Redox reactions
Oxidation is to gain Oxygen, or a Halogen, or rather a non metal. Better definitions will be to loose electrons, or increase of Oxidation number.
Reduction is to gain electrons or reduction in Oxidation number.
Oxidation is loss of electrons.
Reduction is gain of electrons.
As Oxygen forming Oxide of say Potassium ( K ) as K2O Oxygen itself is getting reduced as that atom is gaining electrons from K
The oxidation number is defined as the effective charge on an atom in a compound, calculated according to a prescribed set of rules. An increase in oxidation number corresponds to oxidation, and a decrease to reduction. The oxidation number of a compound has some analogy to the pH and pK measurements found in acids and bases -- the oxidation number suggests the strength or tendency of the compound to be oxidized or reduced, to serve as an oxidizing agent or reducing agent. The rules are shown below. Go through them in the order given until you have an oxidation number assigned.
For atoms in their elemental form, the oxidation number is 0
For ions, the oxidation number is equal to their charge
For single hydrogen, the number is usually +1 but in some cases it is -1
For oxygen, the number is usually -2
The sum of the oxidation number (ONs) of all the atoms in the molecule or ion is equal to its total charge.
For example, in the extraction of iron from its ore:
Fe2O3 + 3 CO -> 2 Fe + 3 CO2
In this Iron ( Fe ) is getting reduced as its oxidation number changes from 3 to 0
While Carbon monoxide is oxidized as it gained a carbon atom.
An oxidising agent is substance which oxidises something else. In the above example, the iron(III) oxide is the oxidising agent.
A reducing agent reduces something else. In the equation, the carbon monoxide is the reducing agent.
Oxidising agents give oxygen to another substance.
Reducing agents remove oxygen from another substance.
Because both reduction and oxidation are going on side-by-side, this is known as a redox reaction.
Cu2+ + Mg -> Cu + Mg2+
In the above reaction Copper atom is gaining electrons ( or oxidation number is changing from +2 to 0 ( zero )) so Copper atom is getting reduced. While
Magnesium atom is loosing electrons so it is getting oxidized. So the magnesium is reducing the copper(II) ions by giving them electrons to neutralise the charge. Magnesium is a reducing agent. the copper(II) ions are removing electrons from the magnesium to create the magnesium ions. The copper(II) ions are acting as an oxidising agent.
Redox reactions, or oxidation-reduction reactions, have a number of similarities to acid-base reactions. Fundamentally, redox reactions are a family of reactions that are concerned with the transfer of electrons between species. Like acid-base reactions, redox reactions are a matched set -- you don't have an oxidation reaction without a reduction reaction happening at the same time. Oxidation refers to the loss of electrons, while reduction refers to the gain of electrons. Each reaction by itself is called a "half-reaction", simply because we need two (2) half-reactions to form a whole reaction. In redox reactions, chemists typically write out the electrons explicitly:
Cu (s) ----> Cu2+ + 2 e-
This half-reaction says that we have solid copper (with no charge) being oxidized (losing electrons) to form a copper ion with a plus 2 charge. Notice that, like the stoichiometry notation, we have a "balance" between both sides of the reaction. We have one (1) copper atom on both sides, and the charges balance as well. The symbol "e-" represents a free electron with a negative charge that can now go out and reduce some other species, such as in the halfreaction:
2 Ag+ (aq) + 2 e- ------> 2 Ag (s)
Next two silver ions (silver with a positive charge) are being reduced through the addition of two (2) electrons to form solid silver. The abbreviations "aq" and "s" mean aqueous and solid, respectively. We can now combine the two (2) half-reactions to form a redox equation:
Cu ----> Cu2+ + 2 e-
2 Ag+ + 2e- -> 2 Ag
____________________________
Cu + 2Ag+ + 2e- -> Cu2+ + 2 Ag + 2e-
Cancel 2 electrons from both sides
We get Cu + 2Ag+ ---> Cu2+ + 2 Ag
If a chemical causes another substance to be oxidized, we call it the oxidizing agent. In the equation above, Ag+ is the oxidizing agent, because it causes Cu(s) to lose electrons. Oxidants get reduced in the process by a reducing agent. Cu(s) is, naturally, the reducing agent in this case, as it causes Ag+ to gain electrons.
Oxidation and reduction in terms of hydrogen transfer
These are old definitions which aren't used very much nowadays. The most likely place you will come across them is in organic chemistry.
Definitions
Oxidation is loss of hydrogen.
Reduction is gain of hydrogen.
Notice that these are exactly the opposite of the oxygen definitions.
For example, ethanol can be oxidised to ethanal:
CH3CH2OH Changes to CH3CHO in this case loss of Hydrogen can ve termed as Oxidation.
