Redox Processes

This chapter discusses Electromotive Force and Cell Potential (part 1) along with an explanation of Electrolytic Processes (part 2).

EMF and the Standard Cell Potential

Electromotive Force and Potential Difference

Figure 1. Diagram of an object with a higher potential energy than its lower counterpart.

Electromotive force (EMF) is the energy supplied by a source divided by the electric charge transported through the source.
EMF = Energy in Joules (J) / Charge in Coulombs (C)
EMF is measured in Volts (Voltage).
Potential difference is the difference in the amount of potential energy in two different points in space or the difference in potential energy that two objects at different points possess.
For example, seen in Figure 1, the higher object has a higher potential energy than the lower one, as the gravitational potential energy of the top one is higher, as the top one has a higher clearance/height from the ground.
The difference in potential energy between them is potential difference.
In terms of circuits and electrons, potential difference is the difference in the amount of electrical energy in two different points in a circuit.
Once a suitable pathway is present, electrons will always flow from the highest PE to the lowest PE.
In a voltaic cell, the EMF is equal to the potential difference for zero current going through the cell. The EMF is the maximum voltage that can be delivered by the cell.
The Youtube video below offers additional knowledge on potential difference using a similar analogy, and further discusses EMF.

The Cell Potential and Electrode Potentials

In a voltaic cell, a cell potential is generated, causing electrons from the anode (negative electrode) to move to the cathode (positive electrode). There is a current when a cell potential is present.
The cell potential of a voltaic cell is therefore defined as the potential difference between the cathode and the anode when the cell is operating (there is a current). Therefore, the cell potential is always less than the maximum voltage that can be delivered by the cell, as there is a current.
The formula for cell potential, or standard cell potential is:

Figure 2. IB Data Booklet Section 24.

E (cathode) represents the standard electrode potential at the cathode.
E (anode) represents the standard electrode potential at the anode.
Standard electrode potentials can be found in Section 24 of the IB Data Booklet. The table is included in Figure 2.
Standard electrode potentials are agreed by the scientific community to be expressed as reduction processes.
Since the cathode is the species with the more positive standard electrode potential value, the reduction equation of the species with the less positive standard electrode potential value (anode) must be reversed when calculating the overall cell equation. The equation must represent an oxidation reaction.
This is because oxidation occurs at the anode, and since all electrode potentials are shown as reduction processes, the electrode potential half equation must be reversed.
This is shown in the sample problem below.

An exam-style question may formulate a similar question differently:

The Standard Hydrogen Electrode

As described above, in a cell with two species, the half equations and electrode potential values of the species are used to derive the cell potential and equation.
However, how are the electrode potential values of the species derived in the first place?
It is impossible to measure the electrode potential of a species using a a cell with only the lone species, as to measure the cell potential there must be a potential energy difference, which requires two species.
This is where the Standard Hydrogen Electrode (SHE) comes into play.
The SHE has been internationally chosen as the reference to which the electrode potentials of all species are measured against. It has an electrode potential of 0 V.
In a galvanic cell, a SHE, which consists of an inert platinum electrode in contact with 1 mol/dm^3 hydrogen ions and hydrogen gas at 100 kPa, is placed at one electrode.
The electrode potential of a chosen species is determined by connecting the half cell, under standard conditions, to the SHE using a connecting wire with a voltmeter to deduce the voltage, and a salt bridge.
The cell potential of the cell, and thus the electrode potential of the chosen species, can then be found.
An example with copper from a useful Youtube video is shown below.
In the example, below, cell potential = electrode potential of the copper species.
Cell potential = 0.34 V - 0 V = 0.34 V

The Youtube video offers more knowledge about the SHE and the practical processes of determining electrode potentials. Click on the image to watch.

Gibbs Free Energy

The textbook mentions another form of the Gibbs Free Energy formula, which relates it to cell potential.

n = amount, in mol, of electrons transferred in the balanced equation.
F = Faraday's constant ( 96500 C/mol, given in Section 2 of IB Data Booklet)
E (cell) = cell potential in voltage
If negative, the reaction is spontaneous (galvanic cell).
If positive, the reaction is non-spontaneous (electrolytic cell).
For example, with the Daniell cell example...

Test Your Understanding

Complete this worksheet on EMF, Cell Potential, SHE, and Gibbs Free Energy to ensure you understand this section of Chapter 19. The worksheet mostly consists of fill-in-the-blank questions and some normal response ones as well. The PDF version is below.

DOWNLOAD 19.1 ...out .pdf

Different types of Electrolytic Cells

Electrolysis of Aqueous NaCl

The figure above shows an idealized drawing of a cell in which an aqueous solution of sodium chloride is electrolyzed.

The Na+ ions migrate toward the negative electrode and the Cl- ions migrate toward the positive electrode. But, now there are two substances that can be reduced at the cathode: Na+ ions and water molecules.

Cathode (-):

Na+ + e- → Na E ° red = -2.71 V

2 H2O + 2 e- → H2 + 2 OH- E ° red = -0.83 V

Because it is much easier to reduce water than Na+ ions, the only product formed at the cathode is hydrogen gas.

Cathode (-): 2 H2O(l) + 2 e- →H2(g) + 2 OH-(aq)

There are also two substances that can be oxidized at the anode: Cl- ions and water molecules.

Anode (+):

2 Cl- → Cl2 + 2 e- E °ox = -1.36 V

2 H2O → O2 + 4H+ + 4 e- E °ox = -1.23 V

The standard-state potentials for these half-reactions are so close to each other that we might expect to see a mixture of Cl2 and O2 gas collect at the anode. In practice, the only product of this reaction is Cl2.

