REDOX EQUILIBRIA
Oxidation and Reduction
Electrochemical Cells and Fuel Cells

History of Battery Development:

Back in the late 1700’s, Luigi Galvani observed the behavior of two dissimilar metals zinc and copper, in contact with the electrolytes (ions in solution) of tissues. These 2 metals produced an electric charge that caused a frog’s leg to contract.But batteries were around long before this discovery of animal electricity. A pre-historic battery is shown below:

AncientBagdad Batteries
Clay jar with iron rod surrounded by copper cylinder. When filled with vinegar + an electrolytic, solution produces 1.1 volts DC (circa. 250 BC to 640 AD). So what causes the frogs leg to jump? Indeed, why do you jump? What caused all this were electrons moving from one place to another.

To understand batteries, we need to know what the chemicals are doing; and they are only just transferring electrons.Electron transfer reactions are redox reactions, shown by half-equations.

Oxidation - loss of electrons. (L.E.O) In the Bagdad battery copper ions gained e-: Cu2++ 2e-Cu(s) (1/2 Reaction)

Reduction is the - gain of electrons (G.E.R). Zn(s)Zn(s) + 2e- (1/2 Reaction)

Whole redox equations can be created by combining two, equations (as above)and balancing the electrons.

Fe2+ Fe3+ + e- oxidation loses electrons (LEO), called a ½ reaction

MnO4- + 8H+ + 5e  Mn2+ + 4H2O reduction gains electrons (GER), called a ½ reaction

All equations must balance, so multiplying all coefficients in the oxidation reaction by 5:

5Fe2+ 5Fe3+ + 5e (means that 5 electrons are gained and five are lost)

overall combined equation: MnO4- + 8H+ + 5Fe2+ Mn2+ + 4H2O + 5Fe3+

A species which can accept electrons from another species is an oxidizing agent. Eg MnO4- is the oxidizing agent above

A species which can donate electrons to another species is a reducing agent. Eg Fe2+ in the above reaction is the RA.

Oxidation Numbers

As you can see electrons are the key to understanding redox equations; we must follow the electrons. We have invented a few rules to make sure we can keep track of electrons, and they are called oxidation numbers.

Rules for Oxidation Numbers.

1)Elements in their standard states, the oxidation number of each atom is zero: This is obviously so because it has not reacted yet! It has not been oxidized or reduced.Ex: Cl2, S, Na and O2

2)The oxidation number of an atomis the charge on the atom if the bonding were ionic.In simple ions, the oxidation number of the atom equals the charge on the ion:Ex: Na+, K+, H+ allhave an oxidation number of +1 ; whereas O2-, S2- all have an oxidation number of -2.

3)In neutral compounds, the sum of the oxidation numbers on the atoms is zero:

Eg SO3; oxidation number of S = +6, oxidation number of O = -2.

+6 + 3(-2) = 0

In polyatomic ions (Cr2O72- ). The sum of the oxidation numbers on atoms is equals the charge on the ion.

Eg : Cr2O72-; oxidation number of Cr = +6, oxidation number of O = -2.

2(+6) + 7(-2) = -2

element / oxidation state / exceptions
Group 1 metals / always +1
Group 2 metals / always +2
Oxygen / usually -2 / except in peroxides and F2O
Hydrogen / usually +1 / except in metal hydrides where it is -1
Fluorine / always -1
Chlorine / usually -1 / except in compounds with O or F

Ok, let’s use what we now know and see if we can follow the paths of electrons in an equation. Hopefully you’ll remember that we first split an equation into 2 reactions (half (½) reactions). Let us work with the following chemicals.
Manganate(VII) ions, MnO4-, oxidize hydrogen peroxide, H2O2, to oxygen gas. The reaction is done with potassium manganate(VII) solution and hydrogen peroxide solution acidified with dilute sulphuric acid.

As with all equations we must keep both reactants and products balanced for charge and for atoms. To balance the charge we will use electrons. If we need to balanceoxygen’s we will use water (H2O has oxygen, this may seem strange at first, but just goalong). To balance hydrogen’s we will use (H+).

All you are allowed to add are:
1)water (H2O)

2)hydrogen ions (H+)

3)electrons (e-)

During the reaction, the manganate(VII), purple coloured ions are reduced to green manganese(II) ions.
Let's start with the hydrogen peroxide half-equation. What we know is: H2O2 O2

The oxygen is already balanced. What about the hydrogen?

All you are allowed to add to this equation are water, hydrogen ions and electrons.

If you add water to supply the extra hydrogen atoms needed on the right-hand side, you will mess up the oxygen’s again - that's obviously wrong!
Add two hydrogen ions to the right-hand side.

