The subject of this module is "REDOX-reactions", where REDOX is an abbreviation of:
RED from REDuctor, a reducing agent
OX from OXidator, an oxydising agent
In every redox reaction, those two are the most important participants.
In a redox reaction always react an oxidator and a reductor.
Simultanously a new oxidator and a new reductor are made.
You should be able (in a class room?) to mix some iron powder and sulfur powder in a test tube and then, carefully heat the mixture above a burner flame.
First you see the sulfur melt and mix with the ironpowder --> it seems te become a black and liquid mixture, but soon the real reaction starts.
It start even to glow; lots of energy are released.
The result - after cooling you can hold it - is a hard, shapeless substance, black colour
and it doesn't look like iron or sulfur anymore.
Here we have an example of a redox reaction with iron as the reductor and sulfur as the oxidator.
What exactly happened?
A short remembrance to the chemical concepts that you will need:
Redox has everything to do with electrons, mostly the valency electrons of certain atoms.
What about that donation or capturing of electrons?
Do you still know:
All particles (atoms, molecules, ions) always contain electrons.
Describe exactly what happens when two particles meet, and one will donate and the other will capture electrons.
How many electrons are present in the following particles:
An oxydator is any particle that can/'want' accept or capture electrons;
[n.b.: 'want' is in brackets, because 'to want something' is not exactly a property of matter. Particles have nothing to 'want'.
it is an electron acceptor
Only living creatures can 'want'.
But nevertheless we use this kind of words in this course, just because it is fun to pretend that substances have that kind of properties.
Besides, we do not know everything about matter!!
A reductor is a particle that can or want donate electrons;
it is an electron donor
Electrons are not just stand alone available. They belong always to one or more particles.
Transfer of electrons, normally does not occur at distance; it occurs in direct contact between particles.
So a reductor cannot shoot an electron and somewhere else it hits an oxydator.
Some rules must be known for substances/particles that participate in a redox reaction:
- There are oxydators and reductors that only react if some auxiliary substances are present.
Neutral elements very often are oxydators (in particular some non metals) or reductors (many metals).
this has everything to do with the fact that metal atoms have few valency electrons, and non metals have a lot.
To study this, you can go to module 03 about the chemical bondings and to module 01 about the atoms.
- Call table X to see the half reactions.
- Extract all neutral metals and write them in order of strenghth.
- Litium is the strongest and Gold is the weakest. Check that.
- Att.: only extract those metals that do not need helping substances like H+.
- Look, now you have got the so called potential series of metals.
- You can do the same for the non metals, under exlusion of those that need auxiliary substances.
Chose the right answer:
Oxygen can oxydise glucose completely. the products of this complete oxydation are:
- Oxygen and Hydrogen
- Carbondioxyde and water
- Carbondioxyde and ureum
- Only water
Lots of redox reactions occur, also in daily life, such as burning of fuels (petrol with Oxygen, or rusting of Iron), but also such as the energy production in the human body. Its all kind of combustion.
Nevertheless, most redox reactions of this course are not as such the combustion reactions.
In daily life, exept from Oxygen, there are many other oxydators (and reductors): Iron, glucose, permanganate, Hydrogen, Alcohol, etcetera.
Explain if the following substances are oxydator or reductor: oxygen, Iron, glucose, permanganate, Hydrogen.
Explain if the following statements are true or false:
- a substance acting as an oxydator, always must contain the element Oxygen.
Iron(s) + Copper(II)sulfate(aq) Copper(s) + Iron(II)sulfate(aq)
is a redox reaction.
In the history of chemistry there has been a development in the definitions of oxydator and reductor.
First they connected the word oxydator to the element Oxygen, but today we know many oxydators having nothing tot do with Oxygen.
We use different definitions now.
Write for yourself (without looking further in this course) how you would define a redox reaction.
What is your formulation?
Which is the right answer? Explain:
During the oxydation of glucose, the molecules of glucose will:
- lose electrons
- H+ lose ions
- H+ catch ions
- accept electrons
A redox reaction is a chemical process
where electrons are transferred from one substance to another.
A redox reaction is a chemical process
in which the oxydation numbers of one or more elements change.
Any substance that accepts electrons during a reaction is an oxydator (Ox)
The opposite: a substance that loses electrons is a reductor (Red)
During a redox reaction always react a reductor and an oxydator.
electrons are transferred from the reductor to the oxydator.
In this transfer, always the valency electrons are involved (the electrons of the outer orbital of the atoms)
Try to remember what again is an electronic formula and give some examples.
The answer is something like:
An electron formula shows all valency electrons as dots or dash (one dash = 2 dots)
When electron forulas change during a reaction, there must have been transfer of electrons.
or: it was a redox reaction.
In the middle example, a sulphur atom donates two electrons and act as a reductor, where sulfate is formed from sulfite.
At rusting of Iron, the iron(III) ion is a product.
- Who is the oxydator or/and the reductor?
- What happens with the valency electrons?
