Mr. Braswell's Chemistry Help

Learning Zone Presents:

                                DEFINITION OF OXIDATION AND REDUCTION

 

Redox (shorthand for reduction-oxidation reaction) describes all chemical reactions in which atoms have their oxidation number (oxidation state) changed.

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

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

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                                       SIMPLE REDOX (SYNTHESIS REACTIONS)

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

H2 + F2 → 2 HF

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

the oxidation reaction:

H2 → 2 H+ + 2 e

and the reduction reaction:

F2 + 2 e → 2 F

Elements, even in molecular form, always have an oxidation number of zero.

In the first half-reaction, hydrogen is oxidized from an oxidation number of zero to an oxidation number of +1.

In the second half-reaction, fluorine is reduced from an oxidation number of zero to an oxidation number of −1.

When adding the reactions together the electrons cancel:

H2

2 H+ + 2 e

F2 + 2 e

2 F


H2 + F2

2 H+ + 2 F

And the ions combine to form hydrogen fluoride:

H2 + F2 → 2 H+ + 2 F → 2 HF

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           SIMPLE REDOX (SINGLE REPLACEMENT REACTIONS)

Displacement reactions (Single replacement reactions)

Redox occurs in single displacement reactions or substitution reactions.

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

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

Fe + CuSO4FeSO4 + Cu

The ionic equation for this reaction is:

Fe + Cu2+ → Fe2+ + Cu

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

Fe → Fe2+ + 2 e

And the copper is reduced:

Cu2+ + 2 e → Cu

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                              COMPLEX REDOX REACTIONS (HONORS ONLY)

Other examples

SOME REDOX REACTIONS OCCUR IN ACIDIC MEDIA

(ALL MECHANISMS NOT SHOWN HERE)

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

Complete oxidation of materials containing carbon produces carbon dioxide.

               http://video.about.com/chemistry/Handheld-Fireballs.htm

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                                OXIDIZING AGENTS AND REDUCING AGENTS

 http://khanexercises.appspot.com/video?v=yp60-oVxrT4

Oxidizing and reducing agents

The chemical way to look at redox processes is that the reductant transfers electrons to the oxidant.

Thus, in the reaction, the reductant or reducing agent loses electrons and is oxidized, and the oxidant or oxidizing agent gains electrons and is reduced.

The pair of an oxidizing and reducing agent that are involved in a particular reaction is called a redox pair.

Oxidizers

Substances that have the ability to oxidize other substances are said to be oxidative and are known as oxidizing agents, oxidants, or oxidizers.

Put another way, the oxidant removes electrons from another substance, and is thus itself reduced.

And, because it "accepts" electrons, it is also called an electron acceptor.

Oxidants are usually chemical substances with elements in high oxidation numbers

Reducers

Substances that have the ability to reduce other substances are said to be reductive and are known as reducing agents, reductants, or reducers.

In other words, the reductant transfers electrons to another substance, and is thus itself oxidized.

And, because it "donates" electrons it is also called an electron donor.

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                                            SIMPLE BALANCING REDOX

                   LEO The Lion Goes GER 
             Loose Electrons, Oxidize                 Gain Electrons, Reduce

How do you know this
is a "Redox" equation?

 

hydrogen plus oxygen yields water

      JUST LOOK AT THE CHANGE IN OXIDATION NUMBERS!

                                              H2   +  O2    →    H2O

                                             H2   +  O2    →    2H+1   +  O-2

BALANCE THE MASS IN EACH HALF REACTION

                                      REDUCTION:  O2    →  O-2

                                      Mass is must be balanced first:  O2    →  2O-2

                                      OXIDATION:  H2   →    2H+1

                                       Mass is balanced already

BALANCE THE CHARGE IN EACH HALF REACTION 

                                   REDUCTION:  O2    →  2O-2

                      Charge must be balanced:  4e-  +   O2    →  2O-2

                                  OXIDATION:  H2   →    2H+1

                    Charge must be balanced:     H2     →    2H+1  + 2e-

 BALANCE THE CHARGE BETWEEN THE TWO HALF REACTION 

                                        4 e- gained:  4e-  +   O2    →  2O-2

                           2 e- lost, so multiply by 2:    2 [H2     →    2H+1  + 2e-]

     gives

                                                    2H2     →    4H+1  + 4e-

 CANCEL ELECTRONS AND COMBINE BALANCED HALF REACTIONS 

2H    +  O2    →  4H+1+  2O-2 

2H2   +  O2    →    2H2O 

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                                                 ELECTROLYTIC CELLS

The Electrolysis of Molten NaCl

An idealized cell for the electrolysis of sodium chloride is shown in the figure below.

A source of direct current is connected to a pair of inert electrodes immersed in molten sodium chloride.

Because the salt has been heated until it melts, the Na+ ions flow toward the negative electrode and the Cl- ions flow toward the positive electrode.

When Na+ ions collide with the negative electrode, the battery carries a large enough potential to force these ions to pick up electrons to form sodium metal.

Negative electrode (cathode): Na+ + e- →  Na

Cl- ions that collide with the positive electrode are oxidized to Cl2 gas, which bubbles off at this electrode.

Positive electrode (anode): 2 Cl-  → Cl2 + 2 e-

The net effect of passing an electric current through the molten salt in this cell is to decompose sodium chloride into its elements, sodium metal and chlorine gas.

Electrolysis of NaCl:   
 Cathode (-): Na+ + e- → Na
 Anode (+): 2 Cl-  → Cl2 + 2 e-

The potential required to oxidize Cl- ions to Cl2 is -1.36 volts and the potential needed to reduce Na+ ions to sodium metal is -2.71 volts.

The battery used to drive this reaction must therefore have a potential of at least 4.07 volts.

This example explains why the process is called electrolysis.

The suffix -lysis comes from the Greek stem meaning to loosen or split up.

