Lesson Explainer: The Faraday Constant Chemistry

In this explainer, we will learn how to use the Faraday constant to calculate the mass or volume of substances liberated during electrolysis.

In order to understand the Faraday constant, we need to properly understand charge, the flow of electrons, and current.

Electrons have a property called charge. A particle with charge will attract or repel other particles with charge, depending on whether the charges have the same sign (+/+ or /) or opposite signs (+/) respectively.

The charge of an electron is extremely small from our perspective. Chemists often do not worry about charge in the way physicists do. We “count” charge as in the “charge of a proton” or “charge of an electron.” For example, a charge of 1+ is “the charge that 1 proton has” and a charge of 2 is “the charge that 2 electrons have.” Of course, electrons and protons have equal and opposite charges, so you can define these charges in multiple ways.

However, this level of simplicity is not appropriate for electrochemistry. We need to use units of charge that connect with other units such as grams, volts, and joules.

The common unit of charge is the coulomb, with symbol C. 1 coulomb is equivalent to the magnitude of the charge of 6.24×10 electrons (about six billion billion electrons). 16240000000000000000coulombmagnitudeofthechargeofelectrons

The charge of 1 electron is exactly 1.602176634×10 coulombs (using the current definition); however, it is usually rounded to 1.602×10 C.

Unfortunately, there is a slight complication in that 1.602176634×10 C is given the symbol 𝑒 (for elementary charge). This is the charge of 1 proton, not one electron, so the collision between chemistry and physics leads us to this: chargeofanelectronchargeofaprotonelementarycharge== or chargeechargep()=()=𝑒.+

This may seem irrelevant, but it is important that we know that 𝑒 (the elementary charge), e (the symbol for an electron), and 𝑒 (the charge of an electron in coulombs) are not the same thing.

Definition: Elementary Charge, 𝑒

The elementary charge, 𝑒, is the charge of 1 proton, in coulombs: 𝑒=1.602176634×10𝑒=1.602×10(3).CCd.p.

Definition: Coulomb (Unit of Charge), C

1 coulomb is equivalent to the charge of 6.2415090744×10 protons: 1=6.2415090744×10𝑒1=6.24×10𝑒(2).CCd.p.

Before 2019, the coulomb was defined based on the current necessary to achieve a given force between parallel wires. It is now defined in other ways that give basically the same result.

Now that we have considered charge and coulombs, we will develop our understanding of current. A current (in a wire) results from the flow of electrons. Current is measured in amperes.

1 ampere (abbreviated to 1 amp) is equivalent to the flow of 6.24×10 electrons every second. Recall that the charge of 6.24×10 electrons is equal in magnitude to 1 coulomb and so 1 amp is equal to a flow of 1 coulomb per second.

Definition: Ampere (amp), A

1 ampere is equivalent to the flow of 1 coulomb per second: 1=1/.ACs

A complication is that current is defined as a flow of positive charge around a circuit, but electrons have negative charges. However, for the purposes of a circuit, a negative charge moving in one direction can be treated like a positive charge moving in the opposite direction. This is shown in the figure below.

When electrons flow around a circuit, it is unlikely to be for just 1 second, so we have another relationship to define, that between the current, time, and total charge of the electrons that pass a given point.

A current is a flow of positive charge and it is given the symbol 𝐼.

The flow of charge corresponding to a current is given the symbol 𝑄.

The time a current flows for is given the symbol 𝑡.

This is the relationship between the two is shown below.

Equation: Charge Delivered by a Current over Time

A current 𝐼 lasting for a time 𝑡 will involve the flow of charge 𝑄: 𝑄=𝐼𝑡.

If we want to work out the number of electrons flowing during that time, we can divide the total charge, 𝑄, by the elementary charge, 𝑒: Numberofelectronsinvolved=𝑄𝑒.

Example 1: Calculating the Total Charge Delivered by a Wire given the Current and the Time

A current of 0.2 A passes through a wire for a period of 60 s. Calculate the total charge delivered by the wire.