- Give the half reactions, and what is the redox couple?
A redox reaction is always built up of two half reactions:
|The equation of couple 1:
||ox 1 + electrons
|The equation of couple 2:
||ox 2 + electrons
|| red 2
||red 1 + ox 2
||ox 1 + red 2
The number of electrons involved must satisfy the rule:
ALWAYS THE SAME NUMBER OF ELECTRONS IS DONATED AS IS CAPTURED
Electrons do not get lost, are not just created from nothing, and so:
You must always confirm and certify the two half reactions in that way: the number of donated electrons must equal the number of captured electrons.
|The equation of couple 1:
||Al3+ + 3e-
||| x 2
|The equation of couple 2:
||I2 + 2e-
||| x 3
||2Al + 3I2
||2Al3+ + 6I-
The total equation is only reliable if the number of electrons in half reaction 1 equals the number of electrons in half reaction 2.
In case of insolubility of the salt Aluminium Iodide, there will be a secondary reaction: the precipitation of an insoluble salt.
Att.: This is not the case in the example; check that in table XI:
solubility of salts in water.
Put an iron nail into a solution of Copper(II)sulfate.
The blue colour of the copper(II)-ions become lighter, can even disappear, and a new solid will appear in the reaction vessel.
Give the two redox couples and the both equations of the half reactions.
Again we need table X with the redox couples, where we meet a column with reductors and a column with oxydators.
Every line, two conjugated particles are joined.
No matter what book about chemistry, always is there a redox table, almost always with some small differences,
for example, the position of reductors and oxydators, the strong and the weak ones.
So be carfull.
Oxidators en Reductors exist as atom, as molecule or as (complex) ion.
Choose from table X an example for every kind (so three different oxidators and three different reductors, and show the half reactions.
Goto answer 10-12
Many oxydators and reductors only act in the presence of certain auxiliary substances.
For example, sulfate only can oxydise in the presence of an acid (ion H+).
Chose another example from tabel X of an oxydator and of a reductor that need auxiliary substances
and write their half reactions.
The same particle (same substance) can occur more than one time in the redox table, on different positions, varying in strenghth.
Att.: this phenomena only occurs when using different auxiliary substances.
In principle always the strongest will react, but of course under the condition that those needed extra substances are really present.
Permanganate appears in some tables three times in the column of oxydators, every time under different conditions.
What will be the half reaction to be used in case of a simple solution of just potassium permanganate?
Any chemist knows a lot of substances from the redox table by heart, not only the formulas, but also the names.
Try hard. You will become more critical when the media come with any statement.
A redox reaction often occurs in a reversible proces of the type: equilibrium (see module 08).
A reductor, after donating electrons, changes in a substance that, in its turn, can gain electrons again: an oxydator.
Ans, an oxidator in a redox reaction always becomes a reductor.
||ox 1 + electrons
||ox 2 + electrons
|| red 2
||red 1 + ox 2
||ox 1 + red 2
Whoever studies (module 08) the equilibrium reactions, knows that an equilibrium very often has a 'one-side-position' (there are many more weaker substances), wether left or right.
Please know that there are in the same way weak oxidators and weak reductors.
The stronger an oxydator, the stronger the tendency of this oxydator to pick up electrons.
and of course:
The stronger a reductor, the stronger the tendency of this reductor to lose electrons.
In the tables you find the oxydators and reductor ordened in strenght.
Compare table X with redoxtables in chemistry books or on the internet.
Check if they are indeed ordered in strenght and if this is done in all tables in the same way.
The redox reaction occurs spontaneously if we join stronger substances that create weaker ones.
The opposite is possible, but only with help from outside (see further on).
So, being spontaneous or not depends directly on the strenght of the reactants.
In general we may say that a redox reaction is an equilibrium:
red1 + ox2 ox1 + red2
This equilibrium responds to the rules for chemical equilibria, for example that the stronger ones react in favor of the weaker ones.
In other words: if you join strong substances with enough energy, than the biggest part of those stronger substances will disappear (react) while the weaker substances appear in a spontaneous process.
Those weak products still maintain a weak tendency to react back in a revers reaction, but - weak as they are - they do not succeed very well.
You can also say: direct redox reactions occur if you join strong substances.
You can predict a redox reaction on ground of the position of the oxydator and reductor in the redox table.
a redoxreaction occurs (spontaneously) if the oxidator has a higher position in table X than the reductor.
Behind this rule, something is hidden:
Of the oxidator is above the reductor, the products (in the next column of the table) are weaker ones.
And we know that they remain in an equilibrium.
- Check now if Iodine will spontaneously react with Lead, yes or no.
- If yes, can this reaction be supported by auxiliar substances?
- Which ones?
If the reactants (red1 and ox2)are strong (stronger that the products ox1 and red2), a spontaneous redox reaction will occur, or: the equilibrium goes to the right, in favor of the products.