Electrolysis literally uses an electric current to split a compound into its elements.

 electrolysis 
2 NaCl(l)2 Na(l) + Cl2(g)

This example also illustrates the difference between voltaic cells and electrolytic cells.

Voltaic cells use the energy given off in a spontaneous reaction to do electrical work.

Electrolytic cells use electrical work as source of energy to drive the reaction in the opposite direction. 

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                                                 ELECTROLYSIS OF NaCl

The Electrolysis of Aqueous NaCl

The figure below shows an idealized drawing of a cell in which an aqueous solution of sodium chloride is electrolyzed.

Once again, the Na+ ions migrate toward the negative electrode and the Cl- ions migrate toward the positive electrode. But, now there are two substances that can be reduced at the cathode: Na+ ions and water molecules.

Cathode (-):   
 Na+ + e- →  Na Eored = -2.71 V
 2 H2O + 2 e-  → H2 + 2 OH-  Eored = -0.83 V

Because it is much easier to reduce water than Na+ ions, the only product formed at the cathode is hydrogen gas.

Cathode (-): 2 H2O(l) + 2 e-  → H2(g) + 2 OH-(aq)

 go to reduction table

There are also two substances that can be oxidized at the anode: Cl- ions and water molecules.

Anode (+):   
 2 Cl-  → Cl2 + 2 e-  Eoox = -1.36 V
 2 H2O  → O2 + 4 H+ + 4 e- Eoox = -1.23 V

The standard-state potentials for these half-reactions are so close to each other that we might expect to see a mixture of Cl2 and O2 gas collect at the anode.

In practice, the only product of this reaction is Cl2.

Anode (+): 2 Cl-  → Cl2 + 2 e-

At first glance, it would seem easier to oxidize water (Eoox = -1.23 volts) than Cl- ions (Eoox = -1.36 volts).

It is worth noting, however, that the cell is never allowed to reach standard-state conditions.

The solution is typically 25% NaCl by mass, which significantly decreases the potential required to oxidize the Cl- ion.

The pH of the cell is also kept very high, which decreases the oxidation potential for water.

Electrolysis of aqueous NaCl solutions gives a mixture of hydrogen and chlorine gas and an aqueous sodium hydroxide solution.

    electrolysis 
2 NaCl(aq) + 2 H2O(l)2 Na+(aq) + 2 OH-(aq) + H2(g) + Cl2(g)

Because the demand for chlorine is much larger than the demand for sodium, electrolysis of aqueous sodium chloride is a more important process commercially.

Electrolysis of an aqueous NaCl solution has two other advantages.

It produces H2 gas at the cathode, which can be collected and sold.

It also produces NaOH, which can be drained from the bottom of the electrolytic cell and sold.

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                                               ELECTROYSIS OF WATER

Electrolysis of Water

A standard apparatus for the electrolysis of water is shown in the figure below.

Diagram of a Hoffman voltameter used for the electrolysis of water to produce hydrogen and oxygen gases

 electrolysis 
2 H2O(l)2 H2(g) + O2(g)

A pair of inert electrodes are sealed in opposite ends of a container designed to collect the H2 and O2 gas given off in this reaction.

The electrodes are then connected to a battery or another source of electric current.

By itself, water is a very poor conductor of electricity.

We therefore add an electrolyte to water to provide ions that can flow through the solution, thereby completing the electric circuit.

The electrolyte must be soluble in water. It should also be relatively inexpensive.

Most importantly, it must contain ions that are harder to oxidize or reduce than water.

2 H2O + 2 e-  H2 + 2 OH-  Eored = -0.83 V
2 H2→  O2 + 4 H+ + 4 e-  Eoox = -1.23 V

The following cations are harder to reduce than water: Li+, Rb+, K+, Cs+, Ba2+, Sr2+, Ca2+, Na+, and Mg2+.

 go to reduction table

Two of these cations are more likely candidates than the others because they form inexpensive, soluble salts: Na+ and K+.

The SO42- ion might be the best anion to use because it is the most difficult anion to oxidize.

The potential for oxidation of this ion to the peroxydisulfate ion is -2.05 volts.

2 SO42-   S2O82- + 2 e-  Eoox = -2.05 V

When an aqueous solution of either Na2SO4 or K2SO4 is electrolyzed in the apparatus shown in the above figure, H2 gas collects at one electrode and O2 gas collects at the other.

What would happen if we added an indicator such as bromothymol blue to this apparatus?

Bromothymol blue turns yellow in acidic solutions (pH < 6) and blue in basic solutions (pH > 7.6).

According to the equations for the two half-reactions, the indicator should turn yellow at the anode and blue at the cathode.

Cathode (-):  2 H2O + 2 e-  H2 + 2 OH-
Anode (+):   2 H2O2 + 4 H+ + 4 e-

 

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                                                      FARADAY'S LAW

Faraday's law of electrolysis can be stated as follows.

The amount of a substance consumed or produced at one of the electrodes in an electrolytic cell is directly proportional to the amount of electricity that passes through the cell.

In order to use Faraday's law we need to recognize the relationship between current, time, and the amount of electric charge that flows through a circuit.

 1 F = The quantity of electricity that is capable of depositing or liberating 1 gram equivalent weight( supply or react with one mole of electrons in a redox reaction) of a substance in electrolysis, approximately 9.6494 × 104 coulombs.

By definition, one coulomb of charge is transferred when a 1-amp current flows for 1 second.

(Around 6.242 × 1018 electrons passing a given point each second constitutes one amp)

in other words:  when of 6.242 × 1018 negative electrons pass a particular point in an electrolytic cell, that represents one coulomb...remember I do not mean 6.242 × 1018  x -1 C/electron, I mean  6.242 × 1018 x 1.602 x 10-19 C/electron.)

these electrons will come from an external power source...a battary.

1 C = 1 amp-s = 6.242 × 1018 electrons/point

and

each electron = 1.602 x 10-19 C

Example: To illustrate how Faraday's law can be used, let's calculate the number of grams of sodium metal that will form at the cathode when a 10.0-amp current is passed through molten sodium chloride for a period of 4.00 hours.

