulate the metal
a) Paint the metal. This is a pure insulating technique and serves merely to keep the reacting species separated. Unfortunately any small scratch will allow the reaction to start up and may even accelerate the process.
b) Oil the metal. (various rust-proofing companies use this technique). The unfortunate problem is that oil is a volatile liquid and eventually either runs off the surface or evaporates away, leaving the metal unprotected. Special surfactants can be added to the oil to make it cling to the metal surface and slow evaporation. You will find these surfactant agents in treatments such as RustCheck and The new Canadian Tire rust-proofing treatments. No matter, these treatments must be repeated regularly (yearly) to maintain protection.
c) anodize the metal: sometimes, the oxides of the metal are very stable and act to insulate the metal from the air. This is most obvious in copper roofing where the (orange) copper metal is quickly oxidized to the green oxide CuO. This oxide is quite stable and protects the underlying metal from further corrosion. This is useful in that small scratches quickly 'seal themselves up' with more oxide and the process is halted. Aluminum is often deliberately anodized in a controlled electrochemical cell to form an aluminum oxide coating that is relatively stable (in neutral solutions).
2) Coat the metal with a material that oxidizes easier than the metal you wish to protect. For example, steel coated with zinc will not oxidize until first all the zinc is oxidized because the zinc will change the electrical potential of the steel and prevent it from being oxidized even if it is exposed to the air and water. This protection will last until the zinc is used up. at which time, the steel will start to oxidize.

Cathodic protection

a) Use a sacrificial anode made of a material that oxidized easier than the metal you wish to protect. This is similar to coating the metal with the material (as in galvanized steel) but is simpler to do since it merely means attaching chunks of the anode material rather than going through a coating process.
Pipelines are often protected this way, the steel pipes are attached by a wire to a large block of zinc buried next to the pipeline in the ground. The zinc serves to electrically charge the steel changing its potential and thus reducing the amount of oxidation.

 

b) Electrically change the metal's potential. Similar to the use of sacrificial anodes, this process involves attaching the negative pole of an electrical system to the metal we need to protect. In cars, for example, the negative pole of the battery is the ground. This serves to charge up the car and change it's potential. Thus, the oxygen has to overcome both the iron's reduction potential as well as the batteries artificial potential to force the reaction to happen. Cars would rust out far quicker it they were grounded positive since that, in effect, would serve to increase the rate of oxidation of the steel.

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Prof. Michael J. Mombourquette.
Copyright (c) 1997
Revised: September 02, 2006

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