Addition of Hydrogen to an atom is Reduction for that Atom.
You would need to use an oxidising agent to remove the hydrogen from the ethanol. A commonly used oxidising agent is potassium dichromate(VI) solution acidified with dilute sulphuric acid.
Ethanal can also be reduced back to ethanol again by adding hydrogen to it. A possible reducing agent is sodium tetrahydridoborate, NaBH4.
So
Oxidising agents give oxygen to another substance or remove hydrogen from it.
Reducing agents remove oxygen from another substance or give hydrogen to it.
Here are the steps to follow to balance a redox equation in acidic medium (add the starred step in a basic medium):
Divide the equation into an oxidation half-reaction and a reduction half-reaction
Balance these
Balance the elements other than H and O
Balance the O by adding H2O
Balance the H by adding H+
Balance the charge by adding e-
Multiply each half-reaction by an integer such that the number of e- lost in one equals the number gained in the other
Combine the half-reactions and cancel
**Add OH- to each side until all H+ is gone and then cancel again**
As a side note, the term "oxidation", with its obvious root from the word "oxygen", assumes that oxygen has an oxidation number of -2. Using this as a benchmark, oxidation numbers were assigned to all other elements. For example, if we look at H2O, and assign the value of -2 to the oxygen atom, the hydrogens must each have an oxidation number of +1 by default, since water is a neutral molecule. As an example, what is the oxidation number of sulfur in sulfur dioxide (SO2)? Given that each oxygen atom has a -2 charge, and knowing that the molecule is neutral, the oxidation number for sulfur must be +4.
What about for a sulfate ion (SO4 with a total charge of -2)? Again, the charge of all the oxygen atoms is 4 x -2 = -8. Sulfur must then have an oxidation number of +6, since +6 + (-8) = -2, the total charge on the ion. Since the sulfur in sulfate has a higher oxidation number than in sulfur dioxide, it is said to be more highly oxidized.
Working with redox reactions is fundamentally a bookkeeping issue. You need to be able to account for all of the electrons as they transfer from one species to another. There are a number of rules and tricks for balancing redox reactions, but basically they all boil down to dealing with each of the two half-reactions individually. Consider for example the reaction of aluminum metal to form alumina (Al2O3). The unbalanced reaction is as follows:
Al + O2 --> Al2O3
Looking at each half reaction separately:
Al --> Al3+ + 3e-
This reaction shows aluminum metal being oxidized to form an aluminum ion with a +3 charge. The half-reaction below shows oxygen being reduced to form two (2) oxygen ions, each with a charge of -2.
O2 + 4e- --> 2 O2-
If we combine those two (2) half-reactions, we must make the number of electrons equal on both sides. The number 12 is a common multiple of three (3) and four (4), so we multiply the aluminum reaction by four (4) and the oxygen reaction by three (3) to get 12 electrons on both sides. Now, simply combine the reactions. Notice that we have 12 electrons on both sides, which cancel out. The final step is to combine the aluminum and oxygen ions on the right side using a cross multiply technique:
2Al + 3O2 become Al2O3
Suppose we have this reaction:
Fe(s) + Cd2+(aq) ------> Fe2+(aq) + Cd(s)
In this reaction iron (Fe) is being oxidized to iron(II) ion, while the cadmium ion (Cd2+) in aqueous solution is being reduced to cadmium solid. The question is: how does this reaction behave in "non-standard" conditions?
The first thing to answer is how does it behave in standard conditions? We need to look at the standard potential for each half-reaction, then combine them to get a net potential for the reaction. The two (2) half-reactions are:
Fe2+ (aq) + 2 e- ------> Fe (s), E° = -0.44 V
Cd2+ (aq) +2 e- ------> Cd (s), E° = -0.40 V
Notice that both half-reactions are shown as reductions -- the species gains electrons, and is changed to a new form. But in the complete reaction above, Fe is oxidized, so the half-reaction needs to be reversed. Quite simply, the potential for the half-reaction of iron is now 0.44 V. To get the potential for the entire reaction, we add up the two (2) half-reactions to get 0.04 V for the standard potential.