1.  We start by calculating the amount of electric charge, in C, that flows through the cell.

10 C/sec x 3600 x 4 = 144,000 C/sec

Before we can use this information,

2.  we need a bridge between this macroscopic quantity and the phenomenon that occurs on the atomic scale.

This bridge is represented by Faraday's constant, which describes the number of coulombs of charge carried by a mole of electrons.

where

each electron = 1.602 x 10-19 C

 go to EMF and Free Energy Change

 

Thus, the number of moles of electrons transferred when 144,000 coulombs of electric charge flow through the cell, due to the 10 amps, can be calculated as follows.

moles of chemical = current x time x conversion factors
moles of chemical=coulombs
second
x seconds x   1 Faraday  
96500 coulombs
x1 mole e-
1 Faraday
x1 mole chemical
# mole e-

 

moles = 144,000 C/sec/96500 C/mol e- 1.49 mole e-

According to the balanced equation for the reaction that occurs at the cathode of this cell, we get one mole of sodium for every mole of electrons.

                                            from the reduction of sodium we get:

Cathode (-): Na+ + e-  Na

Thus, we get 1.49 moles, or 34.3 grams, of sodium in 4.00 hours.

The consequences of this calculation are interesting. We would have to run this electrolysis for more than two days to prepare a pound of sodium.

Example Problem: If you are trying to coat a strip with aluminum and you have a current of 10.0 A (amperes) running for one hour, what mass of Al is formed?
 
The solution of this problem involves a lengthly unit conversion process:
 
see example above:
 

Practice Problem 

Consider the electrolysis of molten barium chloride.

1.  write the half reactions for this process.

2.  how many grams of barium metal can be produced by supplying 0.5 A for 30 minutes.

Practice Problem 

Calculate the amounts of copper metal and liquid bromine produced in 1 hour at inert electodes in a solution of copper II bromide by a current of 4.5 amps.

 

 

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go to practice problems

                                           ELECTROCHEMICAL CELLS

 

http://www.blackgold.ab.ca/ICT/Division4/Science/Div.%204/Voltaic%20Cells/demo.htm

 http://www.blackgold.ab.ca/ICT/Division4/Science/Div.%204/Voltaic%20Cells/Voltaic.htm

http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/galvan5.swf

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/animations/CuZncell.html

http://www.chembio.uoguelph.ca/educmat/chm19105/galvanic/galvanic1.htm

A Galvanic cell consists of two half-cells.

In its simplest form, each half-cell consists of a metal and a solution of a salt of the metal.

The salt solution contains a cation of the metal and an anion to balance the charge on the cation.

In essence the half-cell contains a metal and an aqueous solution.

In a galvanic cell one metal is able to reduce the cation of the other and, conversely, the other cation can oxidize the first metal.

The two half-cells must be physically separated so that the solutions do not mix together.

A salt bridge or porous plate is used to separate the two solutions.

The number of electron transferred in both directions must be the same, so the two half-cells are combined to give the whole-cell electrochemical reaction.

For the Daniell cell, depicted in the figure, the two metals are zinc and copper and the two salts are sulfates of the respective metal.

1.  Zinc is the more reducing metal so when a device is connected to the electrodes, the electrochemical reaction is (see table J)

Zn + Cu2+ → Zn2+ + Cu

The zinc electrode is dissolved and copper is deposited on the copper electrode.

By definition, the cathode is the electrode where reduction (gain of electrons) takes place, so the copper electrode is the cathode.

The cathode attracts cations, so has a positive charge.

In this case copper is the cathode and zinc the anode.

Galvanic cells are typically used as a source of electrical power.

Cu2+/Cu cell:   Cu2++ 2 e is in equilibrium with Cu:             E0 = +0.34 V
Zn2+/Zn cell:  Zn2+ + 2 e is in equilibrium with Zn:             E0 = −0.76 V

Thus the overall reaction is:

Cu2+ + Zn is in equilibrium with Cu + Zn2+

sponteneity of a reaction

Eo = Eo (reduction half reaction) – Eo (oxidation half reaction)

go to practice

standard potential table

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/animations/ZnCutransfer.html

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/animations/PbbatteryV8web.html

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                         EMF and Free Energy Change

EMF and Free Energy Change: 

ΔGo = -nFEo

from before:

ΔGo = ΔH -TΔS

n = number of electrons transferred.

F = Faraday's constant in moles of electrons, where, 1 F =96,500 C/mol or 96,500 J/V-mol. 

See FARADAY'S LAW on this page.

Eo = Eo (reduction half reaction) – Eo (oxidation half reaction)

example:

Calculate the standard free-energy change, for the following reaction:

 

4Ag   +  O2  + 4H+ → 4Ag+ + 2H2O 

1.  From the table of standard reduction values find Eo

       reduction:  O2  + 4H+ 4e-  → 2H2O  Eo = +1.23 V

       oxidation:  4Ag → 4Ag+ + 4e-         Eo = +0.80 V

2.  Find the potential difference: 

      Eo = Eo (reduction half reaction) – Eo (oxidation half reaction)

 

Eo = Eo (1.23 V) – Eo (0.80 V) = 0.43V

remember: from the reduction table, the sign may be changed for the species that undergoes oxidation.

In the example above the charge of the oxidized species has not been reversed...therefore the formula is:

Eo = Eo (reduction half reaction) – Eo (oxidation half reaction)

3.  from the balanced half reactions 4 electrons were transferred, so n = 4

4.  solve Gibb's equation:        ΔGo = -nFEo

     ΔGo = -(4)(96,500 J/V-mol)(0.43V) 

 ΔGo = -170 kJ/mol 

this means that 170 kJ of energy is available to do work. 

F = the quantity of electrical charge (in coulombs) that is contained in 1 mole of electrons (this is the Faraday constant).

