Electrochemistry

An electrochemical cell converts chemical energy into electrical energy using a redox reaction.

Since, metals can be oxidised or reduced depending upon their chemical environment, then such an arrangement as shown below may be set-up.

The Daniell cell, specifically, uses Zn_(s)/Zn²⁺_(aq) and Cu_(s)/Cu²⁺_(aq) reactions. Note that the two rods in the diagram are called electrodes.

The zinc rod is the negative electrode and the copper rod the positive electrode.

Each metal in contact with a solution of its ions is called a half-cell. Half-cells are often represented by half-cell equations, which show the electrode processes:

At the negative electrode: Oxidation

Zn_(s) → Zn²⁺_(aq) + 2e^-

At the positive electrode: Reduction:

Cu²⁺_(aq) + 2e^- → Cu_(s)

So the overall reaction equation is:

Remember this is a REDOX reaction.

Note in the diagram:

1. The electrons flow in a clockwise direction, from the zinc rod to the copper.
2. The salt bridge, completes the circuit by allowing the passage of ions from the copper sulphate solution to the zinc sulphate solution. This salt bridge is usually a strip of filter paper soaked in saturated potassium sulphate.

Electrochemical cells can be made, as long as you pair up two half-cells of different potential so that an electrical current can be produced.

How do we find the potential of any half-cell individually? The answer is that we use a standard hydrogen electrode (s.h.e) and give it a potential of zero.

Therefore, when it is connected to another half-cell, the electromagnetic field (e.m.f) between the s.h.e and the second half-cell is equal to the potential of the second half-cell.

The diagram below outlines the main characteristics and conditions required to establish the s.h.e:

The reaction that takes place is:

2H⁺_(aq) + 2e^- → 2H_{2 (g)}

We can place redox systems in order of their oxidising/reducing ability if we measure their electrode potential against the standard hydrogen electrode, which has a potential of 0.00V.

Note: Standard conditions must be adhered to, so for example if using copper rods they must be immersed into a solution of 1 mol dm^-3 copper ions.

By convention, we define the e.m.f of an electrochemical cell as follows:

Eθ = Eθ right-hand half-cell - Eθ left-hand half-cell.

The s.h.e is always placed on the left-hand side.

The standard electrode potential of a metal is the potantial acquired when the metal is immersed in a 1 mol dm^-3 solution of its ions at a temperature of 25^oC.

It has the symbol Eθ and the units V.

As stated previously, we can tabulate the order of oxidising/reducing ability of a system, this we call the electrochemical series.

The most positive E value is at the top, i.e. the greatest oxidising agent. The most negative value E at the bottom i.e. the greatest reducing agent.

Note: These reactions are in equilibrium, they can go either way.

To calculate the standard cell potential, from the standard electrode potentials we do the following:

Eθ cell = Eθ right hand half-cell - Eθ left-hand half cell.

The left-hand half cell is conventionally the oxidising half-cell

For example:

Eθ cell = Eθ Cu²⁺/Cu - Eθ Zn²⁺/Zn

= +0.34 - (-0.76) V

= +1.1V

This is the shorthand way of representing an electrochemical cell.

For example:

Daniell cell:

The continuous vertical line represents the phase boundary. The broken vertical line represents the salt bridge.

We can predict the likelihood of a reaction if two systems in the electrochemical series are linked by cells. Remember, the system which is lower in the series will lose electrons, and the one higher in the series will gain electrons.

For example:

Cell reaction:

This carbon-zinc dry cell is usually called a battery, however it is in fact a cell, with an e.m.f. of about 1.5V.

They are cheap and portable with a relatively short life, they must be thrown away when used up.

The electrode reactions that are occurring in this cell are as follows:

At the negative electrode:

Zn_(s) → Zn²⁺_(aq) + 2e^-

At the positive electrode

2NH₄⁺_(aq) + 2e^- → 2NH_3(g) + H_2(g)

The H₂ produced is removed by the reaction with manganese (IV) oxide:

H_2(g) + 2MnO_2(s) → MnO_3(s) + H₂O_(l)

Lead-acid batteries are most commonly found in cars. They can be recharged since they are made up of secondary cells. A typical battery is made up of 6 cells, each with an e.m.f of 2V.

Acting as electrodes are plates that separate the cells, immersed in dilute sulphuric acid.

The negative electrodes are plates made from lead, the positive electrodes are plates made from lead coated in lead (IV) oxide.

At the negative electrode:

At the positive electrode:

PbSO₄ is made by both reactions. When the car moves, an electric current is generated. In other words, the PbSO₄ made when current is drawn from the cell is decomposed when the car is running. Hence, the reversible reaction.

Rust is in fact the compound hydrated iron (III) oxide with the general formula, Fe₂O₃.xH₂O.

Rusting occurs when both water and oxygen are present. It is an electrochemical process involving two redox half-equations.

The corrosion starts with the oxidation of iron to Fe²⁺:

Fe_(s) → Fe²⁺_(aq) + 2e^-

E = -0.44V

The iron acts as a negative terminal. The released electrons travel through the iron until they make contact with oxygen. This forms the positive electrode.

O_2(aq) + 2H₂O_(l) + 4e^- → 4OH^-_(aq)

E = +1.23V.

