Friday, February 18, 2011


Here, each of the reactant is taken in a separate container in contact with a rod/sheet of a metallic-conductor (electronic conductor) called an electrode. Electrical contact between the two reactants is established by placing a conducting salt bridge in-between.

Fig: 9.2 - Experimental set up for an indirect redox reaction

The species capable of losing electrons transfers them to the electrode placed in it. This makes the metallic conductor or electrode negatively charged due to the accumulation of electrons. On the other hand, the species capable of gaining electrons accepts them from the metallic conductor (electrode) placed in it. As a result this metallic conductor electrode) gets positively charged due to the deficiency of electrons. When a metallic wire connects the two electrodes, the electrons flow from the negatively charged electrode to the positively charged electrode. Thus, in an indirect redox reaction, electrons flow in a particular direction through an external conducting wire connecting the two electrodes.

This flow of electrons generates electricity, which can be used for doing some useful work.

Thus, in an indirect redox reaction, the decrease in the chemical energy is liberated in the form of electrical energy.

An electrochemical reaction differs from a chemical reaction in the following respects.

Chemical reaction Electrochemical reaction
Electron transfer from one species to another takes place directly in the same medium. Electron transfer from one species to another takes place indirectly through electrodes.
Energy is liberated in the form of heat, light and sound. Energy is liberated in the form electrical energy.
The chemical reaction is instantaneous proceeding at a finite rate. This reaction takes place only on the application of electricity.
Redox reactions take place in the same medium. Redox reaction takes place separately at the Anode and cathode surface.

Electrochemical cell

Spontaneous redox reactions are the metal displacement reactions. Therefore such reactions have been used for producing electricity. This is done by carrying out these reactions in specially designed units called electrochemical cells.

An electrochemical cell is a device used to convert chemical energy of an indirect redox reaction into electrical energy. This is also called Voltaic cell or a Galvanic cell. It is set up by dipping two electrodes (conducting rods) into the same or two different electrolytes. No reaction takes place inside the cell until a conducting wire joins the two electrodes.
  electrochemical cell with one and two electrodes
Fig: 9.3 - An electrochemical cell (a) one electrolyte (b) two electrolytes

A typical metal displacement reaction is the reaction between zinc metal and copper sulphate solution i.e.,

reaction between zinc metal and copper sulphate solution

The electrochemical cell based on this reaction is set up as follows.

Zinc sulphate solution is taken in a beaker and a zinc rod is dipped in to it. Similarly copper sulphate solution is taken in another beaker and a copper strip is dipped in to it. An inverted U tube containing concentrated solutions of inert electrolytes such as KCl, KNO3 etc., connects the two solutions. The two openings of the U tube are plugged with porous materials like glass wool or cotton. This U tube is called as salt bridge as it acts like a bridge connecting the solutions of the two beakers. In place of salt bridge, one can also use either a paper strip, unglazed porcelain or clay porous pot or asbestos fibre for developing electrical contact between the two half-cells. When a key is inserted to complete the outer circuit, the following observations are made.
  • There is a flow of electrical current through the external circuit.
  • The zinc rod loses weight, while the copper rod acquires weight.
  • The concentration of ZnSO4 solution increases, while that of CuSO4 solution decreases.
  • The two solutions in the beakers have electrical neutrality.

The above indications clearly show the overall reaction to be:

Zinc is oxidized to Zn2+ ions and go into the solution during the reaction.

The electrons released at the electrode move towards the other electrode through the outer circuit. These electrons are accepted by Cu2+ ions of CuSO4 solution. Thus the Cu2+ ions are reduced to metallic copper, which get deposited on the copper electrode.

With the onset of the chemical reaction, the zinc plate begins to dissolve and loses weight, electrons get generated and move and finally, the copper plate gains weight. Therefore, this indirect redox reaction is accompanied by the liberation of energy in terms of electric charge, which is the electrical energy. In this way the chemical reaction leads to the production of electrical energy that further helps in doing useful work (deposition of copper metal).