Answer

A current arises from the flow of electrons. An ampere (A), or amp, is equivalent to 1 coulomb (a unit of charge) passing a point each second (a unit of time).

For a current of 0.2 A, 0.2 C is delivered each second, so we can calculate the total charge delivered in 60 s: totalchargedeliveredcurrenttimeAsAsC(𝑄)=(𝐼)×(𝑡)𝑄=0.2×60=12=12.

An ampere-second is equivalent to a coulomb, since an amp is equivalent to a coulomb per second: 1=1/1=1/=1.ACsAsCssC

Electrons do have a negative charge; however, current is considered as the equivalent flow of positive charge in the opposite direction to the electron flow, so 𝑄 is positive.

The total charge delivered by a current of 0.2 A in 60 s is 12 C.

Example 2: Calculating the Charge Delivered by a Set of Currents and Identifying the Current Delivering the Greatest Charge

Which of the following electrical currents would deliver the greatest charge?

  1. 20 A for 0.15 s
  2. 0.25 A for 15 s
  3. 0.05 A for 60 s
  4. 1.0 A for 1.0 s
  5. 0.10 A for 35 s

Answer

The charge (in coulombs, C) delivered by a current is equal to the current (in amps, A) multiplied by the time the current is flowing (in seconds, s).

This can be expressed by the equation chargecurrenttime(𝑄)=(𝐼)×(𝑡).

To work out the charge for the five currents, we multiply the current by the time for all the candidates and get the following charges.

Current (A)Time (s)Charge (C)
A200.153
B0.25153.75
C0.05603
D1.01.01
E0.10353.5

The current that delivers the greatest total charge is current B (0.25 A for 15 s).

The correct answer is B.

Chemists are not used to dealing with 6.24×10 electrons or protons; they prefer to use moles.

1 coulomb is equivalent to the charge of 0.00001036426966 moles of protons (1.04×10).

Definition: Mole, mol

The mole is the SI unit of the amount of a substance (𝑛) 1=6.02214076×101=6.022×10.molentitiesmolentities(rounded)

Entities can be particles, atoms, molecules, ions, and so on.

Example: 1=6.02214076×10molelectronselectrons.

This leads us to the Faraday constant (𝐹), an application of Faraday’s first and second laws of electrolysis. Michael Faraday (1791–1867) was a key figure in the development of electrochemistry.

Initially, Faraday observed that the longer he passed electricity through an electrolyte, the greater the amount of substance that was produced.

Definition: Faraday’s First Law of Electrolysis

The mass, 𝑚, of a substance deposited at an electrode in grams is directly proportional to the amount of charge, 𝑄, passed in coulombs, C.

The Faraday constant (𝐹), allows us to translate between coulombs and moles. In other words, it is the number of coulombs per mole of elementary charges.

We know the value of 1 elementary charge in coulombs, 𝑒=1.602176634×10,C so we can work out the charge of 1 mole of elementary charges by multiplying 1𝑒mol by Avogadro’s number. Avogadro’s number is the number of entities in 1 mole of entities: AvogadrosnumberCchargeofmolincoulombsCCCC=6.02214076×10𝑒=1.602176634×10,1𝑒=1.602176634×10×6.02214076×10=96485.33212331=96485.3321=96500.

Since we know the charge of 1 mol of elementary charges in coulombs, we also know the value per mole. This is the Faraday constant.

Definition: Faraday Constant, 𝐹

The Faraday constant is the charge per mole of elementary charges: 96485.3321/96500/.CmolCmol(rounded)

In calculations, we often round the Faraday constant (𝐹) to 96‎ ‎500 C/mol as this level of precision is often sufficient. If we are going to use 𝐹 in a calculation that could give a result more precise than this, we should use a more precise value of 𝐹.

Faraday’s constant is useful as it gives us a conversion factor between moles of elementary charges and coulombs.