H2O2 O2 + 2 H+
Now all you need to do is balance the charges. You would have to add 2 electrons to the right-hand side to make the overall charge on both sides zero.

H2O2 O2 + 2 H+ + 2e- (loses e- , so is a reduction reaction)
Now for the manganate(VII) half-equation:
You know (or are told) that the manganate(VII) ions turn into manganese(II) ions. Write that down.

MnO4-  Mn 2+
The manganese balances, but you need four oxygen’s on the right-hand side. These can only come from water - that's the only oxygen-containing thing you are allowed to write into one of these equations in acid conditions.

MnO4-  Mn 2+ + 4 H2O
By doing this, we've introduced some hydrogens. To balance these, you will need 8 hydrogen ions on the left-hand side.

8 H+ + MnO4-  Mn 2+ + 4 H2O

Now that all the atoms are balanced, all you need to do is balance the charges. At the moment there are a net 7+ charges on the left-hand side (1- and 8+), but only 2+ on the right. Add 5 electrons to the left-hand side to reduce the 7+ to 2+.

5e- + 8 H+ + MnO4-  Mn 2+ + 4 H2O (gains e- , so is an oxidation reaction)

A species which can accept electrons from another species is an oxidizing agent. Eg MnO4- is the oxidizing agent above

A species which can donate electrons to another species is a reducing agent. Eg H2O2 is the Reducing Agent.


This is the typical sort of half-equation which you will have to be able to work out. The sequence is usually:
Balance the atoms apart from oxygen and hydrogen.
Balance the oxygen’s by adding water molecules.
Balance the hydrogen’s by adding hydrogen ions.
Balance the charges by adding electrons.
Combining the half-reactions to make the ionic equation for the reaction
The two half-equations we've produced are:

So to balance electrons we find the least common denominator.

So now we simplify

ELECTROCHEMICAL CELLS

So let’s follow the electrons now in actual working situations, in batteries.

1.Consider a zinc rod immersed in a solution containing Zn2+ ions (eg ZnSO4)

The Zn atoms on the rod can lose two electrons and move into solution as Zn2+ ions: Zn(s) Zn2+(aq) + 2e

This process would result in an accumulation of NEGATIVE electrons charge on the zinc rod.

These electrons leave this rod and go to then second beaker, there a copper rod is immersed in a solution of copper ions (eg CuSO4), due to the following processes: Cu2+(aq) + 2e Cu(s) - Reduction (rod becomes positive)

Electrochemical cells are spontaneous reactions thatgenerates a current. There are two half-cells connected by a wire, so that electrons flow from one metal (electrode - anode) to another through metal (electrode-cathode)an external circuit and the ions flow through an internal cell connection (or salt bridge). The half-cell in which a half-reaction occurs with a loss of electrons (oxidation) is the anode. The second half-cell in which a half-reaction occurs with a gain of electrons (reduction), is the cathode.

Oxidation-reduction reactions occur when electrons are given up by the substance being oxidized (the reducing agent) and simultaneously gained by the substance being reduced (the oxidizing agent). Consider the following redox reaction to help explain the electron transfer process: Cu2+ + Zn  Zn2+ + Cu

In the diagram below, each beaker represents one of the two half cells for the above reaction. But, because there is no way for electrons to move from one beaker to the other, a redox reaction cannot yet occur.

If Zn and Cu electrodes were connected as in shown in the following diagram, an oxidation-reduction reaction would occur since electrons could flow through the external wire.

Creating an Electrochemical/ Voltaic /or Galvanic Cell

If twodifferent electrodes are connected, the potential difference (Volts) between the two electrodes will cause a current to flow between them. Thus an electromotive force (emf) is established and can generate electrical energy.

The circuit must be completed by allowing ions to flow from one solution to the other. This is achieved by means of a salt bridge(salts are any ionic compound, eg: KCl).

The combination of two electrodes in this way is known as an electrochemical cell, and can be used to generate electricity. The two components which make up the cell are known as half-cells.

ANODE (-ve)CATHODE (+VE)

Thus electrons flow from the zinc electrode to the copper electrode.

(GERC) Reduction takes place at the copper electrode: Cu2+(aq) + 2e  Cu(s) and Cations to Cathode

(LEOA) Oxidation thus takes place at the zinc electrode: Zn(s)  Zn2+(aq) + 2e and Anions to Anode

The overall cell reaction is as follows: Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)

The sulphate ions (anions) flow through the salt bridge from the Cu2+(aq) solution to the Zn2+(aq) solution (anode), to complete the circuit. The cell reaction including spectator ions can thus be written as follows:

CuSO4(aq) + Zn(s)  Cu(s) + ZnSO4(aq).