One Faraday is equal to 96,500 coulombs/mole of e-.

Since 1 volt = 1Joule/coulomb, one Faraday also equals 96,500 J/volt*mole e-

Joules/mol = -(electrons)* (coulombs/mol of electrons) * (Joules/coulomb)

Practice Problem 

Calculate the ΔGo of the following reaction:

Mg(s)       +          Pb+2 (aq)    < = >      Mg+2(s)       +          Pb(s)

and

Calculate the ΔGo of the following reaction:

Br2(l)        +          2I-(aq)     < = >      2Br-(aq)           +          I2(s)

 

go to practice problems

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                Effect of concentration on Cell EMF

What happens when the Daniell cell is used to do work?

  • The copper electrode becomes lighter as copper atoms are oxidized to Cu2+ ions, which go into solution.
  • The silver electrode becomes heavier as Ag+ ions in the solution are reduced to silver metal.
  • The concentration of Cu2+ ions at the anode increases and the concentration of the Ag+ ions at the cathode decreases.
  • Negative ions flow from the salt bridge toward the anode to balance the charge on the Cu2+ ions produced at this electrode.
  • Positive ions flow from the salt bridge toward the cathode to compensate for the Ag+ ions consumed in the reaction.

An important property of the cell is missing from this list.

Over a period of time, the cell runs down, and eventually has to be replaced.

Let's assume that our cell is initially a standard-state cell in which the concentrations of the Cu2+ and Ag+ ions are both 1 molar.

Cu|Cu2+(1.0 M)||Ag+(1.0 M)|Ag

As the reaction goes forward copper metal is consumed and silver metal is produced therefore, the  driving force behind the reaction must become weaker.

Therefore, the cell potential must become smaller.

Ecell = Eocell - (0.0257/n) ln Q

or in terms of log10

Ecell = Eocell - (0.0592/n) log Q

 

In the Nernst equation,

E is the cell potential at some moment in time,

Eo is the cell potential when the reaction is at standard-state conditions,

R is the ideal gas constant in units of joules per mole,

T is the temperature in kelvin,

n is the number of moles of electrons transferred in the balanced equation for the reaction,

F is the charge on a mole of electrons,

Qc is the reaction quotient at that moment in time.

The symbol ln indicates a natural logarithm, which is the log to the base e, where e is an irrational number equal to 2.71828...

Three terms in the Nernst equation are constants: R, T, and F.

The ideal gas constant is 8.314 J/mol-K. The temperature is usually 25oC.

Substituting this information into the Nernst equation gives the following equation.

Example: The standard-state potential for the Daniell cell is 0.46V. Two moles of electrons are transferred from copper metal to Ag+ ions in the balanced equation for this reaction, so n is 2 for this cell.

Because we never include the concentrations of solids in either reaction quotient or equilibrium constant expressions, Qc for this reaction is equal to the concentration of the Cu2+ ion divided by the concentration of the Ag+ ion.

Substituting what we know about the Daniell cell into the Nernst equation gives the following result, which represents the cell potential for the Daniell cell at 25oC at any moment in time.

The concentration of aqueous solutions in the half cells directly impact the EMF of the cell.

E = Eo - (0.0592 V/n) logk  

 Suppose we have this reaction:

Fe(s) + Cd2+(aq) ------> Fe2+(aq) + Cd(s)
 
In this reaction iron (Fe) is being oxidized in a 1.0 M solution of to iron(II) ion, while the cadmium ion (Cd2+) in a 0.5 M aqueous solution is being reduced to cadmium solid.
 
a.  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.

 

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.

E = Eo - (0.0592 V/n) logQ           

E = 0.04 V - (0.0592 V/2) log(1.0/0.5)

    E =  0.031 V      

B.  Suppose we now have a concentration of Cd2+ of 0.005 M, what is its potential?  

E = 0.04 V - (0.0592 V/2) log(1.0/0.005)

E =  0.028 V 

YOU TRY THIS ONE:

 

 

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.

Sample Calculation

Calculate the cell potential for the following system:

Write the half-reactions with the half-cell potentials:

Multiply the reactions to get the lowest common multiple of electrons:

  

                   

When the reaction quotient is very small, the cell potential is positive and relatively large. This isn't surprising, because the reaction is far from equilibrium and the driving force behind the reaction should be relatively large.

When the reaction quotient is very large, the cell potential is negative. This means that the reaction would have to shift back toward the reactants to reach equilibrium.

Practice Problem 

What is the potential of a cell made up of zinc and copper half-cells at the concentrations below:

Zn|Zn2+(0.25 M)||Cu2+(0.15M)|Cu

The cell potential depends on the logarithm of the ratio of the concentrations of the products and the reactants.

As a result, the potential of a cell or battery is more or less constant until virtually all of the reactants have been converted into products.

The Nernst equation can be used to calculate the potential of a cell that operates at non-standard-state conditions.

Practice Problem 9:

Calculate the potential at 25oC for the following cell.

Cu|Cu2+(0.024 M)||Ag+(0.0048 M)|Ag

Click here to check your answer to Practice Problem 9

Click here to see a solution to Practice Problem 9

go to reduction table

 

 

 

Using the Nernst Equation to Measure Equilibrium Constants

For a system in Equilibrium ΔG = 0.  When the voltaic cell reaches equilibrium E = 0.

By using the Nernst equation:

                                   E = Eo - (0.0592 V/n) logk 

the value of k may be determined, where n = number of electrons transferred. 

At equilibrium the equation above becomes:

                                            0 = Eo - (0.0592 V/n) logk 

This equation may be simplified to the following: 

                     logk = nEo/0.0592 V  

Practice Problem

Calculate the equilibrium constant at 25oC for the reaction between zinc metal and acid.