Electrochemical cell based on redox reaction of zinc and  copper sulphate

Fig: 9.4 - Electrochemical cell based on redox reaction of zinc and copper sulphate

Salt Bridge and Its Function

A salt bridge is a low resistance device, which establishes an electrical contact between two electrolytes not in direct contact. This is used to overcome the direct liquid-liquid junction that leads to intermixing. The salt bridge consists of a glass U-tube filled with KCl containing Agar-Agar paste, which sets into a gel. The choice of the electrolyte added to the gel depends upon the nature of the electrolytes used in the cell. However these electrolytes are inert for they should not react chemically or undergo any electrochemical reactions with the electrolytes electrically connected with them. Commonly used salts are, KCl, NH4NO3, KNO3, or K2SO4. The KCl-bridge cannot be used when any salt of lead, or silver is used in the cell because lead chloride and silver chloride are insoluble in water.

Functions of a Salt Bridge

  • When in a galvanic cell two solutions are kept in separate containers, an electrical contact between the two is needed to complete the circuit. A salt bridge completes the circuit by allowing the migration of anions from one container into the salt bridge and from the salt bridge into the other container.
  • The salt bridge prevents the physical transference/diffusion of the electrolytes from one container to the other.
  • A salt bridge helps in maintaining the charge balance in the reactions taking place at the two containers by releasing counter ions into the solution. Other wise due to the accumulation of the respective charges in the two containers there will be no flow of electrons and the cell will stop functioning.
  • A direct liquid-liquid junction is thermodynamically unstable state. The unequal rates of migration of the cations and anions across a liquid-liquid junction give rise to a potential difference across the junction. This potential difference across the liquid-liquid junction is called liquid junction potential. A salt bridge eliminates a direct contact between the two solutions, and thus minimizes the liquid junction potential.

Tuesday, February 15, 2011


Faraday's laws of electrolysis:

Michael Faraday, a pioneer in the properties of electric currents, formulated two basic laws of electrolysis:

FARADAY'S FIRST LAW may be stated as follows:

Faraday's First Law of Electrolysis
"The amount of any substance deposited, evolved, or dissolved at an electrode is directly proportional to the amount of electrical charge passing through the circuit."

The amount of electricity passing through the circuit in a given time is the number of moles of electrons passing through the circuit in that time, and the charge Q is related to the current I by

Q =  It

The charge on the electron is 1.602 x 10-19 C, and Avogadro's number is 6.023 x 1023. It follows that one mole of electrons has a charge of 9.65 x 104 C. This quantity is known as the FARADAY or FARADAY'S CONSTANT (F).

FARADAY'S SECOND LAW may be stated as follows:

Faraday's Second Law of Electrolysis
"The mass of different substances produced by the same quantity of electricity are directly proportional to the molar masses of the substances concerned, and inversely proportional to the number of electrons in the relevant half-reaction."

This means that z moles of electrons are needed to discharge an ion Xz+ or Xz-.

In the apparatus below,

1 Faraday will discharge 9 g Al (1/3 mole), 20 g Ca (1/2 mole) and 23 g Na (1 mole).

The relevant half reactions are:

Al3+ + 3e- Al
Ca2+ +2e- Ca
Na+ + e- Na

Faraday's second law, illustrated


Extraction of aluminium:

Aluminium is obtained by the electrolytic reduction of its molten oxide, alumina (Al2O3). Because alumina has a very high melting point (2045 ºC), the mineral cryolite (Na3AlF6) is added to lower the melting point in order that the electrolysis may be carried out at about 950 ºC. The electrolytic cell has carbon anodes and a carbon cathode (which forms the lining of the tank in which the electrolysis takes place). Carbon dioxide is formed at the anodes, and aluminium at the cathode. It is heavier than the molten alumina/cryolite mixture, and sinks to the bottom of the cell, where it is tapped off. The procedure is known as the Hall-Héroult process.

Extraction of aluminium

Aluminium extraction is very demanding on electrical current (typically, 3-5 V and 100 000 A), and is economical only where power is cheap.

Electrorefining of copper:

When copper is first obtained by reduction of its ores, it is cast as impure slabs or ingots, called blister copper. In the electrorefining process, the blister ingots are used as anodes in an electrolytic cell, where an acid solution of copper (II) sulphate is used as electrolyte. Initially, the cathodes consist of thin sheets of pure copper.