The question now is: what is the total potential (in volts) for a nonstandard reaction? Suppose again that we have the same reaction, except now we have 0.0100 M Fe2+ instead of the standard 1.0 M. We need to use the Nernst equation to help us calculate that value. If you go to the Redox Half-Reaction Calculator, you should notice that the reaction is selected and the appropriate values are entered into the boxes. Since we don't have any species "B" or "D", we have entered zero for their concentrations. The concentration of the solid Fe is 1.0 M (actually, concentrations of solids and solvents (liquids) don't enter into the Nernst equation, but we set them to 1.0 so that the mathematics works out). If you click on the "Evaluate" button, you should learn that the standard potential is -0.44 V, while the nonstandard potential is -0.5 V. If you scroll down on the calculator, you can enter 0.5 as the first half-reaction. We again change the sign since we're actually reversing the Fe reaction
Using the calculator again, we calculate the nonstandard potential of the Cd reaction. Suppose we now have a concentration of Cd2+ of 0.005 M, what is its potential? The calculator should return a standard potential of -0.4 V and a nonstandard potential of -0.47 V. Place this value in the box for the second half-reaction, then click on "Evaluate". You should learn that the net nonstandard potential is 0.03 V, slightly less than the value of the net standard potential. Since this value is less than the net standard potential of 0.04 V, there is less of a tendency for this reaction to transfer electrons from reactants to products. In other words, less iron will be oxidized and cadmium will be reduced than at standard conditions.
Calulate the net standard potential for this reaction:
2 Ag+ (aq, 0.80 M) + Hg (l)------> 2 Ag (s) + Hg2+ (aq, 0.0010M)
Answer: 0.025 V. Since the value is positive, the reaction will work to form the products indicated. Negative values of the potential indicate that the reaction tends to stay as reactants and not form the products. The net standard potential for this reaction is 0.01 V -- since the nonstandard potential is higher, this reaction will form products than the standard reaction.
One of the more useful calculations in redox reactions is the Nernst Equation. This equation allows us to calculate the electric potential of a redox reaction in "non-standard" situations. There exist tables of how much voltage, or potential, a reaction is capable of producing or consuming. These tables, known as standard potential tables, are created by measuring potential at "standard" conditions, with a pressure of 1 bar (≅1 atm), a temperature of 298° K ( or 25° C, or room temperature) and with a concentration of 1.0 M for each of the products. This standard potential, or E°, can be corrected by a factor that includes the actual temperature of the reaction, the number of moles of electrons being transferred, and the concentrations of the redox reactants and products. The equation is:
E = E0 - ( RT/nF ) ln Q
Walther H. Nernst (1864-1941) received the Nobel prize in 1920 "in recognition of his work in thermochemistry". His contribution to chemical thermodynamics led to the well known equation correlating chemical energy and the electric potential of a galvanic cell or battery.
E = cell potential (Volt ) under specific conditions
E0 = cell potential in Volts at standard-state conditions ( at 298 K )
R = ideal gas constant = 8.314 J/mol-K
T = temperature (kelvin), which is generally 25C (298 K)
n = number of moles of electrons transferred in the balanced equation
F = Faraday's constant, the charge on a mole of electrons = 95,484.56 C/mol ( Some authors use the value 96485.309 )
Q is ratio of concentrations in appropriate powers of the equation aA + bB --> cC + dD
ln Q = the natural log of the reaction quotient at the moment in time
Since the temperature is generally 25C (298 K), three of the terms in the above Nernst equation can be considered constants: R, T, and F. Substituting the values of these constants, results in the following equation:
E = E0 - ( 0.02568 / n ) Ln Q
For the cell
Zn | Zn2+ || H+ | H2 | Pt
we have a net chemical reaction of
Zn(s) + 2 H+ = Zn2+ + H2(g)
and the standard cell potential DE° = 0.763.
If the concentrations of the ions are not 1.0 M, and the H2 pressure is not 1.0 atm, then the cell potential DE may be calculated using the Nernst equation:
0.0592 V P(H2) [Zn2+]
DE = DE° - ------- log ------------
n [H+]2
with n = 2 in this case, because the reaction involves 2 electrons. The numerical value is 0.0592 only when T = 298 K. This constant is temperature dependent. Note that the reactivity of the solid Zn is taken as 1. If the H2 pressure is 1 atm, the term P(H2) may also be omitted. The expression for the argument of the log function follows the same rules as those for the expression of equilibrium constants and reaction quotients.
Indeed, the argument for the log function is the expression for the equilibrium constant K, or reaction quotient Q.
When a cell is at equilibrium, DE = 0.00 and the expression becomes an equilibrium constant K, which bears the following relationship:
n DE°
log K = --------
0.0592
where DE° is the difference of standard potentials of the half cells involved. A battery containing any voltage is not at equilibrium.
The Nernst equation also indicates that you can build a battery simply by using the same material for both cells, but by using different concentrations.
Cells of this type are called concentration cells.
Calculate the EMF of the cell
Zn(s) | Zn2+ (0.024 M) || Zn2+ (2.4 M) | Zn(s)
Solution
Zn2+ (2.4 M) + 2 e = Zn Reduction
Zn = Zn2+ (0.024 M) + 2 e Oxidation
--------------------------------------------
Zn2+ (2.4 M) = Zn2+ (0.024 M), DE° = 0.00 - - Net reaction
Using the Nernst equation:
0.0592 (0.024)
DE = 0.00 - ------- log --------
2 (2.4)
= (-0.296)(-2.0)
= 0.0592 V
Understandably, the Zn2+ ions try to move from the concentrated half cell to a dilute solution. That driving force gives rise to 0.0592 V. From here we can also calculate the energy of dilution.
