# File - Leo Burke Academy

March 21, 2018 | Author: Anonymous | Category: Science, Chemistry

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Oxidation Numbers Many chemical reactions involve the loss or gain of electrons by atoms, ions or molecules. The loss or gain of electrons can result in a change in the charge of a species. For example, when sodium atoms lose electrons to chlorine, each species acquires a different net charge. Sodium becomes a 1+ ion and chlorine becomes a 1- ion. Knowing how to draw Lewis diagrams for atoms, ions and covalent compounds can be an asset in this lesson. Introduction to Oxidation Numbers The oxidation number - the actual or hypothetical charge of an atom or ion - is a way of keeping track of the electrons in a particular species. In a redox reaction, two elements undergo a change in oxidation number. The oxidation number of one element will increase as a result of a loss of electrons and the oxidation number of the other element will decrease as a result of a gain of electrons. Consider the example of sodium reacting with chlorine:

Each sodium atom is electrically neutral. When a sodium atom loses its valence electron, the net charge of sodium changes from 0 to 1+.

Since it is electrically neutral (#protons = #electrons), a sodium atom is assigned an oxidation number of 0. Since the sodium ion has one less electron, it is assigned the oxidation number +1. (Note the location of the + sign; the sign always comes first in an oxidation number.)

When a chlorine atom gains an electron, the net charge of chlorine changes from 0 to 1-.

Since the two atoms in Cl2 have the same number of electrons as a neutral chlorine atom, each chlorine atom is assigned a oxidation number of 0. Since each chloride ion has one extra electron, each is assigned the oxidation number -1.

There are at least two ways to assign oxidation numbers to the elements involved in a chemical reaction. One method involves drawing Lewis diagrams for the species and using electronegativity values to assign oxidation numbers. A second method requires the use of a set of rules. You will explore this second method. Assigning Oxidation Numbers There is a well established set of rules for assigning oxidation numbers to elements in a chemical reaction. Mastery of these rules is essential to success. Each rule is presented below. One or more examples of its application is/are provided. Rule 1: Simple Ion Charge = Oxidation Number

Assigning oxidation numbers to most monatomic ions is simply a matter of identifying or determining the ion charge. For example,  the ions of Group 1 metals are assigned the oxidation number +1.  the ions of Group 2 metals are assigned the oxidation number +2.  the ions of Group 17 (or Group 7A) nonmetals are assigned the oxidation number -1. Rule 2: Hydrogen

Hydrogen is assigned the oxidation number +1 in all compounds except in metal hydrides where it is assigned the number -1.  In a water molecule, hydrogen is bonded to oxygen by a polar covalent bond. The electron pair is "closer" to oxygen making hydrogen partially positive (δ+) and oxygen partially negative (δ-). Each hydrogen is assigned the oxidation number -1.  In calcium hydride, CaH2, hydrogen is in the form of the hydride ion, H-. It is assigned the oxidation number -1. The calcium ion is assigned +2. Rule 3: Oxygen

Oxygen is always assigned the oxidation number -2, except in peroxides like hydrogen peroxide where it is assigned -1, and in OF2 where is is assigned a +2 due to the higher electronegativity of the fluorine atom. Consider these examples:  In calcium oxide, CaO, oxide ion has a 2- charge. Its oxidation number is -2.  In water, H2O, the oxygen atom strongly attracts the valence electrons of hydrogen and is assigned an oxidation number of -2.  in hydrogen peroxide, H2O2, each oxygen equally shares one electron pair but strongly attracts the bonding electron of the covalently bonded hydrogen atom. Each oxygen atom is assigned the oxidation number is -1.

Rule 4: The Elements

The oxidation number of a pure element is zero. Consider these examples.  Atoms in pure gold are electrically neutral. The gold atom is assigned the oxidation number of 0.  Fluorine atoms in F2 equally share their bonding electron pair. Each atom is assigned the oxidation number of 0. Rule 5: Compounds

The sum of the oxidation numbers in a compound is zero.  The sum of oxidation numbers in NaCl is (+1) + (-1) = 0  The sum of oxidation numbers in H2O is 2(+1) + (-2) = 0 In compounds that do not contain hydrogen or oxygen, the more electronegative element is assigned the oxidation number it would have in an ionic compound. Consider these examples.  In sulfur dichloride, SCl2, chlorine has a higher electronegativity so it is assigned -1 and sulfur is assigned +2.  In iodine monobromide, IBr, bromine has higher electronegativity than iodine, so it is assigned -1 and iodine is assigned +1. Rule 6: Polyatomic Ions