Electrorefining of copper

During electrolysis, copper passes into solution from the anodes, (leaving the impurities, normally containing silver, gold and platinum) as an anode slime, which sinks to the bottom of the cell. The anode reaction is

Anode reaction: electrorefining of copper

At the cathode, copper (II) ions are discharged and the pure copper sheet becomes coated with an increasingly thick layer of very pure copper:

Cathode reaction: electrorefining of copper


Electroplating consists of depositing a thin layer of a metal on another, either for protection or for the sake of appearance. Typically, a brass or nickel object is coated with a layer of silver by making use of electrolysis of a silver solution, using the object to be coated as the cathode:

The anode consist of pure silver, and the cathode is the object to be plated. The electrolyte is a mixure of silver nitrate with potassium cyanide.

The reactions are:

At the anode: Ag → Ag+ + e-

At the cathode: Ag+ + e- → Ag

The cyanide ensures a low concentration of silver ions, a condition for providing the best plating results.

During the process, the concentration of silver in the electrolyte remains constant, as the rate of reduction at the cathode (which is the rate of deposition of silver on the object) is the same as the rate of reduction at the anode (which is the rate of rate of dissolution of the silver anode).


Electrolytic Cell V/s Galvanic Cell

The main points of difference between an electrolytic cell and a galvanic cell (electrochemical cell) are:

Difference between an electrolytic cell and a galvanic cell

Electrochemical cell (Galvanic Cell) Electrolytic cell
A Galvanic cell converts chemical energy into electrical energy. An electrolytic cell converts electrical energy into chemical energy.
Here, the redox reaction is spontaneous and is responsible for the production of electrical energy. The redox reaction is not spontaneous and electrical energy has to be supplied to initiate the reaction.
The two half-cells are set up in different containers, being connected through the salt bridge or porous partition. Both the electrodes are placed in a same container in the solution of molten electrolyte.
Here the anode is negative and cathode is the positive electrode. The reaction at the anode is oxidation and that at the cathode is reduction. Here, the anode is positive and cathode is the negative electrode. The reaction at the anode is oxidation and that at the cathode is reduction.
The electrons are supplied by the species getting oxidized. They move from anode to the cathode in the external circuit. The external battery supplies the electrons. They enter through the cathode and come out through the anode.


Electrolysis of water:

Water may be electrolyzed in the apparatus shown below. Pure water is however a very poor conductor of electricity, and one has to add dilute sulphuric acid in order to have a significant current flow.

Apparatus for the electrolysis of dilute sulphuric acid

The electrodes consist of platinum foil. The electrolyte is dilute sulphuric acid. Hydrogen gas is evolved at the cathode, and oxygen at the anode.

The ratio, by volume, of hydrogen to oxygen, is exactly 2:1.

Remember that electron flow in the circuit is opposite to the conventional current flow. Thus, while the conventional current flows from the positive pole through the electrolyte and ends up at the negative pole, electrons flow from the negative pole in the reverse direction. The positive pole of a battery accepta electrons from the electrolyte by means of the anode of the electrolytic cell. The reaction at the cathode (tube A) is the reduction of protons:

Reduction of hydrogen ion

Oxidation takes place at the anode (tube B). There are two anions competing to give up their electrons, namely sulphate (SO42-), and hydroxide (OH-) from the ionization of water. The oxidation of OH- according to the reaction

Oxidation of hydroxide ion

has a standard electrode potential of -0.40V, compared to the oxidation of sulphate (-0.17V), and consequently, OH- will be oxidized preferentially. The overall reaction is therefore

Electrolysis of water: overall reaction

and the electrons are reurned to the battery, thus completing the circuit.


Electrolysis of Aqueous Solutions
In the electrolysis of aqueous solutions only one ion is involved in the selective discharge of ions at each electrode during the electrolysis. The ion which is selected for discharge at an electrode depends on a number of factors, including
Position of the ions in the electrochemical series,
For positive ions, the facility of discharge decrease in going from those least electropositive to those most electropositive. For example, if both copper and hydrogen ions are present in solution, it will be the copper ions which take electrons from the cathode to become copper atoms.

For negative ions, the ease of discharge decrease in going from those least electronegative to those most electronegative.

Concentration of the Ions in Solution
Irrespective of the position of the ions in the electrochemical series, there is a tendency to promote the discharge of the most concentrated ion present. For example, in concentrated sodium chloride solution (i.e. brine) , the two cations present are the chlorine ion and the hydroxyl ion. Although the hydroxyl ion is more easily oxidised than the chlorine ion, it is the chlorine ion which will be discharged because its concentration is much greater than that of the hydroxyl ion.
Nature of the Electrode
This is not as important as either of the other two factors, except in certain cases. For example in the electrolysis of molten sodium chloride using a mercury cathode, sodium ions are discharged in preference to hydrogen ions which are lower in the series.