If you write the equation in the reverse direction,
Zn2+ (0.024 M) = Zn2+ (2.4 M), its voltage will be -0.0592 V. At equilibrium concentrations in the two half cells will have to be equal, in which case the voltage will be zero.
Show that the voltage of an electric cell is unaffected by multiplying the reaction equation by a positive number.
Solution
Assume that you have the cell
Mg | Mg2+ || Ag+ | Ag
and the reaction is:
Mg + 2 Ag+ = Mg2+ + 2 Ag
Using the Nernst equation
0.0592 [Mg2+]
DE = DE° - ------ log --------
2 [Ag+]2
If you multiply the equation of reaction by 2, you will have
2 Mg + 4 Ag+ = 2 Mg2+ + 4 Ag
Note that there are 4 electrons involved in this equation, and n = 4 in the Nernst equation:
0.0592 [Mg2+]2
DE = DE° - ------ log --------
4 [Ag+]4
which can be simplified as
0.0592 [Mg2+]
DE = DE° - ------ log --------
2 [Ag+]2
Thus, the cell potential DE is not affected.
The standard cell potential dE° for the reaction
Fe + Zn2+ = Zn + Fe2+
is -0.353 V. If a piece of iron is placed in a 1 M Zn2+ solution, what is the equilibrium concentration of Fe2+?
Solution
The equilibrium constant K may be calculated using
K = 10(n DE°)/0.0592
= 10-11.93
= 1.2x10-12
= [Fe2+]/[Zn2+].
Since [Zn2+] = 1 M, it is evident that
[Fe2+] = 1.2E-12 M.
From the standard cell potentials, calculate the solubility product for the following reaction:
AgCl = Ag+ + Cl-
Solution
There are Ag+ and AgCl involved in the reaction, and from the table of standard reduction potentials, you will find:
AgCl + e = Ag + Cl-, E° = 0.2223 V - - - -(1)
Since this equation does not contain the species Ag+, you need,
Ag+ + e = Ag, E° = 0.799 V - - - - - - (2)
Subtracting (2) from (1) leads to,
AgCl = Ag+ + Cl- . . . DE° = - 0.577
Let Ksp be the solubility product, and employ the Nernst equation,
log Ksp = (-0.577) / (0.0592) = -9.75
Ksp = 10-9.75 = 1.8x10-10
This is the value that you have been using in past tutorials. Now, you know that Ksp is not always measured from its solubility.
In the lead storage battery,
Pb | PbSO4 | H2SO4 | PbSO4, PbO2 | Pb
would the voltage change if you changed the concentration of H2SO4?
The net cell reaction is
Pb + PbO2 + 2 HSO4- + 2 H+ 2 PbSO4 + 2 H2O
and the Nernst equation
DE = DE° - (0.0592/2)log{1/{[HSO4-]2[H+]2}}.
Choose the correct Nernst equation for the cell
Zn(s) | Zn2+ || Cu2+ | Cu(s).
A ) DE = DE° - 0.0296 log([Zn2+] / [Cu2+])
B ) DE = DE° - 0.0296 log([Cu2+] / [Zn2+])
C ) DE = DE° - 0.0296 log(Zn / Cu)
D ) DE = DE° - 0.0296 log(Cu / Zn)
The cell as written has
Reduction on the Right: Cu2+ + 2 e = Cu
oxidation on the left: Zn = Zn2+ + 2 e
Net reaction of cell is Zn (s) + Cu2+ = Cu (s) + Zn2+
The standard cell potential DE° is 1.100 V for the cell,
Zn(s) | Zn2+ || Cu2+ | Cu(s).
If [Zn2+] = 0.01 M, and [Cu2+] = 1.0 M, what is DE or EMF?
A likely wrong result is 1.041 V.
The term that modifies DE is -(0.059/n)log{[Zn2+]/[Cu2+]} (n = 2 in this case).
Understandably, if the concentration of Zn2+ is low, there is more tendency for the reaction,
Zn = Zn2+ + 2 e.
The logarithm of the equilibrium constant, log K, of the net cell reaction of the cell
Zn(s) | Zn2+ || Cu2+ | Cu(s) . . . DE° = 1.100 V
is
A ) 1.100 / 0.0291
B ) -1.10 / 0.0291
C ) 0.0291 / 1.100
D ) -0.0291 / 1.100
E ) 1.100 / 0.0592
Use the Nernst equation in the form
0 = 1.100 - 0.0296 log ([Zn2+] / [Cu2+])
The Nernst equation is useful for the determination of equilibrium constants.