Of red1 and ox2 hardly remains a bit and often we do not even apply the double arrow symbol
The transfer of electrons can take place in two ways: spontaneous or forced (non spontaneous)
If particles that are eager to donate, react, have contact with other particles that will capture, very probably the redox reaction will occur spontaneous.
Of course this phenomena has much to do with strong and weak substances.
Another way to divide redox reactions (also to do with electron transfer): direct or indirect redox reactions
In this chapter we will talk about direct redox reactions and in the next chapter about indirect redox reactions.
Direct redox reactions occur at the moment of direct contact between the involved particles.
Direct redox reactions can only occur spontaneously; there are no forced direct redox reactions.
You can join weak redox substances, but spontaneously nothing will happen.
If, for example, Oxygen molecules and petrol molecules meet each other (both in gas form = free), the result of this meeting can be a spontaneous redox reaction.
In that case the electron transport from the petrol moleucules to the Oxygen molecules will cause a big difference of substances. New substances are formed, products.
Bondings are broken; new bonds are made, as long as product are being formed (water and carbondioxyde).
The pictures below show a reaction between sulfite and permanganate.
Consult the redox table and give the half reactions.
two solids, sodium sulfite and potassium permanganate, are dissolved in water
(one tube with colourless solution and one tube with a purple, transparant solution)
Than we join of both solutions about 1 ml in a third test tube.
A new product is formed, brown and insoluble in water (a brown precipitate is made)
(in the right test tubes some extra water is added to have a better look at colour and precipitate)
Rusting of iron is a direct redox reaction, yes or no? Explain your answer.
Goto answer 10-18
Mind that many reactions are not finished immediately when the products are formed.
It is very well possible that secondary reactions occur, where products (for example the new oxydator and reductor) are involved.
The Nitrate ion
Nitrate ions are very good oxydators, but only under the condition that sufficient acid (H+) is present.
The nitrate has various ways to react, dependent on the concentration of that nitrate:
NO3-+ 2H+ + 1 e-
NO2 + H2O
Moderate and low concentrations
With few acid: NO3-+ 4H+ + 3 e-
NO + 2H2O
With more acid: 2NO3-+ 12H+ + 10 e-
N2 + 6H2O
Very low concentrations
2NO3-+ 10H+ + 8 e-
N2O + 5H2
is the normal Nitrogen gas, colourless and odourless.
NO + NO2
the 'nitrous vapors', yellow brown, suffocating
a colourless gas, in fact a kind of nerve affecting gas.
It causes strange face contractions (kind of laughing) and it helps to anaestethise people (or animals).
It was found in 1860 and applied, for example, by dentists.
The sulfate ion"
Can only act as an oxydator (and rather strong oxydator!) onder the condition of being accompanied by a good concentration of H+.
Sulfuric acid is than a strong oxydator, certainly when concentrated and even more at higher temperature.
The sulfate ion (with H+) can be found in the redox table, not as high as you would expect; it seems not as strong as you would expect.
What could be the reason for that?
Controll the nex (true) statements, and use the redox table; give your comments:
red 1 + ox 2 ox 1 + red 2
- Chlorine is a stronger oxydator than Bromine
- Zn is a stronger reductor than Fe
- Copper(II)- and SilverI)-ions are weak oxydators
- Nitrate and sulfate react only if the oxydator is in acid environment
- Permanganate is in acid environment a stronger oxydator than without acid
- The ion Fe2+ can be both: oxydator and reductor
If the reactants (red1 and ox2)are strong (stronger than the products ox1 and red2), than a spontaneous redox reaction will occur, or: the equilibrium goes far to the right, in favor of the products.
Of red1 and ox2 hardly remains a bit; often we do not even use the equilibrium arrows.
Metals have a very important impact in the redox chemistry; that's why we ask attention for these elements now.
The real applications of the redox come later.
All metals are reductors (they have few valency electrons that can be donated), but not all of them in the same way.
On the contrary, Platinum and God are very weak, Calcium or Sodium are extremely strong.
Those very weak metals, in practice, do not really react; they are inert; the strong ones often can react heavily allready if they touch a bit of water.
Between these two extremes, the metals react more of less with acids.
In industrial areas you can notice sometimes some 'acid rain' when you look at metals and chalk in the neighbourhood.
Normally the metals must be protected against influences of water and acids, and we do so with paint, oil, etcetera. All to avoid corrosion.
The metals that do practically not react are the so called 'noble' and 'half noble metals.
Below find a redox table with metals:
Au3+ + 3e- Au
NO3- + 2H+ + e- NO2 + H2O
Ag+ + 1e- Ag
Fe3+ + 1e- Fe2+
Cu2+ + 2e- Cu
Cu2+ + 1e- Cu+
SO42- + 2H+ + 2e- SO32- + H2O
Pb2+ + 2e- Pb
Ni2+ + 2e- Ni
PbSO4 + 2e- Pb + SO42-
Fe2+ + 2e- Fe
Zn2+ + 2e- Zn
2H2O + 2e- H2 + 2OH-
Al3+ + 3e- Al
Mg2+ + 2e- Mg
Al(OH)4- + 3e- Al + 4OH-
Na+ + 1e- Na
Ba2+ + 2e- Ba
Ca2+ + 2e- Ca
K+ + 1e- K
Lets analyse this table:
In the column with oxidators you find the Gold ion, the stronguest oxydator with its conjugated reductor, the metal Gold, the weakest reductor.