Electrolysis of an Aqueous Copper Sulphate Solution using Copper Electrodes

The electrolysis of an aqueous solution of copper sulphate using copper electrodes (i.e. using active electrodes) results in transfer of copper metal from the anode to the cathode during electrolysis. The copper sulphate is ionised in aqueous solution.

CuSO4 ==> Cu(++) + SO4(-.-)

The positively charged copper ions migrate to the cathode, where each gains two electrons to become copper atoms that are deposited on the cathode.

Cu(++) + 2e(-) ==> Cu

At the anode, each copper atom loses two electrons to become copper ions, which go into solution.

Cu ==> Cu(++) + 2e(-)

The sulphate ion does not take part in the reaction and the concentration of the copper sulphate in solution does not change. The reaction is completed when the anode is completely eaten away. This process is used in electroplating.

Electrolysis of a solution of dilute Sulphuric Acid
The electrolysis of an aqueous solution of dilute sulphuric acid is often carried out in a Hofmann Voltammeter, an apparatus in which the gases evolved at the anode and cathode can be collected in separate graduated tubes. When the solution is electrolyzed hydrogen is produced at the cathode and oxygen at the anode. These gases can be shown to be present in a 2 to 1 ratio and result from the electrolysis of water under acidic conditions.

Sulphuric acid is a strong electrolyte is fully dissociated in aqueous solution.

H2SO4 ==> 2 H(+) + SO4(2 -)

Water is a weak electrolyte and is only slightly dissociated

H2O ==> H(+) + OH(-)

During electrolysis, the hydrogen ions migrates towards the cathode, and are discharged there (i.e. they gain an electron and are converted to hydrogen gas).

2 H(+) + 2 e(-) ==> H2

At the anode the concentration of hydroxyl ions is too low to maintain a reaction and the sulphate ions are not oxidized but remain on in solution at the end. Water molecules must be the species reacting at the anode.

2 H2O ==> O2 + 4 H(+) + 4 e(-)

The overall reaction is

Cathode Reaction :
2 H(+)   +   2e(-)   ==>   H2

4 H(+) + 4e(-) ==> 2H2
Anode Reaction :
2 H2O   ==>   O2  +  4 H(+)  +  4 e(-)

Overall Cell Reaction:
4 H(+)   +   2 H2O   ==>   2 H2   +   O2   +  4 H(+)

For every hydrogen ions discharged at the anode, another hydrogen ion is formed at the cathode. The net result is that the concentration of the sulphuric acid remains constant and this electrolysis consists of the decomposition of water with the overall reaction

2H2O ==> 2H2 + O2


Electrolysis of Molten Lead Bromide

A small quantity of solid lead bromide is taken in a silica crucible and two graphite electrodes (may be obtained from used torch cells) are inserted. A battery consisting of two dry cells is connected to the electrodes through a key and an ammeter.

electrolysis of molten  lead bromide

Fig: 9.11 - Electrolysis of molten lead bromide

When the key is pressed, no current flows through the system. This is because solid lead bromide does not conduct electricity. But, when the crucible containing lead bromide is heated the solid lead bromide melts. Now, on pressing the key, electricity flows through the system and a red brown gas (bromine) evolves at the anode and metallic lead deposits at the cathode.

The following reactions occur at the two electrodes:

electrolysis of lead bromide using graphite electrodes

Thus, electrolysis of lead bromide using graphite electrodes produces lead metal at the cathode and bromine gas at the anode.


An electrolytic cell decomposes chemical compounds by means of electrical energy, in a process called electrolysis; the Greek word lysis means to break up. The result is that the chemical energy is increased. Important examples of electrolysis are the decomposition of water into hydrogen and oxygen, and bauxite into aluminium and other chemicals.


