FFR: Faraday’s Laws

During an electrochemical reaction, one electrode undergoes electrochemical corrosion. That means that the electrode is losing mass as it reacts. Typically, in a metal-metal cell, the anode degrades as it gets oxidized. Let’s take a look at our favorite galvanic cell, the Zinc-Copper cell.

Zinc has a Standard Electrode Potential of -0.7618V. Copper’s is 0.34V. Therefore, copper gets reduced and zinc gets oxidized. The reaction is:

$Cu^{2+}_{(aq)}+Zn_{(s)}\rightarrow Cu_{(s)}+Zn^{2+}_{(aq)}$

As you can see, the copper ions initially in solution become solid copper once getting reduced. Zinc, however, starts out as a solid electrode, and then gets oxidized into zinc ions in solution. Through this process, the zinc electrode loses mass as it gets oxidized, causing a decrease in the mass of the electrode.

So how can we determine how long it takes for the anode to fully degrade? Or perhaps how big a sacrificial anode is needed for cathodic protection.

For this, we turn to the two laws developed by Michael Faraday. The two laws are very similar, with slight differences.

As he originally put it, Faraday’s first law reads: “...that when subjected to the influence of the electric current, a quantity of [water] is decomposed exactly proportionate to the quantity of electricity which has passed”.

In other words, the mass of an electrode affected is directly proportional to the amount of electricity that passes through it. Put simply, the amount of material affected by electrolysis is dependent on how much electricity is used. If we want to apply this to solve problems, we have to use it to formulate an equation.

The law states that mass of the water is exactly proportional to the amount of electricity passed through.

Therefore, we can start with:

$m \propto[Electricity]$

We know that this is exactly proportional, so we just need to work with units to develop our equation. Let’s start with electricity. When Faraday states “quantity of electricity” he refers to the amount of charge that passes through. That is typically represented by $Q$ and given in coulombs.

$m\propto Q$

Okay, so now we need to get from coulombs to units of mass. Let’s say we are working with a half-cell whose equation is:

$A^+_{(aq)}+e^-\rightarrow A_{(s)}$

Now we can say that we are trying to determine the relationship between the mass of electrode affected and the charge passed through the electrode.

$m_A\propto Q$

Charge in electrochemistry is carried by the electrons moving from one electrode to the other. This means we can change $Q$ into the number of electrons. The relationship between that is:

$n_{e^-} = \frac QF$

Where $n_e$ is the number of electrons in moles. Now we need to relate the moles of electrons to the mass of electrode. We turn to the reaction for this.

$A^+_{(aq)}+e^-\rightarrow A_{(s)}$

As you can see, in this particular half-cell, we have a 1:1 molar ratio of electrons:electrode. Keeping with our general terms, we will call the number of electrons $n$.

We can use this relationship to reach mole of electrode affected.

$n_A = \frac{n_{e^-}}n$

So now we are at:

$m\propto \frac QF\cdot\frac 1n$

We are nearly there. We are at the moles of electrode. If we want to convert moles to mass, we just need to use the molar mass of the electrode.

$m_A=M_A\cdot n_a$

This puts us at:

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

As the law states exactly proportional, we can now make this an equality:

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

This equation relates the mass of the electrode degraded or deposited to the the amount of electricity that is passed through the electrode.

Faraday’s first law asserts that a given compound will be predictably affected by electricity depending on the amount of electricity used. His second law generalizes that even further, stating that the same amount of electricity will act upon the same equivalent mass of different compounds.

The big difference between this law and the first law is that the amount of electricity is constant, but the amount of electrode affected is not. The equivalent mass for a compound is the ratio of molar mass to moles of electrons. This is because the same amount of electricity (the same number of electrons) can only react with half the oxygen as it can silver. This is due to their half-reactions:

$O_{2(g)}+4e^-+4H^+\rightarrow2H_2O$

$Ag_{(aq)}^++e^-\rightarrow Ag_{(s)}$

As you can see, silver only needs one electron, but there are two electrons required per oxygen atom, and 4 per oxygen molecule. This means that the same amount of electricity that would degrade one mole of silver would only react with one fourth mole of oxygen.

This law uses the same mathematical formula as the first law, but we are holding different values constant. With the first law, we were varying the charge to solve for the mass. With the second law, we keep charge constant and change the equivalent mass to solve for the mass affected.

Example Problem 8.4.1

Reviewed Zn/Cu galvanic model: initial usable zinc mass is stipulated as 20.00 g. The cell transfers 3.00×10³ C, entirely through Zn→Zn²⁺+2 e⁻ with 100% useful-current efficiency. Use F=96485 C mol⁻¹ and M(Zn)=65.38 g mol⁻¹. Find remaining zinc mass.

One common situation involved in problems regarding Faraday’s laws is the constant-current situation.

Current, represented as $I$, is a measure of the flow of electricity. It’s base unit is the Ampere, $A$, which is the equivalent of $1\frac Cs$. That means that in a cell which outputs $1A$ of current, there will be $1C$ of charge put off over $1s$. Generally, we can use this equation to relate current to charge:

$Q=I\Delta t$

This equation only holds true for constant-current situations, but under those situations, the equation for Faraday’s laws can be modified to:

$m_A=\frac {I\Delta t}{F}\frac{M_A}{n}$

This allows us to predict the lifetime of a cell that undergoes a constant current by determining when the electrodes will be used up. This assumption of a constant current is fairly invalid, but we can think of it as an average current throughout the lifetime of the cell.

Example Problem 8.4.2

Reviewed ideal lead–acid discharge model: 10.0 kg usable Pb and 10.0 kg usable PbO₂ are available with excess acid. Each half-reaction transfers two electrons; current is constant at 5.00×10² A. Assume 100% utilization and Faradaic efficiency, M(Pb)=207.2 and M(PbO₂)=239.2 g mol⁻¹, F=96485 C mol⁻¹. Find active-material-limited lifetime.

Batteries are sometimes rated in terms of the charge they can provide. It is a way of determining the life-time of the battery without requiring a constant current assumption. Battery ratings are typically given in $\text{A}\cdot\text{hr}$. Since we typically deal with charge in coulombs, let’s convert that.

$1\text{A}\cdot\text{hr}=x \text C$

Using the definition of Amps we discussed earlier:

$1\text{A}\cdot\text{hr}=1\frac{\text{C}}{\text{s}}\cdot\text{hr}=x \text C$

$1\text{A}\cdot\text{hr}=1\frac{\text{C}}{\text{s}}\cdot60\text{min}=x\text C$

$1\text{A}\cdot\text{hr}=1\frac{\text{C}}{\text{s}}\cdot3600\text{s}=x\text C$

$1\text{A}\cdot\text{hr}=3600\text{C}$

There’s our conversion factor. This lets us convert $Q$ between $\text A \cdot \text{hr}$ and $\text{C}$.

Example Problem 8.4.3

What is the rating (in $\text A\cdot\text{hr}$) of a zinc-copper battery that starts fresh with 4g of zinc and 6g of copper in its ionic form?

Chapter Problems

1. Reviewed replacement: at 298.15 K, a hypothetical M²⁺/M reduction half-cell is measured at E=−3.220 V relative to a same-temperature SHE, with a(M²⁺)=2.00 and pure M activity 1. Find E°, using E=E°−RT ln Q/(nF), R=8.31446 and F=96485.3 in SI units.
   Review the authored solution

   For M²⁺+2 e⁻→M, n=2 and Q=1/2. E=E°+(RT/2F)ln2. Thus E°=−3.220−0.008904=−3.228904 V≈−3.229 V. The explicit sign, reference, temperature and formal species model supersede the incomplete gallium record.