If we have 1 mole of elementary charges, we can multiply by the Faraday constant to work out the total charge in coulombs: chargemolmolmolCmolC(1𝑒)=1×𝐹=1×96500/=96500.

If we want, we can also calculate the Faraday constant from Avogadro’s constant (𝑁). Avogadro’s constant is slightly different from Avogadro’s number, in that it is the number of entities per mole (rather than just for 1 mole): 𝐹=𝑒×𝑁=1.602176634×10×6.02214076×10=96485.3321/=96500/.CmolCmolCmol(rounded)

We can use the Faraday constant to bridge between the charges associated with a current as well: totalchargenumberofmolesofelectronsthatowedtheFaradayconstant=×𝑄=𝑛𝐹.

There is a unit of charge that also bears Faraday’s name other than the Faraday constant. Be careful to distinguish between the two—confusingly, the letter F is used for both.

Definition: Faraday (Unit of Charge)

The faraday is the total charge of 1 mole of protons: 1=96485.3321=96500.faradayCC(rounded)

For simplification, we use a value of 96‎ ‎500 C for the faraday and a value of 96‎ ‎500 C/mol for Faraday’s constant.

Later, Faraday developed his ideas to take into account the fact that different metals form differently charged ions. He explained why, for example, turning copper ions (Cu2+), which require two electrons, into copper atoms would require more electricity than turning an equivalent quantity of silver ions (Ag+) into atoms.

Definition: Faraday’s Second Law of Electrolysis

The mass, 𝑚, of a substance deposited at an electrode in grams is directly proportional to its chemical equivalent weight.

Definition: Equivalent Weight (Gram Equivalent)

The equivalent weight is the mass of a substance that exactly reacts with a fixed quantity of another substance and corresponds to the atomic weight of the substance divided by the valence.

In the diagram below, we can see how the Faraday constant and the quantities we have been discussing are related to each other. Faraday’s second law of electrolysis means we must be mindful of how many electrons are involved in an oxidation or reduction reaction. From there, we are able and work out more familiar quantities, like the length of time required to perform a particular reaction.

How To: Calculating the Time Needed to Deliver the Electrons Required for a Given Reduction Reaction

Shown below is the reduction of a hydrogen ion, part of the electrochemical splitting of water molecules into hydrogen and oxygen gases: H+eH+212

Each hydrogen ion requires 1 electron, but how many amps are needed to deliver the electrons necessary to make 1 gram of hydrogen gas in 1 minute?

To start with, 1 gram of hydrogen gas consists of 1 mole of hydrogen atoms: molarmassofhydrogengmolHamountofhydrogenmassofhydrogenmolarmassofhydrogenggmolHmolH=1/(𝑛)=(𝑚)(𝑀)=11/=1.

These hydrogen atoms are part of hydrogen molecules (H2), so there are 0.5 moles of hydrogen molecules in 1 gram of hydrogen gas.

For each hydrogen atom produced by the reduction reaction, we must have a hydrogen ion (H+) to start with, so we must begin with 1 mole of hydrogen ions: 𝑛()=𝑛×=1×=1.eH1e1HmolH1e1Hmole++++

1 mole of hydrogen ions requires 1 mole of electrons to be reduced to 1 mole of hydrogen atoms.

In total, we need 1 mole of electrons from our current.

The strength of an electrical current is usually measured in amps. 1 amp is equal to the flow of 1 coulomb per second.

We can use Faraday’s constant to convert from the amount of electrons in moles to the charge our current must deliver: 𝐹=96500/,𝑄=𝑛𝐹=1×96500/=96500.Cmol(rounded)moleCmolC(rounded)

Note that the sign of 𝑄 is opposite to the total charge of the electrons because the conventional current is defined in the opposite direction to the flow of electrons.

We know that in order to generate 1 gram of hydrogen gas, we need the current to deliver 96‎ ‎500 C in 1 minute. We need to convert the time to seconds because amps are defined using the second rather than the minute.