Reduction is to gain electrons or reduction in Oxidation number.
Oxidation is loss of electrons.
Reduction is gain of electrons.
As Oxygen forming Oxide of say Potassium ( K ) as K2O Oxygen itself is getting reduced as that atom is gaining electrons from K
The oxidation number is defined as the effective charge on an atom in a compound, calculated according to a prescribed set of rules. An increase in oxidation number corresponds to oxidation, and a decrease to reduction. The oxidation number of a compound has some analogy to the pH and pK measurements found in acids and bases -- the oxidation number suggests the strength or tendency of the compound to be oxidized or reduced, to serve as an oxidizing agent or reducing agent. The rules are shown below. Go through them in the order given until you have an oxidation number assigned.
For atoms in their elemental form, the oxidation number is 0
For ions, the oxidation number is equal to their charge
For single hydrogen, the number is usually +1 but in some cases it is -1
For oxygen, the number is usually -2
The sum of the oxidation number (ONs) of all the atoms in the molecule or ion is equal to its total charge.
For example, in the extraction of iron from its ore:
Fe2O3 + 3 CO -> 2 Fe + 3 CO2
In this Iron ( Fe ) is getting reduced as its oxidation number changes from 3 to 0
While Carbon monoxide is oxidized as it gained a carbon atom.
An oxidising agent is substance which oxidises something else. In the above example, the iron(III) oxide is the oxidising agent.
A reducing agent reduces something else. In the equation, the carbon monoxide is the reducing agent.
Oxidising agents give oxygen to another substance.
Reducing agents remove oxygen from another substance.
Because both reduction and oxidation are going on side-by-side, this is known as a redox reaction.
Cu2+ + Mg -> Cu + Mg2+
In the above reaction Copper atom is gaining electrons ( or oxidation number is changing from +2 to 0 ( zero )) so Copper atom is getting reduced. While
Magnesium atom is loosing electrons so it is getting oxidized. So the magnesium is reducing the copper(II) ions by giving them electrons to neutralise the charge. Magnesium is a reducing agent. the copper(II) ions are removing electrons from the magnesium to create the magnesium ions. The copper(II) ions are acting as an oxidising agent.
Redox reactions, or oxidation-reduction reactions, have a number of similarities to acid-base reactions. Fundamentally, redox reactions are a family of reactions that are concerned with the transfer of electrons between species. Like acid-base reactions, redox reactions are a matched set -- you don't have an oxidation reaction without a reduction reaction happening at the same time. Oxidation refers to the loss of electrons, while reduction refers to the gain of electrons. Each reaction by itself is called a "half-reaction", simply because we need two (2) half-reactions to form a whole reaction. In redox reactions, chemists typically write out the electrons explicitly:
Cu (s) ----> Cu2+ + 2 e-
This half-reaction says that we have solid copper (with no charge) being oxidized (losing electrons) to form a copper ion with a plus 2 charge. Notice that, like the stoichiometry notation, we have a "balance" between both sides of the reaction. We have one (1) copper atom on both sides, and the charges balance as well. The symbol "e-" represents a free electron with a negative charge that can now go out and reduce some other species, such as in the halfreaction:
2 Ag+ (aq) + 2 e- ------> 2 Ag (s)
Next two silver ions (silver with a positive charge) are being reduced through the addition of two (2) electrons to form solid silver. The abbreviations "aq" and "s" mean aqueous and solid, respectively. We can now combine the two (2) half-reactions to form a redox equation:
Cu ----> Cu2+ + 2 e-
2 Ag+ + 2e- -> 2 Ag
____________________________
Cu + 2Ag+ + 2e- -> Cu2+ + 2 Ag + 2e-
Cancel 2 electrons from both sides
We get Cu + 2Ag+ ---> Cu2+ + 2 Ag
If a chemical causes another substance to be oxidized, we call it the oxidizing agent. In the equation above, Ag+ is the oxidizing agent, because it causes Cu(s) to lose electrons. Oxidants get reduced in the process by a reducing agent. Cu(s) is, naturally, the reducing agent in this case, as it causes Ag+ to gain electrons.
Oxidation and reduction in terms of hydrogen transfer
These are old definitions which aren't used very much nowadays. The most likely place you will come across them is in organic chemistry.
Definitions
Oxidation is loss of hydrogen.
Reduction is gain of hydrogen.
Notice that these are exactly the opposite of the oxygen definitions.
For example, ethanol can be oxidised to ethanal:
CH3CH2OH Changes to CH3CHO in this case loss of Hydrogen can ve termed as Oxidation.
Addition of Hydrogen to an atom is Reduction for that Atom.