An electrolytic cell has three component parts: an electrolyte and two electrodes (a cathode and an anode). The electrolyte is usually a solution of water or other solvents in which ions are dissolved. Molten salts such as sodium chloride are also electrolytes. When driven by an external voltage applied to the electrodes, the electrolyte provides ions that flow to and from the electrodes, where charge-transferring, or faradaic, or redox, reactions can take place. Only for an external electrical potential (i.e. voltage) of the correct polarity and large enough magnitude can an electrolytic cell decompose a normally stable, or inert chemical compound in the solution. The electrical energy provided undoes the effect of spontaneous chemical reactions.

Anode and cathode definitions depend on charge and discharge

Michael Faraday defined the cathode as the electrode to which cations flow (positively charged ions, like silver ions Ag+
), to be reduced by reacting with (negatively charged) electrons on the cathode. Likewise he defined the anode as the electrode to which anions flow (negatively charged ions, like chloride ions Cl
), to be oxidized by depositing electrons on the anode. Thus positive electric current flows from the cathode to the anode. To an external wire connected to the electrodes of a battery, thus forming an electric circuit, the cathode is positive and the anode is negative.

Consider two voltaic cells, A and B, with the voltage of A greater than the voltage of B. Mark the positive and negative electrodes as cathode and anode, respectively. Place them in a circuit with anode near anode and cathode near cathode, so the cells will tend to drive current in opposite directions. The cell with the larger voltage discharges, making it a voltaic cell. Likewise the cell with the smaller voltage charges, making it an electrolytic cell. For the electrolytic cell, the external markings of anode and cathode are opposite the chemical definition. That is, the electrode marked as anode for discharge acts as the cathode while charging and the electrode marked as cathode acts as the anode while charging.


As already noted, water, particularly when ions are added (salt water or acidic water) can be electrolyzed (subject to electrolysis). When driven by an external source of voltage, H+
ions flow to the cathode to combine with electrons to produce hydrogen gas in a reduction reaction. Likewise, OH
ions flow to the anode to release electrons and an H+
ion to produce oxygen gas in an oxidation reaction.

In molten sodium chloride, when a current is passed through the salt the anode oxidizes chloride ions (Cl
) to chlorine gas, releasing electrons to the anode. Likewise the cathode reduces sodium ions (Na+
), which accept electrons from the cathode and deposits on the cathode as sodium metal.

NaCl dissolved in water can also be electrolyzed. The anode oxidizes chloride ions (Cl
), and Cl2 gas is still produced. However, at the cathode, instead of sodium ions being reduced to sodium metal, water molecules are reduced to hydroxide ions (OH
) and hydrogen gas (H2). The overall result of the electrolysis is the production of chlorine gas and aqueous sodium hydroxide (NaOH) solution.

Commercially, electrolytic cells are used in electrorefining and electrowinning of several non-ferrous metals. Almost all high-purity aluminium, copper, zinc and lead is produced industrially in electrolytic cells.


The Nernst equation is used to calculate the voltage of an electrochemical cell or to find the concentration of one of the components of the cell. Here is a look at the Nernst equation and an example of how to apply it to solve a problem.

The Nernst Equation

Ecell = E0cell - (RT/nF)lnQ

Ecell = cell potential under nonstandard conditions (V)
E0cell = cell potential under standard conditions
R = gas constant, which is 8.31 (volt-coulomb)/(mol-K)
T = temperature (K)
n = number of moles of electrons exchanged in the electrochemical reaction (mol)
F = Faraday's constant, 96500 coulombs/mol
Q = reaction quotient, which is the equilibrium expression with initial concentrations rather than equilibrium concentrations

Sometimes it is helpful to express the Nernst equation differently:

Ecell = E0cell - (2.303*RT/nF)logQ

at 298K, Ecell = E0cell - (0.0591 V/n)log Q

Nernst Equation Example
A zinc electrode is submerged in an acidic 0.80 M Zn2+ solution which is connected by a salt bridge to a 1.30 M Ag+ solution containing a silver electrode. Determine the initial voltage of the cell at 298K.