The following step is that the Fe²⁺ and OH^- ions combine to form Fe(OH)_2(s). This is readily oxidised to iron (III) hydroxide, which then partially dehydrates to give rust, Fe₂O₃.xH₂O.

1. Coating the Iron: By adding a layer of paint, oil or grease to the iron, you prevent oxygen and water from coming into contact with the iron.
2. Sacrificial Protection: A most effective way of preventing rust is to coat the iron with zinc, this is called galvanising. It works due to zinc's greater reactivity, i.e. zinc has a greater tendency to form ions, hence any Fe²⁺ present are reduced to Fe atoms.
3. Alloying: Iron can be alloyed with nickel, chromium or carbon. The presence of other elements helps prevent Fe²⁺ forming and electrons been released, hence preventing the rusting process starting.

The process of decomposing a compound using electricity is called electrolysis.

2H₂O_(l) → 2H_2(g) + O_2(g)

Electrolysis is a most important industrial process with a wide application. However, the largest application is that of the manufacture of chlorine and sodium hydroxide from concentrated aqueous sodium chloride in the Chlor-alkali industry.

The cell consists of two conducting rods called electrodes, dipped into a compound in a molten state or in solution, that is able to conduct electricity called the electrolyte.

Electrolysis only works with direct current (the charge flows in only one direction). Direct current causes one electrode to take on a positive charge called the anode. The second electrode takes on a negative charge called the cathode.

Note: Anions (negative ions) are attracted to the anode. Cations (positive ions) are attracted to the cathode.

An example of electrolysis:

If we place molten sodium chloride into the electrolysis cell and allow current to flow, the following reactions occur at the electrodes:

At the cathode: Reduction

Na⁺_(l) + e^- → Na_(l)

At the anode: Oxidation

2Cl^-_(l) → Cl_2(g) + 2e^-

These two half equations give the overall reaction:

2Na⁺_(l) + 2Cl^-_(l) → Na_(l) + Cl_2(g)

A redox reaction has occured.

In the example above, of molten sodium chloride, there was one cation Na⁺ and one anion Cl^-. Therefore, the products were easily calculated.

However, if a compound is in an aqueous solution, then we have two more ions to deal with, OH^- and H⁺.

So how do we decide with ions will appear at the electrodes and which remain in solution?

In the case of the cathode, we must decide which of the two cations, H⁺ or Na⁺ is reduced most easily back to their atoms. Hydrogen ions are most easily reduced so the following cathode reaction occurs:

2H⁺_(aq) + 2e^- → H_2(g)

In the case of the anode, we must decide which of the two anions, OH^- or Cl^- will be most easily oxidised back to atoms. Chloride ions are most easily oxidised, hence the following anode reaction:

2Cl^-_(aq) → Cl₂ + 2e^-

This leaves Na⁺ _(aq) and OH^-_(aq) in solution i.e. sodium hydroxide.

We can refer to the electrochemical series, redox series or reactivity series for this information.

In 1832, Michael Faraday deduced that:

The quantity of electricity passed is proportional to the amount of substance discharged at the electrode.

Quantity of electricity (charge) = current x time

The units used are, coulombs for charge, amps for current and seconds for time.

Note 1: One mole of electrons has a charge of 96500C. This quantity of charge is called the Faraday constant, F. Thus, the Faraday constant is related to Avagadro's constant, L, and the charge on the electron, e.

F = L x e

Note 2: The number of moles of electrons required to discharge 1 mole of ions is equal to the charge on the ion.

For example, ions with a double charge, such as Cu²⁺ - it will take two moles of electrons to deposit one mole of copper.

This industry involves the production of chlorine and the alkali sodium hydroxide from the electrolysis of concentrated aqueous sodium chloride (brine).

The electrolysis cell used for this reaction is called the diaphram cell.

As previously stated, four ions are present Na⁺, Cl^-, OH^-, H⁺.

At the cathode:

2H⁺_(aq) + 2e^- → H_2(g)

At the anode:

2Cl^-_(aq) → Cl_2(g) + 2e^-

The ions remaining in solution are OH^- and Na⁺.

In the diaphragm cell, a porous asbestos partition, is placed between the electrodes to prevent the sodium hydroxide making contact with the chlorine (sodium hydroxide tends to concentrate near the cathode).

Purified brine solution is fed into the anode side and the level kept above that of the cathode. This allows the sodium chloride solution to seep into the cathode compartment and also prevents OH^- ions migrating to the anode.

Redox reactions involve oxidation and reduction occurring simultaneously.

Oxidation is the loss of electrons

Reduction is the gain of electrons

As stated previously, these two reactions cannot occur in isolation.

The two equations are added to provide the reaction equation:

In terms of oxidation number:

• For the oxidation: there has been an increase of +2
• For the reduction: there has been an increase of -2

Displacement reactions occur if for example you place a more reactive metal in a solution containing ions of a less reactive metal.

For example:

If we now compare this with the following reaction:

We see that metals can be both oxidised and reduced, we write:

The &ruldhar; sign shows that the reaction can go both ways. It is the sign for dynamic equilibrium.

If a piece of metal is placed into a solution of its own ions, two occurrences may appear.

Metal atoms may leave the solid and become metal ions in solution:

Oxidation

In this case the electrons that are released stay on the surface of the solid, which causes it to gain a negative charge.

Or, metal ions in solution may become metal atoms on the surface of the solid.

Reduction

In this case, electrons are attracted into the solution.