You would need to use an oxidising agent to remove the hydrogen from the ethanol. A commonly used oxidising agent is potassium dichromate(VI) solution acidified with dilute sulphuric acid.
Ethanal can also be reduced back to ethanol again by adding hydrogen to it. A possible reducing agent is sodium tetrahydridoborate, NaBH4.
So
Oxidising agents give oxygen to another substance or remove hydrogen from it.
Reducing agents remove oxygen from another substance or give hydrogen to it.
Here are the steps to follow to balance a redox equation in acidic medium (add the starred step in a basic medium):
Divide the equation into an oxidation half-reaction and a reduction half-reaction
Balance these
Balance the elements other than H and O
Balance the O by adding H2O
Balance the H by adding H+
Balance the charge by adding e-
Multiply each half-reaction by an integer such that the number of e- lost in one equals the number gained in the other
Combine the half-reactions and cancel
**Add OH- to each side until all H+ is gone and then cancel again**
As a side note, the term "oxidation", with its obvious root from the word "oxygen", assumes that oxygen has an oxidation number of -2. Using this as a benchmark, oxidation numbers were assigned to all other elements. For example, if we look at H2O, and assign the value of -2 to the oxygen atom, the hydrogens must each have an oxidation number of +1 by default, since water is a neutral molecule. As an example, what is the oxidation number of sulfur in sulfur dioxide (SO2)? Given that each oxygen atom has a -2 charge, and knowing that the molecule is neutral, the oxidation number for sulfur must be +4.
What about for a sulfate ion (SO4 with a total charge of -2)? Again, the charge of all the oxygen atoms is 4 x -2 = -8. Sulfur must then have an oxidation number of +6, since +6 + (-8) = -2, the total charge on the ion. Since the sulfur in sulfate has a higher oxidation number than in sulfur dioxide, it is said to be more highly oxidized.
Working with redox reactions is fundamentally a bookkeeping issue. You need to be able to account for all of the electrons as they transfer from one species to another. There are a number of rules and tricks for balancing redox reactions, but basically they all boil down to dealing with each of the two half-reactions individually. Consider for example the reaction of aluminum metal to form alumina (Al2O3). The unbalanced reaction is as follows:
Al + O2 --> Al2O3
Looking at each half reaction separately:
Al --> Al3+ + 3e-
This reaction shows aluminum metal being oxidized to form an aluminum ion with a +3 charge. The half-reaction below shows oxygen being reduced to form two (2) oxygen ions, each with a charge of -2.
O2 + 4e- --> 2 O2-
If we combine those two (2) half-reactions, we must make the number of electrons equal on both sides. The number 12 is a common multiple of three (3) and four (4), so we multiply the aluminum reaction by four (4) and the oxygen reaction by three (3) to get 12 electrons on both sides. Now, simply combine the reactions. Notice that we have 12 electrons on both sides, which cancel out. The final step is to combine the aluminum and oxygen ions on the right side using a cross multiply technique:
2Al + 3O2 become Al2O3
Suppose we have this reaction:
Fe(s) + Cd2+(aq) ------> Fe2+(aq) + Cd(s)
In this reaction iron (Fe) is being oxidized to iron(II) ion, while the cadmium ion (Cd2+) in aqueous solution is being reduced to cadmium solid. The question is: how does this reaction behave in "non-standard" conditions?
The first thing to answer is how does it behave in standard conditions? We need to look at the standard potential for each half-reaction, then combine them to get a net potential for the reaction. The two (2) half-reactions are:
Fe2+ (aq) + 2 e- ------> Fe (s), E° = -0.44 V
Cd2+ (aq) +2 e- ------> Cd (s), E° = -0.40 V
Notice that both half-reactions are shown as reductions -- the species gains electrons, and is changed to a new form. But in the complete reaction above, Fe is oxidized, so the half-reaction needs to be reversed. Quite simply, the potential for the half-reaction of iron is now 0.44 V. To get the potential for the entire reaction, we add up the two (2) half-reactions to get 0.04 V for the standard potential.
The question now is: what is the total potential (in volts) for a nonstandard reaction? Suppose again that we have the same reaction, except now we have 0.0100 M Fe2+ instead of the standard 1.0 M. We need to use the Nernst equation to help us calculate that value. If you go to the Redox Half-Reaction Calculator, you should notice that the reaction is selected and the appropriate values are entered into the boxes. Since we don't have any species "B" or "D", we have entered zero for their concentrations. The concentration of the solid Fe is 1.0 M (actually, concentrations of solids and solvents (liquids) don't enter into the Nernst equation, but we set them to 1.0 so that the mathematics works out). If you click on the "Evaluate" button, you should learn that the standard potential is -0.44 V, while the nonstandard potential is -0.5 V. If you scroll down on the calculator, you can enter 0.5 as the first half-reaction. We again change the sign since we're actually reversing the Fe reaction
Using the calculator again, we calculate the nonstandard potential of the Cd reaction. Suppose we now have a concentration of Cd2+ of 0.005 M, what is its potential? The calculator should return a standard potential of -0.4 V and a nonstandard potential of -0.47 V. Place this value in the box for the second half-reaction, then click on "Evaluate". You should learn that the net nonstandard potential is 0.03 V, slightly less than the value of the net standard potential. Since this value is less than the net standard potential of 0.04 V, there is less of a tendency for this reaction to transfer electrons from reactants to products. In other words, less iron will be oxidized and cadmium will be reduced than at standard conditions.