Zn(s) + 2 H+(aq Zn2+(aq) + H2(g)

Click here to check your answer to Practice Problem 10

Click here to see a solution to Practice Problem 10

Practice Problem

Use the following standard-state cell potentials to calculate the solubility product at 25oC for Mg(OH)2.

Mg(OH)2 + 2 e-   = > Mg + 2 OH- Eored = -2.69 V
Mg2+ + 2 e-   = >Mg Eored = -2.375 V

Click here to check your answer to Practice Problem 12

Click here to see a solution to Practice Problem 12

 

Practice Problem 

Calculate the Kc of the following reaction:

Mg(s)       +          Pb+2 (aq)    < = >      Mg+2(s)       +          Pb(s)

 and

Calculate the Kc of the following reaction:

Br2(l)        +          2I-(aq)     < = >      2Br-(aq)           +          I2(s)

 

 

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                                     BALANCING REDOX IN ACIDIC SOLUTION

Balancing Redox Reactions Using the Half Reaction Method

Many redox reactions occur in aqueous solutions or suspensions.

In this medium most of the reactants and products exist as charged species (ions) and their reaction is often affected by the pH of the medium.

The following provides examples of how these equations may be balanced systematically.

The method that is used is called the ion-electron or "half-reaction" method.

Example 1 -- Balancing Redox Reactions Which Occur in Acidic Solution

Organic compounds, called alcohols, are readily oxidized by acidic solutions of dichromate ions.

The following reaction, written in net ionic form, records this change.

The oxidation states of each atom in each compound is listed in order to identify the species that are oxidized and reduced, respectively.

An examination of the oxidation states, indicates that carbon is being oxidized, and chromium, is being reduced.

To balance the equation, use the following steps:

First, divide the equation into two halves,

an oxidation half-reaction and reduction half- reaction by grouping appropriate species.

(red.)   (Cr2O7)-2 ----> Cr+3  

(ox.)    C2H6O ---->   C2H4O

Second, if necessary, balance both equations by inspection.

In doing this ignore any oxygen and hydrogen atoms in the formula units.

In other words, balance the non-hydrogen and non-oxygen atoms only.

By following this guideline in the example above, only the reduction half-reaction needs to be balanced by placing the coefficient, 2 , in front of Cr+3 as shown below.

 (red.)  (Cr2O7)-2  ---->  2 Cr+3  

(ox.) C2H6O ----> C2H4O   (as there are equal numbers of carbon atoms on both sides of this equation, skip this step for this half-reaction.

Remember, in this step, one concentrates on balancing only non-hydrogen and non-oxygen atoms)

The third step involves balancing oxygen atoms.

To do this, one must use water (H2O) molecules.

Use 1 molecule of water for each oxygen atom that needs to be balanced.

Add the appropriate number of water molecules to that side of the equation required to balance the oxygen atoms as shown below.

(red.)   Cr2O7-2   ---->  2 Cr+3  +  7 H2O   
(ox.)    C2H6O      ---->  C2H4O (as there are equal numbers of 

oxygen atoms,skip this step for this half-reaction)

The fourth step involves balancing the hydrogen atoms.

To do this one must use hydrogen ions (H+).

Use one (1) H+ ion for every hydrogen atom that needs to balanced.

Add the appropriate number of hydrogen ions to that side of the equation required to balance the hydrogen atoms as shown below

(red.)   14 H+ + (Cr2O7)-2   --->   2 Cr+3 + 7 H2O

(as there are 14 hydrogen atoms in 7 water molecules, 14 H+ ions must be added to the opposite side of the equation)

(ox.)    C2H6O ---> C2H4O + 2 H+

(2 hydrogen ions must be added to the "product" side ofthe equation to obtain a balance)

The fifth step involves the balancing of positive and negative charges.

This is done by adding electrons (e-).

Each electron has a charge equal to (-1).

To determine the number of electrons required, find the net charge of each side the equation.

The electrons must always be added to that side which has the greater positive charge as shown below.

note: the net charge on each side of the equation does not have to equal zero.

The same step is repeated for the oxidation half-reaction.

As there is a net chargae of +2 on the product side, two electrons must be added to that side of the equation as shown below.

At this point the two half-reactions appear as:

(red)   6e-  +  14 H+  +  (Cr2O7)-2  ------->  2 Cr+3 + 7 H2O   

 (ox)    C2H6O  ------>  C2H4O  +  2 H+  +  2e-

 The reduction half-reaction requires 6 e-, while the oxidation half-reaction produces 2 e-.

The sixth step involves multiplying each half-reaction by the smallest whole number that is required to equalize the number of electrons gained by reduction with the number of electrons produced by oxidation.

Using this guideline, the oxidation half reaction must be multiplied by "3" to give the 6 electrons required by the reduction half-reaction.

 (ox.)   3 C2H6O ---> 3 C2H4O + 6 H+ + 6e-

 The seventh and last step involves adding the two half reactions and reducing to the smallest whole number by cancelling species which on both sides of the arrow.

 6e-  +  14 H+  +   (Cr2O7)-2  ----->   2 Cr+3 + 7 H2O                    

 3 C2H6O   ----->   3 C2H4O + 6 H+ + 6e-

 adding the two half-reactions above gives the following:

6e- + 14H+ + (Cr2O7)-2 + 3C2H6O --> 2Cr+3 + 7H2O + 3C2H4O + 6H+ + 6e-

Note that the above equation can be further simplified by subtracting out 6 e- and 6 H+ ions from both sides of the equation to give the final equation.

Note: the equation above is completely balanced in terms of having an equal number of atoms as well as charges.

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                                     BALANCING REDOX IN BASIC SOLUTION

The active ingredient in bleach is the hypochlorite (OCl-) ion.

This ion is a powerful oxidizing agent which oxidizes many substances under basic conditions.

A typical reaction is its behavior with iodide (I-) ions as shown below in net ionic form.

I- (aq) + OCl-(aq) ------> I2 + Cl- + H2O

Balancing redox equations in basic solutions is identical to that of acidic solutions except for the last few steps as shown below.