(Unless you've done some serious memorizing, you'll need to consult the standard reduction potential table, which will give you the following information):

E0red: Zn2+aq + 2e- → Zns = -0.76 V

E0red: Ag+aq + e- → Ags = +0.80 V

Ecell = E0cell - (0.0591 V/n)log Q

Q = [Zn2+]/[Ag+]2

The reaction proceeds spontaneously so E0 is positive. The only way for that to occur is if Zn is oxidized (+0.76 V) and silver is reduced (+0.80 V). Once you realize that, you can write the balanced chemical equation for the cell reaction and can calculate E0:

Zns → Zn2+aq + 2e- and E0ox = +0.76 V

2Ag+aq + 2e- → 2Ags and E0red = +0.80 V

which are added together to yield:

Zns + 2Ag+aq → Zn2+a + 2Ags with E0 = 1.56 V

Now, applying the Nernst equation:

Q = (0.80)/(1.30)2

Q = (0.80)/(1.69)

Q = 0.47

E = 1.56 V - (0.0591 / 2)log(0.47)

E = 1.57 V

Monday, February 14, 2011



Illustration of a redox reaction

Redox (shorthand for oxidation-reduction) reactions describe all chemical reactions in which atoms have their oxidation number (oxidation state) changed. This can be either a simple redox process, such as the oxidation of carbon to yield carbon dioxide (CO2) or the reduction of carbon by hydrogen to yield methane (CH4), or a complex process such as the oxidation of sugar (C6H12O6) in the human body through a series of complex electron transfer processes.

The term comes from the two concepts of reduction and oxidation. It can be explained in simple terms:

  • Oxidation is the loss of electrons or an increase in oxidation state by a molecule, atom, or ion.
  • Reduction is the gain of electrons or a decrease in oxidation state by a molecule, atom, or ion.

Though sufficient for many purposes, these descriptions are not precisely correct. Oxidation and reduction properly refer to a change in oxidation number — the actual transfer of electrons may never occur. Thus, oxidation is better defined as an increase in oxidation number, and reduction as a decrease in oxidation number. In practice, the transfer of electrons will always cause a change in oxidation number, but there are many reactions that are classed as "redox" even though no electron transfer occurs (such as those involving covalent bonds).

Non-redox reactions, which do not involve changes in formal charge, are known as metathesis reactions.

The two parts of a redox reaction
Rusting iron
A bonfire. Combustion consists of redox reactions involving free radicals.

Oxidizing and reducing agents

In redox processes the reductant transfers electrons to the oxidant. Thus, in the reaction, the reductant or reducing agent loses electrons and is oxidized, and the oxidant or oxidizing agent gains electrons and is reduced. The pair of an oxidizing and reducing agent that are involved in a particular reaction is called a redox pair.


Substances that have the ability to oxidize other substances are said to be oxidative or oxidizing and are known as oxidizing agents, oxidants, or oxidizers. Put another way, the oxidant removes electrons from another substance, and is thus itself reduced. And, because it "accepts" electrons, it is also called an electron acceptor.

Oxidants are usually chemical elements or substances with elements in high oxidation numbers (e.g., H2O2, MnO
, CrO3, Cr2O2−
, OsO4) or highly electronegative substances/elements that can gain one or two extra electrons by oxidizing an element or substance (O, F, Cl, Br).


Substances that have the ability to reduce other substances are said to be reductive or reducing and are known as reducing agents, reductants, or reducers. That is, the reductant transfers electrons to another substance, and is thus itself oxidized. And, because it "donates" electrons, it is also called an electron donor. Electron donors can also form charge transfer complexes with electron acceptors.

Reductants in chemistry are very diverse. Electropositive elemental metals, such as lithium, sodium, magnesium, iron, zinc, aluminium, carbon, are good reducing agents. These metals donate or give away electrons readily. Hydride transfer reagents, such as NaBH4 and LiAlH4, are widely used in organic chemistry,[1][2] primarily in the reduction of carbonyl compounds to alcohols. Another method of reduction involves the use of hydrogen gas (H2) with a palladium, platinum, or nickel catalyst. These catalytic reductions are used primarily in the reduction of carbon-carbon double or triple bonds.

Examples of redox reactions

A good example is the reaction between hydrogen and fluorine in which hydrogen is being oxidized and fluorine is being reduced:

H2 + F2 → 2 HF

We can write this overall reaction as two half-reactions:

the oxidation reaction:

H2 → 2 H+ + 2 e

and the reduction reaction:

F2 + 2 e
→ 2 F

Analyzing each half-reaction in isolation can often make the overall chemical process clearer. Because there is no net change in charge during a redox reaction, the number of electrons in excess in the oxidation reaction must equal the number consumed by the reduction reaction (as shown above).