The external connection must be made of a metallic wire in order to allow electrons to flow. The salt bridge must be made of an aqueous electrolyte to allow ions to flow oppositely and internally.By allowing 2 chemical reagents to be connected electrically, but not chemically; chemical energy is thus converted into electrical energy.

Designing electrochemical cells

Half-cells do not necessarily have to consist of a metal immersed in a solution of its own ions..If the half-reaction does not contain a metal, an inert electrode must be used. Platinum, or carbon, is used, as they are inert. If a gas is involved, it must be bubbled through the solution in such a way that it is in contact with the electrode.Examples are shown below:

ION SOLUTION HALF CELLS

a) Cr2O72-(aq) + 14H+(aq) + 6e2Cr3+(aq) + 7H2O(l) (Note the reactants have no solids for the condiut of electrons

A platinum electrode is used, immersed in a solution containing Cr2O72-, H+ and Cr3+ ions:

ION SOLUTION HALF CELLS

b)2H+(aq) + 2e  H2(g)

A platinum electrode is used again, immersed in a solution containing H+ ions. Hydrogen gas is bubbled through the solution, in contact with the electrode:

Standard conditions

The electrode potential depends on the conditions used, including temperature, pressure and concentration of reactants.It is therefore necessary to specify the conditions used when measuring electrode potentials. These conditions are normally set at a temperature of 298 K, a pressure of 1 atm and with all species in solution having a concentration of 1.0 mol/L. Electrode potentials measured under these conditions are known as standard electrode potentials -denoted by Eo.

Reference electrodes (THE SHE)

The emf of electrochemical cells is easy to measure, but the individual electrode potentials themselves cannot actually be measured at all; it is only possible to measure the potential difference between two electrodes.

It is therefore only possible to assign a value to a half-cell if one half-cell is arbitrarily allocated a value and all other electrodes are measured relative to it. An electrode used for this purpose is known as a reference electrode. The electrode conventionally used forthis purpose is the standard hydrogen electrode.

The gas pressure is fixed at 1 atm, the temperature is 25oC and the H+ ions have a concentration of 1.0 moldm-3.

This electrode is assigned a value of 0.00V.Using this electrode, it is possible to assign an electrode potential to all other half-cells.

Eg if the standard Zn2+(aq) + 2eZn(s) electrode is connected to the standard hydrogen electrode and the standard hydrogen electrode is placed on the left, the emf of the cell is -0.76V.

The Zn2+(aq) + 2eZn(s) half-cell thus has an electrode potential of -0.76V.

Eg if the Cu2+(aq) + 2e  Cu(s) electrode is connected to the standard hydrogen electrode and the standard hydrogen electrode is placed on the left, the emf of the cell is +0.34V.

The Cu2+(aq) + 2e  Cu(s) half-cell thus has an electrode potential of +0.34V.

The standard electrode potential of a half-reaction can be defined as follows:

"The standard electrode potential of a half-reaction is the emf of a cell where the left-hand electrode is the standard hydrogen electrode and the right-hand electrode is the standard electrode in question".

The equation emf = ERHS - ELHS can be applied to electrochemical cells in two ways:

Conventional Representation of Cells

As it is cumbersome and time-consuming to draw out every electrochemical cell in full, a system of notation is used which describes the cell in full, but does not require it to be drawn.

Half-cells are written as follows:

-the electrode is placed on one side of a vertical line.

-the species in solution, whether solid, liquid, aqueous or gas, are placed together on the other side of the vertical line.

-if there is more than 1 species in solution, and it is on different sides of the ½ equation, they are separated by a comma

Eg Fe3+(aq) + e  Fe2+(aq)

When two half-cells are connected to form a full electrochemical cell, the cell is written as follows:

-the positive electrode is always placed on the right

-the two half-cells are placed on either side of two vertical broken lines (which represent the salt bridge)

-the electrodes are placed on the far left and far right

-on the left (oxidation)(LEO) is written, on the right (reduction) GER

EgCell reaction: Ag+(aq) + Fe2+(aq)  Ag(s) + Fe3+(aq)

Rechargeable and non-rechargeable cells

Electrochemical cells are the basis for all batteries. Batteries contain two separate half-cells. The solutions are connected by a salt bridge which allows ions to flow through without allowing mixing of the solutions.

Reactions taking place in the half-cells are irreversible the battery is non-rechargeable and called a PRIMARY CELL

If the reactions taking place in the half-cells are reversible, the battery is rechargeable,& called a SECONDARY CELL.

Fuel Cells

A fuel cell is a cell in which a chemical reaction between a fuel and oxygen is used to create a voltage. The fuel and oxygen flow into the cell continuously and the products flow out of the cell. Therefore the cell does not need to be recharged.The most widely used fuel cell is the hydrogen-oxygen fuel cell:

A fuel cell, like a regular electrochemical cell, consists of two half-cells connected by a semi-permeable membrane. An aqueous solution of sodium hydroxide is used as the electrolyte.

