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Redox and Corrosion Basics

  1. Nov 29, 2011 #1
    I have some questions concerning the tendency of a system containing two different metals and water to corrode. This is coming from several similar discussions on another forum, concerning water cooling of computer components. There seems to be a lot of repeated dogma, which may include misinformation or critical omissions. I think I should check with some people who actually know about chemistry! Personally, I took a class in high school.

    First, a basic question on half reaction charts and electrode potential: If you have to use different multiples of the basic half reaction on each side to balance the electrons, do you scale the potential E° as well?

    Now, the main dogma among the WC enthusiasts is that it is best to use distilled water, with a tiny amount of copper sulfate or a piece of bulk silver (or both) added as a biocide. This circulates in a closed system containing copper and brass components. Customers and vendors were both surprised when nickle-plated components corroded, although that is exactly what you would expect. The vendor was assuming the use of a cooling fluid that contains anti-corrosives, while the enthusiasts eschew pre-made fluids, insisting that DI water is best, and then adding copper salts to it! This report is humorous in not knowing how to spell peel, but otherwise explains the situation well enough for me to follow, with some dim memory of half reactions.

    Now I'm wondering about some more details. Given two different metals and a conductive fluid between them, what is the significance of the pre-existing ions in the solution? I see that existing Cu2+ ions plate themselves onto the cathode, and Ni+2 replace them in the solution, destroying the anode. So, is it especially significant and critical that the solution already had the same kind of ions that the cathode is made out of, to allow corrosion to occur?

    What happens after time, when all the Cu from the CuSO4 is attached to the copper components?

    In general, if the solution is conductive but not made from the same stuff as the cathode, how does the reaction work? Does the material from the anode plate itself onto the cathode? Does that mean there is another E° value for mixed materials that needs to be consulted?

    Cases that are especially interesting involve Cu being the anode: Aluminum or silver being present in contact with the fluid.

    Many thanks,
  2. jcsd
  3. Nov 30, 2011 #2


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    Staff: Mentor

    This is a huge topic, not something that can be answered in a short post. But I will try to at least give you some basic theory behind, so that you should be able to build on it.


    Note that just knowing both redox system is only part of the story. Electrode potentials listed in tables are for standard conditions - that is, activity of each ion taking part in the reaction equal to 1. As a first approximation activity equals concentration, so you need all concentrations to equal 1M (mol/L). Potential observed in the real solution, where concentrations are what they are, is called a formal potential, and can be calculated from the Nernst equation:

    [tex]E = E_0 + \frac {RT}{nF} \ln Q[/tex]

    where Q is a reaction quotient for the oxidation reaction. So for example if you have a reaction like

    [tex]aRed + bB \leftrightarrow cOx + dD + ne^-[/tex]

    (where Red and Ox are reduced and oxidized forms of the substance) reaction quotient is

    [tex]Q = \frac{[Ox]^c[D]^d}{[Red]^a^b}[/tex]

    (this is a general form - if there are more or less substances taking part in the reaction, simply add or remove terms; also note there are two conventions used - sometimes it is an oxidation reaction quotient, sometimes it is a reduction equation quotient, depending on which one it is sign in the Nernst equation changes, so in the end result is identical).

    Note that n in the Nernst equation and n in the redox reaction is the same n. In practice we rather use simplified version of the Nernst equation, which holds for 25°C:

    [tex]E = E_0 + \frac {0.059V}{n} \log Q[/tex]

    where log is base 10.

    When you have a solution in contact with a solid, like Cu/Cu2+, there is no such thing as a "concentration of solid copper" - its activity is by definition just 1. Metallic copper is a reduced form, Cu2+ is the oxidized form, reaction is

    [tex]Cu \leftrightarrow Cu^{2+} + 2e^-[/tex]

    and Nernst equation is

    [tex]E = 0.337 + \frac {0.059 V}{2} \log [Cu^{2+}][/tex]

    (0.337V is a standard potential for copper, taken from tables).

    Now, when there are two redox systems, they will react till they get to equilibrium, and at equilibrium both potentials will be identical. You can write Nernst equation for each system separately:

    [tex]E_1 = E_{0,1} + \frac {0.059 V}{n_1} \log Q_1[/tex]

    [tex]E_2 = E_{0,2} + \frac {0.059 V}{n_2} \log Q_2[/tex]

    and after reaction ended E1 = E2. As in both reactions exactly the same number of electrons has to be passed (it doesn't mean n1 = n2!) you can easily calculate what is ratio of moles of substances reacting in both half reactions. In the end it gives you a set of equations that - once solved - give the final concentrations of all ions present.

    Right. In this particular case math is pretty simple:

    [tex]Cu \leftrightarrow Cu^{2+} + 2e^-[/tex]
    [tex]Ni \leftrightarrow Ni^{2+} + 2e^-[/tex]

    [tex]E_{Cu} = 0.337 + \frac {0.059V}{2} \log [Cu^{2+}][/tex]
    [tex]E_{Ni} = -0.25 + \frac {0.059V}{2} \log [Ni^{2+}][/tex]

    at equilibrium ECu=ENi, or

    [tex]0.337 + \frac {0.059V}{2} \log [Cu^{2+}] = -0.25 + \frac {0.059V}{2} \log [Ni^{2+}][/tex]

    or (approximately):

    [tex]\frac{[Ni^{2+}]}{[Cu^{2+}]} = 10^{20}[/tex]

    If you add any amount of divalent copper to the solution being in contact with metallic nickel, it will react:

    [tex]Ni+Cu^{2+} \rightarrow Ni^{2+}+Cu[/tex]

    till there is almost no copper present in the solution. Concentration ratio will be even more dramatic in the case of aluminum (although this is a more complicated case, as Al gets passivated, so it is not that fast to react with copper solution; but eventually it will).
  4. Nov 30, 2011 #3
    What does this mean physically to this system?

    So the copper ions are totally replaced by nickle ions. Where does the copper go? You imply that the neutral copper atoms are not soluable. Do they drop out at the point where the nickle was picked up? Or join with the bulk copper samples?

  5. Dec 1, 2011 #4


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    Staff: Mentor

    That copper ions are quantitatively replaced by nickel.

    Copper will be deposited on the nickel surface, in the place where it was reduced. In general copper sticks very good to most metals, so I would expect it to stay on the nickel surface. But this is chemistry, and reality likes to play tricks.
  6. Dec 2, 2011 #5
    You mean the ions already present due to the copper sulfate?
    What happens if the solution is conductive but not specifically loaded with copper? My original question concerned the role of the solution to the ability to corrode.
  7. Dec 2, 2011 #6


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    Staff: Mentor


    Conductivity of the solution doesn't matter much - that is, if solution is conductive, usually corrosion will be faster, but still if there is no redox pair that can be coupled conductivity is not enough. Instead of adding copper sulfate (copper(II) is a mildly oxidizing agent) much better approach is to add some reducing agent - like sodium sulfite. It will neutralize any oxidizers than can enter the solution (first of all - atmospheric oxygen).
  8. Dec 3, 2011 #7
    If there is nothing special about the fluid, what happens? Two half reactions are possible. Do they "get together" somehow?
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