Elements, even in molecular form, always have an oxidation number of zero. In the first half-reaction, hydrogen is oxidized from an oxidation number of zero to an oxidation number of +1. In the second half-reaction, fluorine is reduced from an oxidation number of zero to an oxidation number of −1.

When adding the reactions together the electrons cancel:

H2 2 H+ + 2 e
F2 + 2 e
2 F

H2 + F2 2 H+ + 2 F

And the ions combine to form hydrogen fluoride:

H2 + F2 → 2 H+ + 2 F → 2 HF

Displacement reactions

Redox occurs in single displacement reactions or substitution reactions. The redox component of these types of reactions is the change of oxidation state (charge) on certain atoms, not the actual exchange of atoms in the compounds.

For example, in the reaction between iron and copper(II) sulfate solution:

Fe + CuSO4FeSO4 + Cu

The ionic equation for this reaction is:

Fe + Cu2+ → Fe2+ + Cu

As two half-equations, it is seen that the iron is oxidized:

Fe → Fe2+ + 2 e

And the copper is reduced:

Cu2+ + 2 e
→ Cu

Other examples

  • The oxidation of iron(II) to iron(III) by hydrogen peroxide in the presence of an acid:
Fe2+ → Fe3+ + e
H2O2 + 2 e → 2 OH
Overall equation:
2 Fe2+ + H2O2 + 2 H+ → 2 Fe3+ + 2 H2O
2 NO3 + 10 e + 12 H+ → N2 + 6 H2O
Iron rusting in pyrite cubes
  • Oxidation of elemental iron to iron(III) oxide by oxygen (commonly known as rusting, which is similar to tarnishing):
4 Fe + 3 O2 → 2 Fe2O3

Balancing redox reactions

Describing the overall electrochemical reaction for a redox process requires a balancing of the component half-reactions for oxidation and reduction. For reactions in aqueous solution, this generally involves adding H+, OH, H2O, and electrons to compensate for the oxidation changes.

Acidic media

In acidic media, H+ ions and water are added to half reactions to balance the overall reaction.

For example, when manganese(II) reacts with sodium bismuthate:

Unbalanced reaction: Mn2+(aq) + NaBiO3(s) → Bi3+(aq) + MnO4 (aq)
Oxidation: 4 H2O(l) + Mn2+(aq) → MnO
(aq) + 8 H+(aq) + 5 e
Reduction: 2 e
+ 6 H+ + BiO
(s) → Bi3+(aq) + 3 H2O(l)

The reaction is balanced by scaling the two half-cell reactions to involve the same number of electrons (multiplying the oxidation reaction by the number of electrons in the reduction step and vice versa):

8 H2O(l) + 2 Mn2+(aq) → 2 MnO
(aq) + 16 H+(aq) + 10 e
10 e
+ 30 H+ + 5 BiO
(s) → 5 Bi3+(aq) + 15 H2O(l)

Adding these two reactions eliminates the electrons terms and yields the balanced reaction:

14 H+(aq) + 2 Mn2+(aq) + 5 NaBiO3(s) → 7 H2O(l) + 2 MnO
(aq) + 5 Bi3+(aq) + 5 Na+(aq)

Basic media

In basic media, OH ions and water are added to half reactions to balance the overall reaction.

For example, in the reaction between potassium permanganate and sodium sulfite:

Unbalanced reaction: KMnO4 + Na2SO3 + H2OMnO2 + Na2SO4 + KOH
Reduction: 3 e
+ 2 H2O + MnO4MnO2 + 4 OH
Oxidation: 2 OH + SO32−SO42− + H2O + 2 e

Balancing the number of electrons in the two half-cell reactions gives:

6 e
+ 4 H2O + 2 MnO4 → 2 MnO2 + 8 OH
6 OH + 3 SO32− → 3 SO42− + 3 H2O + 6 e

Adding these two half-cell reactions together gives the balanced equation:

2 KMnO4 + 3 Na2SO3 + H2O → 2 MnO2 + 3 Na2SO4 + 2 KOH




The purpose of this experiment is to introduce the student to electrochemistry and oxidation-reduction reactions through the construction and operation of a simple galvanic cell.


This experiment is appropriate for a general or first-year college-prep course. An inexpensive cell is made from materials obtained locally. Students test the cell by connecting it to a 1.5-volt flashlight lamp and observing whether or not the lamp will light. In addition, four or more of these cells can be connected in series to form a battery that can power a small radio or other device that operates on direct current.


