Corrosion

As we covered in the chapter on Faraday’s Laws, many times in an electrochemical reaction, one of the solid electrodes gets corroded over time. Recall the equation developed for this situation was:

$m_A=\frac QF \frac{M_A}n$

This section will expound upon some applications of this idea.

A battery’s current depends on cell potential, state of charge, internal resistance and the connected load. A marked voltage or capacity does not mean it supplies one fixed current in every circuit. For a Faraday-law calculation, specify the actual current and how it changes with time.

At the end of a battery’s useful life, it becomes unable to produce electricity, and it becomes useless at providing power to a circuit.

Parallel cells share a terminal voltage, but their individual currents depend on their voltages, resistances and the load. The original fixed-current rule cannot determine a circuit current after a cell becomes depleted.

Series cells carry the same circuit current, while their signed voltages and internal resistances contribute to the circuit. Depletion changes cell behavior; it does not simply turn the cell into a zero-resistance wire.

For ideal AP charge-capacity calculations use Q=∫I dt, or Q=It when I is constant. Specify active mass, electron stoichiometry and utilization/current efficiency. Circuit-design predictions require additional data and are retained enrichment.

Reviewed capacity task: one ideal cell has 600 mA h of usable charge capacity and supplies constant current 200 mA. Determine its lifetime. The original multi-battery circuit is not part of this replacement.

Reviewed Faraday task: one ideal Zn/Cu cell has 5.00 g of usable Zn, excess Cu²⁺ and constant current 0.500 A. Assume Zn→Zn²⁺+2 e⁻, 100% utilization/current efficiency, M(Zn)=65.38 g mol⁻¹ and F=96485.3 C mol⁻¹. Find lifetime.

As we discussed before, cathodic protection involves the addition of a sacrificial metal to a larger metal object, such that this sacrificial metal will corrode instead of the main metal of the object corroding.

The most common two examples of this type of corrosion protection are on ships and pipelines.

Ships are the simpler of the two, as typically they are simply blocks of zinc or some other metal whose half-reaction has a very low $E^\circ$. This ensures that if any redox reaction were to occur while the ship was perched on the ocean, the sacrificial metal would provide the anode to the main metal’s cathode. Typically, steel or copper is protected through the use of this technique. Steel can be represented by the half-reaction:

$Fe^{2+}+2e^-\rightarrow Fe_{(s)}$

Copper, as we’ve mentioned often before, is:

$Cu^{2+}+2e^-\rightarrow Cu_{(s)}$

Both of these have a higher $E^\circ$ than the Zinc reduction reaction:

$Zn^{2+}+2e^-\rightarrow Zn_{(s)}$

This means that the ship, having electrical contact between the Zinc and the protected metal, would undergo corrosion when it is asea. However, the corrosion would lead to the destruction of the sacrificial zinc, instead of the copper or iron. If the zinc wasn’t present, then the iron, having a lower $E^\circ$ than copper, would corrode away, potentially leading to leaks.

So what kind of problems can you expect to see about cathodic protection? Well, one ship engineers face is to identify the amount of zinc a ship needs to sustain its protection. In order to add the right amount of zinc to a ship, a few items need to be addressed. First, if we are to use Faraday’s law’s equation, then we need to determine what we can say about $Q$. The easiest assumption to make is the constant-current assumption, and that requires figuring out what kind of current a ship might expect to pass between its hull and the zinc while it is in the electrolyte we call the ocean.

Another problem is that in order for cathodic protection to work, it requires some sort of electrical contact between the two metals. However you connect the metal to the ship, the metal is always at a risk of losing electrical contact before being fully used up. Therefore, we need to figure out how long a sacrificial play can be used by a ship. One useful metric to measure this is the maximum % degradation a sacrificial play can undergo before it loses its effectiveness. Depending on the connection style, this can range anywhere from 20% up to 80%.

Example Problem 9.2.3

Reviewed ideal sacrificial-anode capacity task: a 5.00 kg Zn anode has 80.0% usable mass and supplies constant current 0.0500 A. Assume Zn→Zn²⁺+2 e⁻, 100% useful-current efficiency, M(Zn)=65.38 g mol⁻¹ and F=96485.3 C mol⁻¹. Estimate the capacity-limited lifetime.

Now the natural current generated by the galvanic cell formed by two metals in water is typically very low. In fact, this current cannot carry electrons that far across the hull, so for a larger ship, sometimes an artificial current needs to be generated. These artificial currents are generated with a DC generator.

The use of an artificially induced current can overcome many of the trials of passive cathodic protection, however, the increased current can also cause shorter lifespans of the sacrificial anodes. The larger currents are needed, however, in cathodic protection of structures that are not immersed in the salt water, which is a pretty darn good electrolyte. Pipelines are typically required ro be protected from corrosion, as a pipeline leak can cause a lot of environmental damage and/or just be really expensive for the owner of the pipeline.

Inducing a current on these larger structures doesn’t pass the generated current across the entire surface. Instead, it creates what is called a current density, which refers to the current associated with a certain area of the cathode to be protected.

Let’s say you have an induced current across a long pipeline. If a current density is created of 600mA/m2, then that means that for every square meter of pipeline, 600mA of current are present.

You used impressed current cathodic protection to protect a bridge which had a surface area of 1400m2, and the current density on the bridge was 800mA/m2. How can you determine the current that runs through your anode? Well, you’ll need to figure out how much of the surface you anode is using. If you had a single anode for the entire bridge, then the anode would be working on 1400m2 of bridge, passing a current of 1400*800=1120A. That is a huge current. In order to mitigate that, you use multiple anodes, spread evenly across the bridge. If you had 1 anode for each 10m2 of bridge, then your current is only 800mA/m2 * 10 m2 = 8A. Quite a bit lower.

Example Problem 9.2.4

Reviewed ideal capacity task: a 10.0 m long cylindrical pipe segment has diameter 5.00 m and stipulated uniform current density 0.300 A m⁻² over its lateral area. Its Zn anode has 10.0 kg usable mass. Assume Zn→Zn²⁺+2 e⁻, 100% utilization/current efficiency, M(Zn)=65.38 g mol⁻¹ and F=96485.3 C mol⁻¹. Estimate lifetime.