We can now work out the current using 𝑄=𝐼𝑡: 𝑄=𝐼𝑡𝐼=𝑄𝑡=9650060=1608.33/=1608.33=1610.CsCsAA(rounded)

In order to produce 1 g of H2 in 1 minute, by the reduction of H+, we would need to use a minimum current of 1‎ ‎610 A. In the real world, a higher current would be needed to compensate for inefficiencies.

Example 3: Calculating the Mass of Aluminum Produced from a Hall–Heroult Cell given the Current and Aluminum’s Molar Mass

The industrial extraction of aluminum using Hall–Heroult cells requires a very large electrical current of 120 kA. How much aluminum is produced per hour, taking the molar mass of aluminum to be 27.0 g/mol and one faraday of charge as 9.65×10 C? Give your answer in kilograms, to 2 decimal places.

Answer

To answer this question, we firstly have to assume that 100% of the current used in the cell goes into producing aluminum (so there are no losses or inefficiencies in the system).

The Hall–Heroult process involves the electrolysis of aluminum oxide (mixed with a few additives that reduce the melting point). So, we can construct the overall equation 2AlO()4Al()+3O()232llg and the two half-equations Al+3eAlOO+2e3+2212

The half-equation shows it takes 3 electrons to reduce one aluminum ion to an aluminum atom.

We can now work out how many atoms of aluminum we can expect to be produced in 1 hour, using a current of 120 kA.

We could work out the number of atoms directly, but it is easier if we recall that one faraday of charge is equivalent to the magnitude of the charge of 1 mole of electrons. Therefore, for each mole of aluminum ions we want to reduce, we will need our current to deliver 3 faradays of charge.

Therefore, we will first work out how many faradays of charge are delivered by a current of 120 kiloamps in 1 hour: chargecurrenttime(𝑄)=(𝐼)×(𝑡).

For this equation, current is in amps (A) and time is in seconds (s), so the first thing we need to do is calculate the number of seconds in 1 hour: secondsinhourhminhsmins1=1×601×601=3600.

Next we can multiply the current by the time to get the charge delivered by the current: 𝑄=120000×3600=432000000.AsC

The coulomb (C) is equivalent to the ampere-second (A⋅s): 1()9.65×10.faradayFisequivalenttoC(rounded)

Using this, we can determine how many faradays are equivalent to the charge we calculated: 432000000×196500=4476.68.CFCF

Since it takes 3 faradays to generate each mole of aluminum atoms, we need to divide this figure by 3: 4476.68×13=1492.23.FmolAlFmolAl

The last step is to convert the amount of aluminum in moles to the mass of aluminum in kilograms. The molar mass of aluminum has been given as 27.0 g/mol, so let’s get the mass in grams first: 1492.23×27.01=40290.2.molAlgmolAlg

Converting this to kilograms gives 40290.2×11000=40.2902.gkggkg

Therefore, the final answer for the mass of aluminum produced in 1 hour, in kilograms, to 2 decimal places is 40.29 kg.

Key Points

  • A coulomb (C) is a unit of charge that can be expressed as a number of elementary charges (𝑒, the charge of a proton): 1=6.24×10𝑒.C(rounded)
  • A current is a flow of charge.
  • An amp (A) is the conventional unit for current; 1 amp is equivalent to a flow of 1 coulomb per second: 1=1/.ACs
  • The total charge, 𝑄, delivered by a current is equivalent to the current, 𝐼 (in amps), multiplied by the time it flows, 𝑡 (in seconds): 𝑄=𝐼𝑡.
  • The Faraday constant is the amount of charge per mole of elementary charges: TheFaradayconstantCmol(rounded)=96500/.
  • A faraday (F) is a unit of charge; 1 faraday is equivalent to the charge of 1 mole of elementary charges: 1=96500.FC(rounded)
  • Using the Faraday constant, molar masses, and our knowledge of how many electrons are required to reduce particular ions, we can calculate the charge required to produce amounts of material by electrolysis.

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