First, divide the equation into two halves; an oxidation half-reaction and reduction-reaction by grouping appropriate species.

(ox)    I-   ----> I2   
(red)   OCl- ----> Cl- + H2O 

Second, if needed, balance both equations, by inspection ignoring any oxygen and hydrogen atoms.

(The non-hydrogen and non-oxygen atoms are already balanced,hence skip this step) 

Third, balance the oxygen atoms using water molecules .

(The hydrogen and oxygen atoms are already balanced; hence, skip this step also.

Fourth, balance any hydrogen atoms by using an (H+) for each hydrogen atom.

(ox)    2 I- ----> I2 
(as no hydrogens are present, skip this step for this half-reaction)
(red)   2 H+ + OCl- -----> Cl- + H2O
(two hydrogen ions must be added to balance the hydrogens in the water molecule).

Fifth, use electrons (e-) to equalize the net charge on both sides of the equation.

Note; each electron (e-) represents a charge of (-1).

Sixth, equalize the number of electrons lost with the number of electrons gained by multiplying by an appropriate small whole number.  

(ox)   2 I-  ---->  I2  +  2e- 
(red)  2e-  +  2 H+  +  OCl-  ---->  Cl-  +  H2O 

(as the number of electrons lost equals the number of electrons gained, skip this step)

 Add the two equations, as shown below.

2 e-  +  2 I-  +  2 H+  +  OCl-  ----> I2  +  Cl-  +  H2O  +  2e-

and subtract "like" terms from both sides of the equation. Subtracting "2e-" from both sides of the equation gives the net equation:

To indicate the fact that the reaction takes place in a basic solution, one must now add one (OH-) unit for every (H+) present in the equation.

The OH- ions must be added to both sides of the equation as shown below.  

2 OH- + 2 I- + 2 H+ + OCl-  ----->  I2 + Cl- + H2O + 2 OH-

Then, on that side of the equation which contains both (OH-) and (H+) ions, combine them to form H2O.

Note, combining the 2 OH- with the 2 H+ ions above gives 2 HOH or 2 H2O molecules as written below.  

2 H2O + 2 I- + OCl-  ---->  I2 + Cl- + H2O + 2 OH-

Simplify the equation by subtracting out water molecules, to obtain the final, balanced equation.

Note that both the atoms and charges are equal on both sides of the equation, and the presence of hydroxide ions (OH-) indicates that the reaction occurs in basic solution.

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                        Balancing REDOX in Acid Solution

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 Practice Problems for Balancing Redox Equations in Acid/Base

Redox Reactions in Acidic Solution 

1.  I¯ (aq)  +  ClO¯(aq)  →  I3¯(aq)   +   Cl¯(aq)

a.  write half reactions

b.  balance mass

c.  balance oxygens

d.  balance hydrogens

e.  balance charge in each half reaction

f.  balance charge between the 2 different half reactions.

g.  add the balanced half reactions.

h.  cancel intermediates.

 

2.  As2O3 (s)   +   NO3¯ (aq)   →   H3AsO4 (aq)   +   NO (g)

 

3.  Br¯ (aq)   +   MnO4¯ (aq)   →   Br2 (l)   +   Mn2+ (aq)

 

4.   CH3OH (aq)    +   Cr2O72- (aq)   →     CH2O(l)   +    Cr3+(aq)

 

5.  Mn2+(aq)   +   BiO3¯ (aq)   →      Bi3+(aq)  +   MnO4¯ (aq)

 

6.  S8(s)   +   NO3¯ (aq)    →      SO32-(aq)    +    NO(g)

 

7.  H3AsO4(aq)   +   Zn(s)   →         AsH3(g)   +   Zn2+(aq)

 

8.  P4(s)   +   Cr2O72-(aq)   →         H3PO4(aq)   +   Cr3+(aq) 

 

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                   Balancing REDOX in Basic Solution

1.  Al(s)   +   MnO4¯ (aq)   →     MnO2(s)    +    Al(OH)4¯ (aq)

 

2.  NO2¯ (aq)   +   Al(s)   →       NH3(aq)   +   AlO2¯ (aq)

 

3.  Cr(s)   +   CrO42-(aq)    →      Cr(OH)3(s)            

Note:  Cr(OH)3 is found in BOTH half reactions! 

 

4.  MnO4¯ (aq)  +  S2-(aq)   →      MnO2(s)     +    SO32-(aq)

 

5.  Cl2(aq)   +    Br2(l)         →       OBr¯ (aq)      +   Cl¯ (aq)

 

6.  H2O2(aq)   +   I¯ (aq)     →      IO3¯ (aq)  

            Note:  IO3¯ is found in both half reactons!

 

7.  NO3¯ (aq)  +   NH3(aq)        →         NO2¯ (aq)

 

8.  S8(aq)   +   MnO4¯ (aq)       →       SO42-(aq)  +  MnO2(s)

 

ANSWER KEY NOT NEEDED!  IF THE EQUATION MASS AND CHARGE BALANCES, IT CAN’T BE WRONG!

 

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                                  Voltaic Cells

1. Balance the following redox reactions, calculate Eo and predict whether each of the reactions will occur as written. Label the anode and cathode half reactions. The reactions are in acidic solution at standard conditions.


2. Given the following:
 

a) Write the shorthand notation for the cell that uses these half reactions.


b) Write the net equation for the cell (Hint: cancel spectators)


c) Calculate Eo for the cell


d) Calculate for the cell

2b.  Use the standard reduction potentials listed below to calculate the standard free energy change ΔGo, for the following reaction.

4Ag(s) + O2(g) + 4H+(aq)→ 4Ag+(aq) + 2H2O(l)

Ox:   O2(g) + 4H+(aq)→  2H2O(l)                           Eo =  +1.23 V

Red:   4Ag(s) → 4Ag+(aq)                                           Eo =  +0.80 V

a) Calculate Eo for the cell


b) Calculate for the cell

3.   a) Use the electrode potentials to calculate the emf of a standard cell that uses the reaction:


b) What is for the cell?


c) if for the reaction is -93.09 kJ what is for the reaction?