Oxygen is pumped into one of the half-cells:

O2(g) + H2O(l) + 4e-4OH-(aq)E0 = +0.40 V

Hydrogen is pumped into the other half-cell:

H2O(l) + 2e-H2(g) + 2OH-(aq)E0 = -0.83 V (flip this)

The oxygen half-cell is more positive and therefore undergoes reduction. The H2 half-cell is more negative and oxidizes

O2(g) + H2O(l) + 4e- 4OH-(aq)reduction

H2(g) + 2OH-(aq)  H2O(l) + 2e-oxidation

______

O2(g) + 2H2(g)  2H2O(l)overall cell reaction, emf = 1.23 V

There are a number of advantages of fuel cells as a way of producing energy:

- The hydrogen-oxygen fuel cell produces water as the only product.

- Fuel cells are more efficient than combustion engines. Typically fuel cells are approximately 50% efficient but combustion engines are approximately 20% efficient.

However there are also a number of limitations of fuel cells as a way of producing energy:

-Hydrogen is flammable. It is therefore both difficult and dangerous to store and transport.

-Fuel cells use toxic chemicals in their manufacture and Fuel cells have a limited lifetime

Grade 11 Cambridge |Stops here|

Electrolysis

Electrolysis: the process in which electrical energy is used to bring about a non-spontaneous chemical change. In both types of cells, the electrode at which the reduction occurs is the cathode and oxidation occurs is the anode.

In an electrolytic cell, the flow of electrons is being ‘pushed’ by an outside source such as a battery. The cathode is called the negative electrode of the electrolytic cell.The anode in the electrolytic cell is called the positive electrode.). We SWITCH the signs

The differences between galvanic and electrolytic cells can be summarized in a table.

Galvanic/Voltaic Cells / Electrolytic Cells
chemical energy electrical energy / electrical energy chemical energy
two half-cells with separate electrolytes and a salt bridge / electrodes in the same electrolyte
chemical reaction is spontaneous Eototal is positive / chemical reaction is not spontaneous Eototal is negative
anode - negative terminal : cathode - positive terminal
oxidation always occurs at the anode
/ anode - positive : Cathode - negative
oxidation occurs at the anode

Electrolysis of Molten Salts

Electrolysis is a process where electrons are forced through a chemical cell, thus causing a chemical reaction.

Reduction always takes place at the cathode. In the electrolysis of molten salt, NaCl, the cathode and anode reactions are:

Anode (oxidation): 2 Cl- Cl2 + 2 e- Cathode (reduction): Na+ + e- Na

Anode oxidation / Cathode reduction
2 Cl- Cl2 + 2 e- / 2 Na+ + 2 e- -> 2 Na
2 Cl- + 2 Na+ Cl2 + 2 Na

If one mole of electrons (96485 C or 1 Faraday) passes from the anode to the cathode, one mole of Na (23 g) will be deposited, and half a mole of chlorine gas Cl2 (or one mole of Cl atoms) will be collected at the anode.

Electrolysis of Water

Pure water does not conduct electricity, because the numbers of H+ and OH- ions are small (10-7 mol/L each). In the presence of an acid, water can be decomposed.

A potential of -2.06 V is the standard cell potential for,

Pt | H2O, [H+] = 1 M | O2 || H2O [OH-] = 1 M | H2 | Pt

And when a potential greater than 2.06 V is applied, the following reactions take place.

Anode oxidation / V / Cathode reduction
H2O 4 H+ + 4 e + O2 / -2.06 V / 4 H2O + 4 e  2 H2 + 4 OH-
2 H2O  2 H2 + O2

The Hall-Heroult process

Aluminum (Al) is the third the most abundant elements on Earth crust, in the form of bauxiteAl2O3. Because Al it is very reactive, this metal remained unknown to mankind until 1827, bound up inside bauxite. In 1886, two young men working in two continents apart electrolyzed molten cryolite Na3AlF6 (melting point 1000° C); their discovery is now known as the Hall-Heroult process, which is a commercial process.

Faraday’s Stoichiometry

Electrolysis CREATES chemical reactions. The following examples illustrate the stoichiometry of electrolysis. The charge on a mole of electrons is called the Faraday. The best estimate of the value of a Faraday, according to the National Institute of Standards and Technology (NIST), is 9.65x104 coulombs per mole of electrons. The charge on an electron based is 1.60 x 10-19 coulombs per electron. If you divide the charge on a mole of electrons by the charge on a single electron you obtain a value of Avogadros number of 6.022 x 1023 particles per mole.