Two to three labs periods.


0.5 M copper(II) sulfate (dissolve 125 g CuSO4·5H2O in distilled or deionized water and dilute to 1.00 liter)*
0.5 M sodium sulfate (dissolve 71 grams of Na2SO4 in distilled or deionized water and dilute to 1.00 liter)
zinc, copper, aluminum, and magnesium strips (approximately 1 cm x 10 cm)*
250-mL beaker*
dialysis tubing*
hook-up wire or bell wire
1.5-volt flashlight lamp with less than 100 milliamp rating
alligator clamps
crimping tool
soldering iron and resin core solder
clamps(for dialysis tubing)
sandpaper or steel wool
*See Modifications/Substitutions


Avoid contact with solutions. Goggles must be worn throughout experiment.


  1. Copper sulfate pentahydrate is available as root killer at garden supply stores.
  2. Aluminum, from aluminum cans, may be used if it is sanded well on both sides. Aluminum gutter nails may also be used.
  3. Copper tubing or fittings may be used in place of copper strips.
  4. Zinc can be obtained from old dry cell battery casings. This should be done carefully to avoid contact with caustic chemicals in battery.
  5. Sausage casings can be used in place of dialysis tubing but diffusion is extremely rapid. Dialysis tubing is readily available in biology labs.
  6. Baby food jars or other open glass jars may be used instead of beakers.


  1. Prepare a test lamp by soldering wire test leads to a 1.5-volt flashlight lamp. Attach alligator clips to the ends of the leads.
  2. Obtain strips of magnesium and copper, each strip should be 2.5 cm longer than the height of the beaker being used. Sand the strips until they are shiny.
  3. Tie a knot with string or use a non-reactive clamp in one end of dialysis tubing that has been soaked in distilled water. The length of tubing should be long enough so that it overlaps the edge of the beaker by 2.5 cm.
  4. Fill the dialysis tubing with the prepared copper(II) sulfate solution. Place the copper strip in this piece of dialysis tubing that is now filled with copper(II) sulfate solution and use string or a rubber band to secure the top of the dialysis tubing around the copper. Leave 2.5 cm of copper sticking out of the tubing.
  5. Fill the beaker with the prepared sodium sulfate solution, place both the dialysis tubing, containing the copper and copper(II) sulfate solution, and the magnesium strip in the beaker. Place the magnesium strip as far away as possible from the dialysis tubing. Secure both the magnesium strip and the copper strip in the dialysis tubing in the beaker by bending them over the edge of the beaker.
  6. Complete the circuit by attaching one alligator clip to the copper strip and the other to the magnesium strip.
  7. Observe and record any activity taking place at the metal strips, the solution (especially color changes), and the test lamp.
  8. Connect several cells in series and operate a small radio or some other direct current device.
  9. Try making cells with zinc or aluminum in place of the magnesium.


All solutions may be flushed down the drain with water.


The following discussion applies when copper and magnesium strips are used as the metal strips. If other metals are used, similar explanations can be used.
Mg(s) + Cu2+(aq) " src="file:///F:/galvanic%20experiment_files/rightarrow.htm" align="BOTTOM" border="0" hspace="4"> Mg2+(aq) + Cu(s) + 1.5v
This reaction is an oxidation-reduction reaction in which the magnesium is oxidized by the copper to Mg2+ ions. The result of this reaction is a transfer of electrons which provides the power required to light the flashlight lamp.

Students performing this experiment should be familiar with the following terms: anode, cathode, oxidizing agent, reducing agent, and electromotive series. Students should be presently studying a unit on electrochemistry and oxidation-reduction reactions in order to fully understand the "chemistry" of this experiment.


  1. The metal strips used in this experiment should have a large surface area to minimize resistance within the cell.
  2. Use a voltmeter instead of the light bulb to eliminate the need for soldering.
  3. Aluminum strips may not work unless sanded thoroughly and acid treated with 6.0 M HCl.
  4. Soldering techniques should be mastered by teacher so that any soldering can be


Summerlin, L.R., and Ealy, J.L. Jr, Chemical Demonstrations-A Sourcebook for Teachers, American Chemical Society, Washington D.C, 1985, p. 115. This experiment is adapted from this source.