                                       hint:  ΔGo = ΔH -TΔS

    T = 298 K (from STAP, standard ambient temperature and pressure)


 

4. Use the standard electrode potentials to calculate the equilibrium constant at 25°C for the reaction:


5.    

a) Calculate the emf of a cell formed from a Zn2+(aq)/Zn(s) half reaction in

 which [Zn2+] = 0.0500M and Cl2(g)/Cl-(aq) half reaction in which [Cl-] =

 0.0500M and the pressure of Cl2(g) is 1.25 atm

        b) Write the equation for the cell reaction.

        c) Which electrode is negative?

     


6. What is the concentration of Cd2+(aq) in the cell:

Zn|Zn2+(aq, 0.0900M)||Cd2+(aq, ?M)|Cd(s)


if the emf of the cell is 0.400V?


 

7. An acidic solution containing Pb2+ ions is electrolyzed and PbO2(s) plated out on the anode.

(a) Write the chemical equation for the anode reaction.

(b) If a current of 0.750 A is used for 25.0 min., what mass of PbO2(s) plates out?

(c) If a solution contains 2.50g of Pb2+, how many minutes will it take to plate out all of the lead, as PbO2(s) using a current of 0.750 amps?


8. How many minutes will it take to plate out 6.00 g of Cd from a Cd2+ solution using a current of 6.00A?


 

9. For the cell:

U(s)|U3+(aq)||Ag+(aq)|Ag(s)

Eo is 2.588 V. Use the emf of the cell and Eo for the Ag+/Ag reaction to calculate Eo for the U3+/U half reaction.

Practice Problem 14:

Determine the oxidation number of the chromium in an unknown salt if electrolysis of a molten sample of this salt for 1.50 hours with a 10.0-amp current deposits 9.71 grams of chromium metal at the cathode.

Click here to check your answer to Practice Problem 14

Click here to see a solution to Practice Problem 14

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___________________________________

SOLUTIONS:

2.

    A. Ag(s), I-(aq)|AgI(s)||Ag+|Ag(s)


    B.
    C. = 0.951 V
    D. = -918 kJ

3.

    A. = 0.292 V
    B. = -56.4 kJ
    C. = 0.502 J/K

4. K = 5.2 x 103

5.

    a) E = 2.239 V
    b)

6. [Cd2+] = 2.02 M

7.

    a)
    b) 1.39 g PbO2(s)


    c) 51.7 min

8. 28.6 miN

9. Eo = -1.788; U3+ + 3e- -----> U(s)

10.  Calculate the volume of O2 gas at 25oC and 1.00 atm that will collect at the cathode when an aqueous solution of Na2SO4 is electrolyzed for 3.00 hours with a 25.0-amp current.

 

 

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___________________________________

 

 

                       Electrochemical Cell (Regents)

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  1. Calculate the standard cell potential produced by a voltaic cell consisting of a nickel electrode in contact with a solution of Ni2+ ions and a silver electrode in contact with a solution of Ag+ ions.

 

 

  1. What is the voltage produced by a voltaic cell consisting of an aluminum electrode in contact with a solution of Al3+ ions and an iron electrode in contact with a solution of Fe2+ ion?

 

 

  1. Calculate the standard cell potential produced by a voltaic cell consisting of a sodium electrode in contact with a solution of Na+ ions and a copper electrode in contact with a solution of Cu2+ ions.

 

  1. What is the voltage produced by a voltaic cell consisting of a calcium electrode in contact with a solution of Cu2+  ions

 

  1. A voltaic cell is constructed using electrodes based on the following half reactions:

Pb2+ (aq) + 2e- -> Pb(s)

Au3+(aq) +3e- -> Au(s0

a.        Which is the anode and which is the cathode in this cell?

b.       What is the standard cell potential?

 

  1. Calculate the standard cell potential produced by a voltaic cell consisting of a nickel electrode in contact with a solution of Ni2+ ions and a copper electrode in contact with a solution of Cu2+  ions

 

 

  1. A voltaic cell is constructed using electrodes based on the following half reactions:

Mn2+ (aq) + 2e- -> Mn(s)

Cu2+ (aq) +2e- -> Cu(s)

Which is the anode and which is the cathode in this cell?

 

What is the standard cell potential?

 

 

  1. What is the voltage produced by a voltaic cell consisting of a lead electrode in contact with a solution of Pb2+ ions and an iron electrode in contact with a solution of Fe2+? Which is anode and which is the cathode?

 

 

  1. What is the voltage produced by a voltaic cell consisting of a zinc electrode in contact with a solution of Zn2+  ions and a silver electrode in contact with a solution of Ag+ ions?

 

 

  1. Calculate the standard cell potential produced by a voltaic cell consisting of a gold electrode in contact with a solution of Au3+ ions and a silver electrode in contact with a solution of Ag+ ions. Which is the anode and which is the cathode? 