In practice Gold has not the least tendency to react, to donate electrons. It is a noble metal.
Generally spoken, we may say that a reaction between reductor and oxydator will occur spontaneously when in this table the reductor has a position above th oxidator.
Than Silver will react with nitric acid, but not with sulfuric aced. And Calcium react spontaneously with water, but Iron does not.
Don't forget here that the table data count with standard temperatures of 20 Ó 25oC.
The position of a substance in these tables can change if the temperature changes.
General rule: to have a spontaneous redox reaction, the oxydator must be above the reductor.
In the table three non-metals are included, namely the three solutions/liquids: nitric acid, sulfuric acid and water.
Explain if it would be wise to allow contact of Al with water.
We can notice that various metals can react in two ways: with or without auxiliary substances.
Example: Lead react much better in the presence of sulfate ions.
Gold and Silver react difficultly. What substance could help to improve their reactivity?
Some particles can be found in both columns of the redox tables.
This means that they can be oxidator as well as reductor, that they can capture as well as donate electrons.
Wether they will go one way or the other depends on the circumstances and of the presence of certain helping substances.
In general we know that a substance wil react as an oxydator when it meets a strong reductor.
and vice versa: react as e reuctor in contact with a strong oxydator.
Tin en tin-ions can be met in three ways: Sn Sn2+ Sn4+
There you have one: Sn2+ can capture as well as donate.
The same for Manganese: Mn2+ MnO2 MnO4-
In the middle you find MnO2 with the oxidation number of Mn: +4 (Mn4+); this one can capture and donate.
Even odder it becomes when one and the same substance simultaneously is reacting as oxydator and as reductor, or:
this substance reacts with itself.
electrons are transferred internally.
Sn2+-ions can react with one another, where one ion will donate electrons (becoming Sn4+) and the other ion will capture and becomes neutral Sn).
Such particular situations we call "autoredox reactions".
Taken from a redox table:
The substance Hydrogen(hy)per oxyde can act as an oxydator and as a reductor.
|H2O2+ 2H+ + 2e-
|O2 + 2H+ + 2e-
Suppose that this (H2O2) is in an aqueous and acid solution.
It can be an oxydator: H2O2 + 2H+ + 2e- 2H2O
it can be a reductor: H2O2 O2 +2H+ + 2e-
In the table you find this oxydator above the reductor, or: the substance reacts spontaneously, is strong enough.
This means that it wan't be easy to keep hydrogen peroxyde in the bottle for a long time.
It has the tendency to be transformed (via this auto redox reaction) into water and oxygen.
Give the two half reactions + the total reaction of the autoredoxreaction of the Iron(II)ions.
Study well the following halfreactions:
Which ion is in both columns? Can this ion participate in an autoredoxreaction? Explain your answer.
|Cu+ + e-
|Cu2+ + e-
Study well the following half reactions:
One ion that participates in the autoredox is the Lead(II)ion, that can capture 2 electrons and also can donate 2 electrons.
|PbO2(s) + SO42-+ 4H+ + 2e-
||PbSO4(s) + 2 H2O
||Pb(s) + SO42-
This reaction is applied in the 'Leas-battery'.
- Which one of the three halfreactions occurs in the Leas battery?
- Why is it, in practice, important here to give attention to the phase (s) of this half reaction?
Apart from the direct, there are also indirect redox reactions;
this means: there is no direct contact between the particles of the oxydator and the reductor. The transfer of the electrons is realised via-via, mostly via metal wiring or other conductive material.
Exteriorly the conducting wires take care for the electron transport from RED to OX, without direct contact between the reactants. They do not meet; they do not collide.
The half reactions take place at the surface of electrodes (mostly a metal or graphite).
There are two kinds of indirect redox reactions:
In both cases, that means: in all indirect redoxreactions, electrodes are applied.
- spontaneous indirect redox reactions, when strong substances react to produce weaker ones. These reaction mostly ar exothermic.
- forces indirect redox reactions, when weak substances must (are forced to) react to produce strong ones. These reaction mostly are endothermic.
Indirect redox reactions only take place if in between the electrodes is an environment that is conductive for electric current (meaning that charged particles are present that can freely move around, like the ions of dissolved or molten salts).
You could also say that the circuit in this system must be a closed circuit. Part of the circuit is made by conductive metal wires an electrodes with free electrons, and the other part is made of a solution of molten or dissolved electrolyte with free ions.
Those free ions can freely move between the positive and the negative electrode.
To make this movement possible, the electrode compartments must have contact.