Anode (+): 2 Cl- → Cl2 + 2 e-

At first glance, it would seem easier to oxidize water (E °ox = -1.23 volts) than Cl- ions (E °ox = -1.36 volts). It is worth noting, however, that the cell is never allowed to reach standard-state conditions. The solution is typically 25% NaCl by mass, which significantly decreases the potential required to oxidize the Cl- ion. The pH of the cell is also kept very high, which decreases the oxidation potential for water. The deciding factor is a phenomenon known as overvoltage, which is the extra voltage that must be applied to a reaction to get it to occur at the rate at which it would occur in an ideal system.

Under ideal conditions, a potential of 1.23 volts is large enough to oxidize water to O2 gas. Under real conditions, however, it can take a much larger voltage to initiate this reaction. (The overvoltage for the oxidation of water can be as large as 1 volt.) By carefully choosing the electrode to maximize the overvoltage for the oxidation of water and then carefully controlling the potential at which the cell operates, we can ensure that only chlorine is produced in this reaction.

In summary, electrolysis of aqueous solutions of sodium chloride doesn't give the same products as the electrolysis of molten sodium chloride. Electrolysis of molten NaCl decomposes this compound into its elements.

Electrolysis

2 NaCl(l) → 2 Na(l) + Cl2(g)

Electrolysis of aqueous NaCl solutions gives a mixture of hydrogen and chlorine gas and an aqueous sodium hydroxide solution.

Electrolysis

2 NaCl(aq) + 2 H2O(l) → 2 Na+(aq) + 2 OH-(aq) + H2(g) + Cl2(g)

Because the demand for chlorine is much larger than the demand for sodium, electrolysis of aqueous sodium chloride is a more important process commercially. Electrolysis of an aqueous NaCl solution has two other advantages. It produces H2 gas at the cathode, which can be collected and sold. It also produces NaOH, which can be drained from the bottom of the electrolytic cell and sold.

The dotted vertical line in the above figure represents a diaphragm that prevents the Cl2 produced at the anode in this cell from coming into contact with the NaOH that accumulates at the cathode. When this diaphragm is removed from the cell, the products of the electrolysis of aqueous sodium chloride react to form sodium hypo-chlorite, which is the first step in the preparation of hypochlorite bleaches, such as Chlorox.

Cl2(g) + 2 OH-(aq) → Cl-(aq) + OCl-(aq) + H2O(l)

Electrolysis of Copper Sulfate

The figure above shows the electrolysis of copper(II) sulfate solution using graphite electrodes. Close inspection of the cathode reveals a pinky-brown coating of copper caused by the reduction of copper(II) ions to form solid copper according to the following equation:

Cu2+ (aq) + 2e- → Cu (s)

The electrolysis of copper(II) sulfate produces pure copper at the cathode.

At the anode, water is oxidised to form oxygen gas:

2H2O (l) → O2 (g) + 4H+ (aq) + 4e-

If the graphite electrodes are replaced by copper electrodes, the reaction taking place at the cathode remains the same - copper(II) ions are reduced - but at the anode the copper electrode itself undergoes oxidation:

Cu (s) → Cu2+ (aq) + 2e−

This reaction is used in the refinement of impure copper. The anode is made from impure copper and pure copper is deposited on the cathode. Effectively, there is a net transfer of copper from the impure sample to the cathode via the copper(II) sulfate solution. The overall reaction that takes place is as follows:

Cu (s) + Cu2+ (aq) → Cu2+ (aq) + Cu (s)

Electrolysis of water

A standard apparatus for the electrolysis of water is shown.

Electrolysis:

2 H2O(l) → 2 H2(g) + O2(g

A pair of inert electrodes are sealed in opposite ends of a container designed to collect the H2 and O2 gas given off in this reaction. The electrodes are then connected to a battery or another source of electric current.

By itself, water is a very poor conductor of electricity. We therefore add an electrolyte to water to provide ions that can flow through the solution, thereby completing the electric circuit. The electrolyte must be soluble in water. It should also be relatively inexpensive. Most importantly, it must contain ions that are harder to oxidize or reduce than water.

2 H2O + 2 e- → H2 + 2 OH- E ° red = -0.83 V

2 H2O→ O2 + 4 H+ + 4 e- E ° ox = -1.23 V

The following cations are harder to reduce than water: Li+, Rb+, K+, Cs+, Ba2+, Sr2+, Ca2+, Na+, and Mg2+. Two of these cations are more likely candidates than the others because they form inexpensive, soluble salts: Na+ and K+.

The SO42- ion might be the best anion to use because it is the most difficult anion to oxidize. The potential for oxidation of this ion to the peroxydisulfate ion is -2.05 volts.

2 SO4 2- → S2O8 2- + 2 e- E ° ox = -2.05 V

When an aqueous solution of either Na2SO4 or K2SO4 is electrolyzed in the apparatus shown in the above figure, H2 gas collects at one electrode and O2 gas collects at the other.

What would happen if we added an indicator such as bromothymol blue to this apparatus? Bromothymol blue turns yellow in acidic solutions (pH < 6) and blue in basic solutions (pH > 7.6). According to the equations for the two half-reactions, the indicator should turn yellow at the anode and blue at the cathode.

Cathode (-): 2 H2O + 2 e- → H2 + 2 OH-

Anode (+): 2 H2O → O2 + 4 H+ + 4 e-

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For Review you can download the notes I made