 

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___________________________________

                                                 Simple Voltaic Cell Practice

Analysis

1. Label the diagrams below to show the three identified voltaic cells.  Identify electrodes, electrolytes, electron flow and direction of ion movement.  Label the cathode, anode, and write the net cell reaction equations for each. (note:  you decide which metals make up the electrodes)

2. Indicate whether the following processes occur at the cathode or at the anode of a voltaic cell.

    a. reduction half-reaction _____________________________________

    b. oxidation half-reaction _____________________________________

    c. reaction of the strongest reducing agent ______________________

    d. reaction of the strongest oxidizing agent ______________________

3. Use the tool to design the voltaic cell with the greatest cell potential.
    Explain why you chose this combination.

 

                   STANDARD REDUCTION POTENTIALS

       STANDARD OXIDATION AND REDUCTION POTENTIALS

               

Table 1.  Standard Electrode Reduction and Oxidation Potential Values

Anodic - exhibits greater tendency to lose electrons

Reduction Reaction

Eo (V)

Oxidation Reaction

Eo (V)

Li+ + e- → Li

-3.04

Li → Li+ + e-

3.04

K+ + e- → K

-2.92

K → K+ + e-

2.92

Ba2+ + 2e- → Ba

-2.90

Ba → Ba2+ + 2e-

2.90

Ca2+ + 2e- → Ca

-2.87

Ca → Ca2+ + 2e-

2.87

Na+ + e- → Na

-2.71

Na → Na+ + e-

2.71

Mg2+ + 2e- → Mg

-2.37

Mg → Mg2+ + 2e-

2.37

Al3+ + 3e- → Al

-1.66

Al → Al3+ + 3e-

1.66

Mn2+ + 2e- → Mn

-1.18

Mn → Mn2+ + 2e-

1.18

2H2O + 2e- → H2 + 2 OH-

-0.83

H2 + 2 OH- → 2H2O + 2e-

0.83

Zn2+ + 2e- → Zn

-0.76

Zn → Zn2+ + 2e-

0.76

Cr2+ + 2e- → Cr

-0.74

Cr → Cr2+ + 2e-

0.74

Fe2+ + 2e- → Fe

-0.44

Fe → Fe2+ + 2e-

0.44

Cr3+ + 3e- → Cr

-0.41

Cr → Cr3+ + 3e-

0.41

Cd2+ + 2e- → Cd

-0.40

Cd → Cd2+ + 2e-

0.40

Co2+ + 2e- → Co

-0.28

Co → Co2+ + 2e-

0.28

Ni2+ + 2e- → Ni

-0.25

Ni → Ni2+ + 2e-

0.25

Sn2+ + 2e- → Sn

-0.14

Sn → Sn2+ + 2e-

0.14

Pb2+ + 2e- → Pb

-0.13

Pb → Pb2+ + 2e-

0.13

Fe3+ + 3e- → Fe

-0.04

Fe → Fe3+ + 3e-

0.04

Arbitrary Neutral : H2

Reduction Reaction

Eo (V)

Oxidation Reaction

Eo (V)

2H+ + 2e- → H2

0.00

H2 → 2H+ + 2e-

0.00

Cathodic - exhibits greater tendency to gain electrons

Reduction Reaction

Eo (V)

Oxidation Reaction

Eo (V)

S + 2H+ + 2e- → H2S

0.14

H2S → S + 2H+ + 2e-

-0.14

Sn4+ + 2e- → Sn2+

0.15

Sn2+ → Sn4+ + 2e-

-0.15

Cu2+ + e- → Cu+

0.16

Cu+ → Cu2+ + e-

-0.16

SO42+ + 4H+ + 2e- → SO2 + 2H2O

0.17

SO2 + 2H2O → SO42+ + 4H+ + 2e-

-0.17

AgCl + e- → Ag + Cl-

0.22

Ag + Cl-  →  AgCl + e-

-0.22

Cu2+ + 2e- → Cu

0.34

Cu → Cu2+ + 2e-

-0.34

ClO3- + H2O + 2e- → ClO2- + 2OH-

0.35

ClO2- + 2OH- → ClO3- + H2O + 2e-

-0.35

2H2O + O2 + 4e- → 4OH-

0.40

4OH- → 2H2O + O2 + 4e-

-0.40

Cu+ + e- → Cu

0.52

Cu → Cu+ + e-

-0.52

I2 + 2e- → 2I-

0.54

2I- → I2 + 2e-

-0.54

O2 + 2H+ + 2e- → H2O2

0.68

H2O2 → O2 + 2H+ + 2e-

-0.68

Fe3+ + e- → Fe2+

0.77

Fe2+  → Fe3+ + e-

-0.77

NO3- + 2H+ + e- → NO2 + H2O

0.78

NO2 + H2O → NO3- + 2H+ + e-

-0.78

Hg2+ + 2e- → Hg

0.78

Hg → Hg2+ + 2e-

-0.78

Ag+ + e- → Ag

0.80

Ag → Ag+ + e-

-0.80

NO3- + 4H+ +3 e- → NO + 2H2O

0.96

NO + 2H2O → NO3- + 4H+ +3 e-

-0.96

Br2 + 2e- → 2Br-

1.06

2Br- → Br2 + 2e-

-1.06

O2 + 4H+ + 4e- → 2H2O

1.23

2H2O → O2 + 4H+ + 4e-

-1.23

MnO2 + 4H+ + 2e- → Mn2+ + 2H2O

1.28

Mn2+ + 2H2O → MnO2 + 4H+ + 2e-

-1.28

Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O

1.33

2Cr3+ + 7H2O → Cr2O72- + 14H+ + 6e-

-1.33

Cl2 + 2e- → 2Cl-

1.36

2Cl- → Cl2 + 2e-

-1.36

Ce4+ + e- → Ce3+

1.44

Ce3+ → Ce4+ + e-

-1.44

Au3+ + 3e- → Au

1.50

Au → Au3+ + 3e-

-1.50

MnO4- + 8H+ + 5e- → Mn2+ + 4H2O

1.52

Mn2+ + 4H2O → MnO4- + 8H+ + 5e-

-1.52

H2O2 + 2H+ + 2e- → 2H2O

1.78

2H2O → H2O2 + 2H++ 2e-

-1.78

Co3+ + e- → Co2+

1.82

Co2+ → Co3+ + e-

-1.82

S2O82- + 2e- → 2SO42-

2.01

2SO42- → S2O82- + 2e-

-2.01

O3 + 2H+ + 2e- → O2 + H2O

2.07

O2 + H2O → O3 + 2H+ + 2e-  

-2.07

F2 + 2e- → 2F-

2.87

2F- → F2 + 2e-

-2.87