The sum of the oxidation numbers of the elements in a polyatomic ion must equal the ion charge. Consider these examples.  Carbonate ion,CO3 2-, Oxygen is assigned the oxidation number -2. There are three oxygen atoms in the formula so the total negative charge is 6-. Since the carbonate ion has a charge of 2-, the oxidation number of carbon must be +4.  Sulfite, SO3 2- , Oxygen is assigned the oxidation number -2. There are three oxygen atoms in the formula so the total negative charge is 6-. Since the sulfite ion has a charge of 2-, the oxidation number of sulfur must be +4.  Sulfate, SO4 2- , Oxygen is assigned the oxidation number -2. There are four oxygen atoms in the formula so the total negative charge is 8-. Since the sulfate ion has a charge of 2-, the oxidation number of sulfur must be +6.  Ammonium ion, NH4 + , Hydrogen is assigned the oxidation number +1. There are four hydrogen atoms in the formula so the total positive charge is 4+. Since the ammonium ion has a charge of 1+, the oxidation number of nitrogen must be -3.  Nitrate ion, NO3 - , Oxygen is assigned the oxidation number -2. There are three oxygen atoms in the formula so the total negative charge is 6-. Since the nitrate ion has a charge of 1-, the oxidation number of nitrogen must be +5. Rules Summary

Rules 5 and 6 are used to assign oxidation numbers to elements not covered by Rules 1-4. By reviewing the examples above, you should see that an element can be assigned one of several oxidation numbers. Sulfur for example was assigned -2, +4, and +6 in three different examples!

Sample Exercise 1 Determine the oxidation number of the elements in potassium hydrogen sulfate, KHSO4. Answer

Step 1: Apply the rules to assign oxidation numbers to K, H and O.  potassium is a 1+ ion in this ionic compound, so it is assigned +1  hydrogen is assigned +1  oxygen is assigned -2 Step 2: Calculate the oxidation number of sulfur. Let x = oxidation number of S

Step 3: Communicate. The oxidation numbers of K, H, S, and O are +1, +1, +6, and -2 respectively. Oxidation Numbers and Redox Reactions You are probably wondering...."what is the purpose of these rules? What's the point?" The point is that not all chemical reactions are redox reactions. By assigning oxidation numbers to the elements in a balanced chemical equation, you can tell whether reacting species have been oxidized/reduced. In other words, you can identify redox reactions by applying the oxidation number rules. Consider the reaction between silver nitrate and sodium chloride:

Now consider it with oxidation numbers assigned to the elements in this total ionic equation:

What do you notice about the oxidation numbers of the elements? None of them changed. No electrons have been gained or lost by the elements involved in the reaction. When none of the species in a system undergo a change in oxidation number, the reaction is not a redox reaction. Now consider the reaction between copper metal and silver nitrate:

What do you notice about the oxidation numbers in the total ionic equation for the reaction?

Copper undergoes a change in oxidation number from 0 to +2 by losing two electrons.

Copper is oxidized. As each silver atom becomes a silver ion, it undergoes a change from +1 to 0 by gaining an electron.

Silver is reduced. The sum of the half reactions is the redox reaction.

The terms oxidation and reduction can now be defined in another way:  oxidation is an increase in oxidation number.  reduction is a decrease in oxidation number. Sample Exercise 2 Determine whether the combustion of ethene, C2H4, is a redox reaction. If it is, identify the oxidizing agent and the reducing agent. Answer

Step 1: Write the balanced chemical equation.

Step 2: Assign the oxidation numbers.

   

C2H4, hydrogen is assigned +1. Since each carbon is polar covalently bonded to two hydrogens, each carbon atom is assigned -2. O2: each oxygen is nonpolar covalently bonded, so each is assigned 0. CO2: each oxygen is assigned -2, so carbon is assigned +4. H2O: oxygen is assigned -2 and each hydrogen is assigned +1.

Step 3: Compare the oxidation numbers of each element. The oxidation numbers of carbon and oxygen atoms have changed, so the reaction can be classified as a redox reaction. Step 4: Identify the oxidizing agent. The oxidation number of oxygen has changed from 0 to -2; therefore, oxygen was reduced. It is the oxidizing agent. Step 5: Identify the reducing agent. The oxidation number of carbon has changed from -2 to +4; therefore, carbon was oxidized. It is the reducing agent. Step 6: Communicate. The combustion of ethene is a redox reaction in which carbon is oxidized and oxygen is reduced. Sample Exercise 3 Determine whether the reaction of zinc metal and lead(II) nitrate is a redox reaction, and if so, identify the oxidizing and reducing agents and their half reactions.