This contact will sometimes be realised with a (semi)porous barrier, of else - if the two compartments are separated - those compartments must be connected with an ion bridge, or better: a salt bridge. That is a kind of tube filled with salty gel (gel with ions).
Below you see a schema of a simple chemical cell, voltaic cell, with copper/zincum electrodes.
Consider well this scheme:
In the whole complex you can discover a sufficiently strong oxydator (Cu2+) and also a rather strong reductor (Zn)
The (spontaneous) reactions dominating the process are:
Zn(s) Zn2+ + 2e-
Cu2+ + 2e- Cu(s)
You should know that realising a redox reaction at the surface of an electrode almost always cost a certain activating energy.
Such a reaction must me started up.
Certainly if gases are formed during this reaction, this activating energy can be considerable.
Suppose that a copper(II)ion arrives at an electrode, and makes there a copper atom; enough activating energy for that is available,
- the electrode will become positive or negative?
- Explain your answer;
- Give the half-reaction;
- What can you observe?
Indirect redox reactions only take place if, between the electrodes, there is an environment conductive for electric current (meaning: charged particles must be present that can freely move around, like ions of dissolved salts).
You can also say that the current circuit must be closed; part of the circuit are the conductive metal wires and the electrodes with free electrons, and the other part is the solution or melt with free ions.
Those free inons can freely move between the postive and the negative electrode.
To mak this movement possible, the electrode compartments must have a contact.
This can be a semipermeable (porous) barrier, or else - if the two electrode compartments do not have direct contact - those compartments must be connected via an ion bridge, a salt bridge. This is some kind of tube filled with salty gel (a gel with ions).
Study the next scheme and explain all processes taking place there.
An electrode can have lack of electrons;
than the electrode is positive: anode
An electrode can have extra electrons;
than the electrode is negative:cathode
Cathode-Negative // Anode-Positive (CNAP)
Anions and cathions
- What could be that?
- How are they charged? explain your answer.
Electrodes are mad of conductive materials: mostly a metal, maybe graphite.
Those substances contain free electrons in a metal lattice. These free electrons can freely move within the electrodes, coming from the negative and going to the positive electrode (one-way traffic!)
Very special is the liquid electrode. How does this work?
Look at the figure and
- explain its functioning.
- What advantages could have this electrode?
Goto answer 10-30
- Explain why metals and graphite conduct electricity
- Explain why copper- or iron electrodes react as reductors
- Give an example of a molten substance that can conduct electricity
- Explain if destilled water can conduct electricity
Inert and participating electrodes:
If the electrode material is of the type: 'very weak' (as reductor or as oxydator he is nothing), than this electrode will be INERT.
The only activity is transporting electrons, but it does not take part in the real redox reaction.
Examples: gold, platinum, graphite.
On the other hand, we use electrodes made of materials that really participate in the redox reaction, for example, in the case of an iron or zincum electrode.
An iron bar can very well serve as an electrode, but will send iron ions into the solution, while electrons remain behind in the barr.
Really, after some time such an electrode will slowly be consumed; it becomes thinner and disappears.
The iron suffers the following half reaction:
- Fe is the iron of the electrode
- The Fe2+ -ions go into the solution to move there freely in the direction of the negative electrode.
- The electrons staying behind in the iron electrode will participate in the electron transfer and the electron transport.
A participating (so not inert) electrode always is made of neutral, non noble metals; they serve as reductors, or: they can donate electrons.
- Explain why non-metals and electrolytes cannot serve as material for electrodes, and why metals and graphite will do.
- Why is it that an corrosive metal Sodium never is used as electrode in a salt solution?
Explain why a participating electrode only can act as anode and never as cathode.
The word "electrode compartment" has been used a few times now;. that is the space directly around an anode or cathode, its surface included.
There the half reactions of the indirect redox reaction take place.
The Cathode and anode compartments may not be separated physically, and everything takes place in one solution or liquid.
To keep the situation equal in all places, stirring of the solution is needed continuously.
Another option is: the two processes occur in separated compartments, but connected with semiporous barriers of with salt bridges.
Without one of these, there is no closed circuit and current is impossible (nor any transfer of electrons)
In the case of a salt bridge, take care that the ions in that salt bridge themselves do not participate in the redox reaction.
A choice pro or contra separated electrode compartments has all to to with what you aims with the redox reaction in charge.
Keeping apart mostly aims to keep de products separated.
These product could possibly react (when in contact), and - for example - produce a precipitate or a gas.
Onthe other hand, someone can also wish that happen.
A solution (or a molten substance) conducts electricity only under the condition of the presence of charged and movable particles (ions in this case).
The conductibility has everything to do with the movability of those ions:
The more movalbe, the better they conduct.
That movability depends on some factors:
- how many ions are present?
- are these ions big or small or hidrated?
- Somtimes an ion (certainly the H+-ion in water) has special techniques to improve their movability.
Explain why a metal or graphite cannot serve as a salt bridge.