Calulate the net standard potential for this reaction:
2 Ag+ (aq, 0.80 M) + Hg (l)------> 2 Ag (s) + Hg2+ (aq, 0.0010M)
Answer: 0.025 V. Since the value is positive, the reaction will work to form the products indicated. Negative values of the potential indicate that the reaction tends to stay as reactants and not form the products. The net standard potential for this reaction is 0.01 V -- since the nonstandard potential is higher, this reaction will form products than the standard reaction.
One of the more useful calculations in redox reactions is the Nernst Equation. This equation allows us to calculate the electric potential of a redox reaction in "non-standard" situations. There exist tables of how much voltage, or potential, a reaction is capable of producing or consuming. These tables, known as standard potential tables, are created by measuring potential at "standard" conditions, with a pressure of 1 bar (≅1 atm), a temperature of 298° K ( or 25° C, or room temperature) and with a concentration of 1.0 M for each of the products. This standard potential, or E°, can be corrected by a factor that includes the actual temperature of the reaction, the number of moles of electrons being transferred, and the concentrations of the redox reactants and products. The equation is:
E = E0 - ( RT/nF ) ln Q
Walther H. Nernst (1864-1941) received the Nobel prize in 1920 "in recognition of his work in thermochemistry". His contribution to chemical thermodynamics led to the well known equation correlating chemical energy and the electric potential of a galvanic cell or battery.
E = cell potential (Volt ) under specific conditions
E0 = cell potential in Volts at standard-state conditions ( at 298 K )
R = ideal gas constant = 8.314 J/mol-K
T = temperature (kelvin), which is generally 25C (298 K)
n = number of moles of electrons transferred in the balanced equation
F = Faraday's constant, the charge on a mole of electrons = 95,484.56 C/mol ( Some authors use the value 96485.309 )
Q is ratio of concentrations in appropriate powers of the equation aA + bB --> cC + dD
ln Q = the natural log of the reaction quotient at the moment in time
Since the temperature is generally 25C (298 K), three of the terms in the above Nernst equation can be considered constants: R, T, and F. Substituting the values of these constants, results in the following equation:
E = E0 - ( 0.02568 / n ) Ln Q
For the cell
Zn | Zn2+ || H+ | H2 | Pt
we have a net chemical reaction of
Zn(s) + 2 H+ = Zn2+ + H2(g)
and the standard cell potential DE° = 0.763.
If the concentrations of the ions are not 1.0 M, and the H2 pressure is not 1.0 atm, then the cell potential DE may be calculated using the Nernst equation:
0.0592 V P(H2) [Zn2+]
DE = DE° - ------- log ------------
n [H+]2
with n = 2 in this case, because the reaction involves 2 electrons. The numerical value is 0.0592 only when T = 298 K. This constant is temperature dependent. Note that the reactivity of the solid Zn is taken as 1. If the H2 pressure is 1 atm, the term P(H2) may also be omitted. The expression for the argument of the log function follows the same rules as those for the expression of equilibrium constants and reaction quotients.
Indeed, the argument for the log function is the expression for the equilibrium constant K, or reaction quotient Q.
When a cell is at equilibrium, DE = 0.00 and the expression becomes an equilibrium constant K, which bears the following relationship:
n DE°
log K = --------
0.0592
where DE° is the difference of standard potentials of the half cells involved. A battery containing any voltage is not at equilibrium.
The Nernst equation also indicates that you can build a battery simply by using the same material for both cells, but by using different concentrations.
Cells of this type are called concentration cells.
Calculate the EMF of the cell
Zn(s) | Zn2+ (0.024 M) || Zn2+ (2.4 M) | Zn(s)
Solution
Zn2+ (2.4 M) + 2 e = Zn Reduction
Zn = Zn2+ (0.024 M) + 2 e Oxidation
--------------------------------------------
Zn2+ (2.4 M) = Zn2+ (0.024 M), DE° = 0.00 - - Net reaction
Using the Nernst equation:
0.0592 (0.024)
DE = 0.00 - ------- log --------
2 (2.4)
= (-0.296)(-2.0)
= 0.0592 V
Understandably, the Zn2+ ions try to move from the concentrated half cell to a dilute solution. That driving force gives rise to 0.0592 V. From here we can also calculate the energy of dilution.