Penlite's and batteries; electrochemical cells, galvanic elements
Spontaneous redox reactions normally occur in direct contact between the particles of the oxidator and the reductor, but can be realised in an indirect way, at a distance, at the surfaces of electrodes.
For example in the case of the Cu/Zn-electrode.
The zincum dissolves spontaneously, forming ions Zn2+, leaving behind the electrons in the Zincum bar that, because of that, collects a negative charge.
The copper ions in the solution capture electrons from the copper bar, become Cu atoms that immediatly join with the copper bar. This copper bar wil thus be charges positively.
The Zincum electrode becomes slowly thinner and de Copper electrode will grow a bit.
In the electrochemical cells (galvanic elements) occurs a spontaneous change of chemical energy onto electrical energy.
Strong substances react (oxidator and reductor) through electrodes (i.e. indirectly), while the weaker products are formed.
The electrons are transported and tranmitted via external conducting connections (wires and electrodes).
The movement of the electrones through wires is called: electric current.
We all know the applications as penlites and batteries.
Within the voltaic cell you must always distinguish and separate well the electrode compartments. Why?
Consider the following equilibria at two electrodes:
(at the anode:) red 1 (Cu)
ox 1 (Cu2+) + electrons (a)
(at the cathode) ox 2 (Zn2+) + electrons
red 2 (Zn) (b)
If a dry battery delivers current (sends electrons), those electrons come from negative electrodes and move through the exterior to the positive electrode.
At equilibrium (a) the formation of Copper dominates (the equilibrium stays left).
As soon as the electrons arrive at this pole, the equilibrium can move even more tot the left.
At equilibrium (b) dominates the formation of Zincum ions.
As soon as electrons go away, this equilibrium tries to dislocate the equilibrium further, also to the left.
These two processes continue for a good time, as long as one of the reactants (Copper bar or Zincum bar) is gone.
We say than: the battery is 'empty'.
De batterij noemen we dan 'leeg'.
a penlite must have a strong oxidator and a strong reductor that do not have direct contact inside the battery. (they may not react directly)
The transfer of electrons must happen through the exterior, indirectly, so via the outside wiring; otherwise it is useless.
Compare the two images of the same penlite and describe the differences.
Explain how an old fashioned car battery funcions (PbSO4//PbSO4/PbO2); do you think that this battery is rechargebale?
Check the redox table (tabel X) and chose the oxydator and the reductor with sufficient force to serve as electrochemical cell, voltaic cell).
Do you think that you could make such a cell yourself?
Normally, a chemical equilibrium will shift to the weak substances (spontaneous reaction).
The other way is also possible: we can force a redox reaction between weak substances to occur and for that we use external forces:
An electric source, like a battery, connected to the electrodes in an electrochemical system.
Than we talk about electrolysis.
Again, of course, there are two electrodes.
At one of them the half reaction takes place of the reductor and at the other, the half reaction of the oxydator.
The external source has a positive and a negative pole.
The negative pole is connected with an electrode, and than?
That negative pole of the source send many electrons to that electrode; this one turns to be negative (cathode).
The positive pole of the source is connected with the other electrode; this one turns to be positive, because it has to lose electrons (anode).
Thes two charged electrodes - if the potential is sufficient - will now function as follows:
The cathode (negative) will try to lose electrons, will look for particles around capable to pick up electrons (those particles must be oxydators)
The anode (positive) will try to catch electrons and will look for particles capable to donate electrons.
(those must be reductors)
If there are particles / substances in an electrode compartment that want to donate or catch electrons - and even weak reductors or oxydators are welcome - than this will happen; if at least the external source is strong enough.
If not, if the source is not strong enough, you always can increase the voltage. The weaker the present oxydator or reductor, the higher the needed voltage of the external source.
Those oxydators and reductors can be molecules, atoms or ions.
Att.: complex ions, in particular those with Oxygen, like sulfate, have some difficulty to participate in this kind of electrolysis reaction, because of their activating energy (here OH- is an exeption).
And again: the redox tables show/determine who will act as reductor and who as an oxidator.
A solution of Calcium nitrate suffers electrolysis with Pt-electrodes.
Give the equations of the two half reactions occurring at the electrodes.
During electrolysis, the redox half reactions occur separately and a total reaction is not always interesting of important.
But a total reaction becomes important when calculations are involved:
Than a total reaction is absolutely needed.
- How much must be dissolved?
- How much of this or that product can be formed? etcetera.
But to understand well the electrolysis, it is sufficient to know what exactly happens at each electrode.
Suppose that you have wet a piece of filter paper with a KI-solution (with some drops of an indicator, fenolftale´ne).
Two electrodes are connected to a dry cell are connected with this wet paper, at about 1 cm from each other.
An electrolysis takes place (forces redox reaction).
- Give the half reactions
- What can you observe during the process?
Electrodes make indirect redox reactions possible. The transfer of the electrons does not occur in direct contact between the reacting particles, but via conducting matter.
In electrolysis, the electrode can be made of inert material (graphite, platina etc) or they participate in the electrolysis process, as, for example, an electrode of Zincum or Copper.
Remind that also the liquids between the electrodes must be conductive (must contain free ions).
The electrolysis explained in another way:
Imagine that the electrodes will force the present particles to electron transfer.
The anode is positive, so has shortage of electrons, wants to capture electrons, is in search of nearby particles that can donate electrons and - if the voltage is enough - will force those particles, that substance to do so.
At the cathode occurs exactly the opposite.
- Formulate exactly what happens at the cathode.
- Will the anode attract oxydators or reductors? Explain your answer.
Have a close look at the following scheme and explain it:
Suppose the two electrodes are two persons: The hand of those two persons hold the electric source, and in this contact, the man is filled with electrons where the woman is emptied by the same source.
Both they don't like this situation; they want to change, to act.
The man is now going to act as a cathode and the woman becomes the anode:
The woman feels great shortage of electrons and, because she stands with her feet in the solution, she will try to find and catch there electrons.
And yes, she notices there any substance that can donate electrons (a reductor), no matter if that substance is willing to donate. She will force that substance to give her electrons.
This will happen, exactly at the moment such a particle touches the woman.
The man however is bursting of an overdose of electrons, but has no direct contact with the woman. He stands alone with his feet in the solution where any substance can take over some electrons, wether the particles like it or not.
That substance is forced to pick up electrons, so the man finds some satisfaction and rest.
This will happen as soon as the particles touche the feet of the man.
Easy to foresee what will happen when man and woman would meet directly. A serious short-circuiting would be the case.
Try to draw the electrolysis according to the above animation.
You already know the next image of the voltaic cell.
Such an electrochemical cell aims to create electrical energy with strong chemical substances (oxydators and reductors).
We always need two redox couples (to be found in the redox tables)
The (half)potential of any redox couple can be measured, and its values you can find in most tables; they are based on standard circumstances: 25oC and 1 atm. gas pressure or a concentration of 1M.
The stronger an oxydator or reductor, the higher the delivered potential.
Joining two couples in a complete cell, there will be a difference in potentials, a voltage, between the two redox couples.
In the case of a penlite that voltage will be about 1.5V.
Here we have two halfreactions of two redox couples:
Red: CuCu2+ + 2e-V(standard) = +0,34 V
Ox: Zn2+ + 2e- Zn
V (standard) = -0,76 V
The final voltage, under standard conditions, is: (red - ox) =0.34 - (-0.76) = 1.10 Volt [att.: you must always subtract the smallest from the biggest]
This battery wil create a voltage of 1.1 volt.
Changing the conditions will change the voltage. A higher concentration of Zn-ions (more OX) probably will create a higher voltage.
The oxydation number is the number of electrons that
- completely or partly -
is donated or captured by one or another particle.
The above diagram shows the changing oxydation numbers of Sulphur during the production process of sulfuric acid.
The process can roughly be described as follows:
S SO2 SO3
Affirmation: Graphic D is the only one that shows the oxydation numbers in the right way.
Explain if that is right or wrong.
- Neutral elements always get an oxydation number with vaulur 0 (the atoms have not lost or gained any electron).
- Hydrogen is neutral, so has an oxydation number 0, but in compounds it is almost always +1.
- Oxygen as an element has oxydation number 0, but in compounds mostly is -2.
What oxydation number have the elements in the following substances:
Potassium dichromate; Sodium permanganate; Silver; Water; Hydrogenperoxyde?
Which of the next reactions are redoxreactions?
(explain your answer with oxydation numbers)
- H2(g) + Br2(aq) 2HBr(aq)
- NH4NO3 N2O+ 2H2O
- NH4Cl NH3+ HCl
2K2CrO4 + H2SO4
K2Cr2O7+ K2SO4 + H2O
- H3BO3 + 4HF HBF4+ 3H2O
- Fe(s) + S(s) FeS(s)
With oxydation numbers, investigate who in the following reactions is oxydator or reductor:
Go to answer 10-46
SO2 + Br2 + 2H2O
2HBr + H2SO4
Mg + H2SO4
MgSO4 + H2
Cu + 2H2SO4
CuSO4 + SO2 + 2H2O
3I2 + 6KOH
KIO3 + 5KI + 3H2O
The below scheme divides the redox reactions in three (green) parts, result of joining direct and indirect redox with weak and strong substances.
This shows spontaneous and non spontaneous redox reactions:
dry and wet batteries
If we want to solve redox reactions, not with oxydation numbers, but with the half reaction equations, than we must follow some rules:
- what kind of reaction?
- what particles participate?
- what are the half reaction equations?
- and the total equation?
- what conclusions can we draw and what observations can be done?
The following scheme shows 6 steps that always return in solving a redox reaction, now and then dependent on the presence of electrodes and wether it is an indirect or direct reaction.
1. design a scheme of what happens, of what is done.