If you write the equation in the reverse direction,
Zn2+ (0.024 M) = Zn2+ (2.4 M), its voltage will be -0.0592 V. At equilibrium concentrations in the two half cells will have to be equal, in which case the voltage will be zero.
Show that the voltage of an electric cell is unaffected by multiplying the reaction equation by a positive number.
Solution
Assume that you have the cell
Mg | Mg2+ || Ag+ | Ag
and the reaction is:
Mg + 2 Ag+ = Mg2+ + 2 Ag
Using the Nernst equation
0.0592 [Mg2+]
DE = DE° - ------ log --------
2 [Ag+]2
If you multiply the equation of reaction by 2, you will have
2 Mg + 4 Ag+ = 2 Mg2+ + 4 Ag
Note that there are 4 electrons involved in this equation, and n = 4 in the Nernst equation:
0.0592 [Mg2+]2
DE = DE° - ------ log --------
4 [Ag+]4
which can be simplified as
0.0592 [Mg2+]
DE = DE° - ------ log --------
2 [Ag+]2
Thus, the cell potential DE is not affected.
The standard cell potential dE° for the reaction
Fe + Zn2+ = Zn + Fe2+
is -0.353 V. If a piece of iron is placed in a 1 M Zn2+ solution, what is the equilibrium concentration of Fe2+?
Solution
The equilibrium constant K may be calculated using
K = 10(n DE°)/0.0592
= 10-11.93
= 1.2x10-12
= [Fe2+]/[Zn2+].
Since [Zn2+] = 1 M, it is evident that
[Fe2+] = 1.2E-12 M.
From the standard cell potentials, calculate the solubility product for the following reaction:
AgCl = Ag+ + Cl-
Solution
There are Ag+ and AgCl involved in the reaction, and from the table of standard reduction potentials, you will find:
AgCl + e = Ag + Cl-, E° = 0.2223 V - - - -(1)
Since this equation does not contain the species Ag+, you need,
Ag+ + e = Ag, E° = 0.799 V - - - - - - (2)
Subtracting (2) from (1) leads to,
AgCl = Ag+ + Cl- . . . DE° = - 0.577
Let Ksp be the solubility product, and employ the Nernst equation,
log Ksp = (-0.577) / (0.0592) = -9.75
Ksp = 10-9.75 = 1.8x10-10
This is the value that you have been using in past tutorials. Now, you know that Ksp is not always measured from its solubility.
In the lead storage battery,
Pb | PbSO4 | H2SO4 | PbSO4, PbO2 | Pb
would the voltage change if you changed the concentration of H2SO4?
The net cell reaction is
Pb + PbO2 + 2 HSO4- + 2 H+ 2 PbSO4 + 2 H2O
and the Nernst equation
DE = DE° - (0.0592/2)log{1/{[HSO4-]2[H+]2}}.
Choose the correct Nernst equation for the cell
Zn(s) | Zn2+ || Cu2+ | Cu(s).
A ) DE = DE° - 0.0296 log([Zn2+] / [Cu2+])
B ) DE = DE° - 0.0296 log([Cu2+] / [Zn2+])
C ) DE = DE° - 0.0296 log(Zn / Cu)
D ) DE = DE° - 0.0296 log(Cu / Zn)
The cell as written has
Reduction on the Right: Cu2+ + 2 e = Cu
oxidation on the left: Zn = Zn2+ + 2 e
Net reaction of cell is Zn (s) + Cu2+ = Cu (s) + Zn2+
The standard cell potential DE° is 1.100 V for the cell,
Zn(s) | Zn2+ || Cu2+ | Cu(s).
If [Zn2+] = 0.01 M, and [Cu2+] = 1.0 M, what is DE or EMF?
A likely wrong result is 1.041 V.
The term that modifies DE is -(0.059/n)log{[Zn2+]/[Cu2+]} (n = 2 in this case).
Understandably, if the concentration of Zn2+ is low, there is more tendency for the reaction,
Zn = Zn2+ + 2 e.
The logarithm of the equilibrium constant, log K, of the net cell reaction of the cell
Zn(s) | Zn2+ || Cu2+ | Cu(s) . . . DE° = 1.100 V
is
A ) 1.100 / 0.0291
B ) -1.10 / 0.0291
C ) 0.0291 / 1.100
D ) -0.0291 / 1.100
E ) 1.100 / 0.0592
Use the Nernst equation in the form
0 = 1.100 - 0.0296 log ([Zn2+] / [Cu2+])
The Nernst equation is useful for the determination of equilibrium constants.