2a. Make a list of all present substances / particles, including electrodes
2b. Define the reductor and the oxydator (underline them)
3. give the half reaction equations
3. give the half reaction equations
OX reacts at the positive electrode
RED reacts at the negative electrode
4. make all equations certain
4. make all electrode equations certain
5. check possible secondary reactions
6. Write down your observations and conclusions
N.B.: the use of the redox table is needed.
In the chemical lab we put a small piece of the metal sodium into distilled water (in a kind of aquarium).
The Sodium is a soft metal; we just can cut a piece of it, but be careful:
they conserve Sodium for a reason under petroleum; it reacts spontaneously with air (Oxygen) and with water.
We dry the piece of Sodium (with filter paper we remove the petroleum) and put is carefully on the water surface.
Immediately it begins to react heavily, a gas is formed,
the piece of Sodium moves fastly on the water surface, burning with yellow flames.
This goes on until the metal has disappeared completely.
We join a couple of drops of indicator to the water; that shows us that the water became basic.
This is obviously a direct reaction: the Sodium particles collide directly with the water particles, and there are no electrodes.
This reaction is very spontaneous.
Now we obey the six steps:
A sketch of the situation:
the present particles/substances are: Sodium and water, nothing more.
We look in the table and must conclude that water is an oxydator as well as a reductor (in both cases very weak)
But Sodium is extremely strong reductor, so it makes sense to think that water in this case will act as an oxydator.
The reductor has a lower position than the oxydator, so the reaction is spontaneous.
The equations of the half reactions:
||Na+ + e-
||2H2O + 2e-
||H2 + 2OH-
NB: the number of electrons in both half reactions must be the same.
2Na + 2H2O 2Na+ + H2 + 2OH-
2Na(s) + 2H2O(l)2NaOH(aq) + H2(g) ΔH < 0
Problably there will be one secondary reaction: the Hydrogen gas (a product) is responsible for the fire; it reacts with oxygen from the air and causes the flames.
Observations and conclusions:
- The reaction is spontaneous, energy comes free in the form of warmth and light; so it is an exothermic reaction.
- A gas is formed (Hydrogen) that easily will start to burn; you can see that.
- The product NaOH creates a basic environment (ionen OH-); we can check that with an indicator: the colour becomes violet.
Below you find 6 redox reactions to solve according to the 6 rules of the scheme.
- Calcium(s) with water(l)
- Copper(s) met concentrated nitric acid
- Potassium permanganate(aq) with oxalic acid(aq) (in the presence of diluted sulfuric acid)
- Copper(s) with Zincum sulfate(aq)
- Bromium(aq) with Potassium Iodide(aq)
- Potassium permanganate(aq) with Sodium sulfite(aq)
- What can we observe in the reaction of Potassium with water?
- Will this reaction be more of less violent, compared to the reaction with Sodium?
In this module we do not enter in all kind of applications, because there will be a separate module for that (module 15).
But these two examples of redox in practice are shown as an exeption: they deal with 'blow at the traffic controll' and 'the black-and-white photography'.
Consumed alcohol enters the blood, partly in the lung and other organs, (including brains). It will affect the reaction power of the driver. In case of accidents, such a driver automatically is guilty. The police have different possibilities to check; one is based on a redoxreaction in the 'blow pipe'.
In that instrument is put potassium dichromate in acid environment with a yellow colour. If during the blowing, alcohol is oxydised by the yellow bichromate, a green substance (with Cr(III)-ions) is produced. If that is seen by the policeman, he wan't let you continue driving your car. The other product that is formed is acetic acid.
- Do you think this controle is useful and just, in the scientific as well as in moral sense?
- Give the half reaction equations and the total equation of the reaction and explain the observations.
In the old days (last century) photographers and hobbyists printed their own photos in their own dark room on photographic paper.
It still exists and still can be done!!
That paper contains a layer of very fine cristals of Silver bromide.
At the moment that light radiation (energy) reaches such a cristal, it is - in a way - affected:
AgBr(s) Ag(s) + Br2(g)
Ag(s) is black, and Br2
(escapes as a gas)
On the paper, immediatly after the lightning, you still can see nothing of this process, but that affection occurred on many places of the paper.
Then, you must submerge this paper in a developer bath (1).
Every affected cristal (has already a littlebit of neutral silver) wil start to react with the substances in that bath: hydroxyde and dihydroxy benzene.
2AgBr(s) + 2KOH(aq) + H2O + OH-
2Ag(black) + more substances
The more light fell on the cristals, the more silver will appear and the more black will be the result.
At the end there is still a lot of unused silverbromide and that - of course - may not stay there.
That would cause problems at the moment the film of the photopaper enters a light room.
The remained silverbromide must be taken away (fixed) with a solution of ammoniumtio sulfate.
AgBr(s) + 2(NH4)2S2O3(aq)
NH4Br(aq) + (NH4)3(AgS4O6)(aq)
At the end it is only a question of drying paper and film.
Check of all reactions in 6.2 the changes of oxydation numbers.