Step 1: Write the total ionic equation.

Step 2: Assign the oxidation numbers.  Zn: Zinc metal consists of neutral zinc atoms, so it is assigned 0.  Pb2+: Lead(II) ion has a 2+ charge, so it is assigned +2.  NO3 - ,is a spectator ion, so it is not necessary to assign oxidation numbers to its elements because they do not change.  Pb: Lead metal consists of neutral lead atoms, so it is assigned 0.  Zn2+: Zinc ion has a 2+ charge, so it is assigned +2.

Step 3: Compare the oxidation numbers of each element. The oxidation numbers of zinc and lead have changed, so the reaction can be classified as a redox reaction. Step 4: Identify the oxidizing agent. The oxidation number of lead has changed from +2 to 0; therefore, lead was reduced. It is the oxidizing agent.

Step 5: Identify the reducing agent. The oxidation number of zinc has changed from 0 to +2; therefore, zinc was oxidized. It is the reducing agent.

Step 6: Communicate. The single replacement reaction involving zinc metal and aqueous lead(II) ions is a redox reaction. Zinc atoms lose electrons to reduce lead(II) ions. Sample Exercise 4 Given these equations:

identify the oxidation and reduction half-reactions by balancing the number of electrons lost and gained. Answer

Step 1: Write the chemical equation showing the assigned oxidation numbers.

Step 2: Write the half reactions to show the gain/loss of electrons.

Zinc is oxidized, oxygen is reduced.

Chlorine is oxidized, potassium is reduced. General Rules for Identifying Redox Reactions Double replacement reactions are never redox reactions. All combustion, single replacement, formation and decomposition reactions are redox reactions. Oxidizing agents and reducing agents are always reactants, never products.

Balancing Redox Reactions All redox reaction equations are combinations of oxidation and reduction half-reactions. The number of electrons lost in oxidation equals the number of electrons gained in reduction. There are various ways of balancing redox reaction equations. For simple reactions, balancing a redox equation can be achieved by balancing the numbers of electrons in the half-reactions. Consider the example of reacting zinc metal with aqueous gold(III) cyanide to produce pure gold. Gold ions in solution gain three electrons to become gold atoms:

The electrons gained by gold ions are those lost by zinc metal:

Obviously, the number of electrons in the oxidation half-reaction does not equal the number of electrons in the reduction half-reaction. To balance them, the equations are multiplied through by coefficients.

The half-reactions are added to produce the redox reaction equation:

By inspection, you can see that the number of gold atoms/ions are balanced as are the numbers of zinc atoms/ions.

Using a Table of Standard Reduction Potentials Most redox reactions are more complex than formation, decomposition and single replacement reactions. The half reactions for some of these reactions can be complex. The Standard Reduction Potentials of Half-Cells table can be used to produce the half-reaction equations. You will use this table for another purpose in the next section, so for now you can ignore the "reduction potentials" part of the table and focus on the half-reactions only. The table is called a table of reduction potentials because each reactant is shown gaining electrons. Each reactant therefore is an oxidizing agent. The strongest oxidizing agent (electron remover) is listed at the top of the table (F2). The species listed on the products side of each equation is the species that loses electrons in an redox reaction (i.e. the reducing agent). The weakest reducing agent is located at the top of the table (F-). and the strongest at the bottom (Li). For any chemical system, the redox reaction involves the strongest oxidizing agent and the strongest reducing agent. Sample Exercise 1 Write the redox reaction equation for the reaction between zinc metal and aqueous chromium(III) sulfate. Answer

Step 1: List the reactant species present.

Step 2: Label the oxidizing agents (O.A.) and reducing agents (R.A.). Sulfate and water are treated as a pair in the table.

Step 3: Identify the strongest oxidizing agent and the strongest reducing agent.

Step 4: Write the half reactions

Step 5: Balance the half reactions so each shows the same number of electrons.

Redox Reactions in Acidic or Basic Solutions Many redox reactions occur in acidic or basic solutions. Hydrogen ions are involved when solutions are acidic and hydroxide ions are involved in basic solutions. The methods for balancing redox equations in acidic or basic solutions are similar. For redox reactions in acidic solutions: 1.Write separate half-reaction equations for oxidation and reduction. 2. Balance all elements except H and O 3. Balance O by adding H2O molecules. 4. Balance H by adding H+ ions 5. Balance charges by adding electrons. 6. Balance the electrons 7. Add the half reactions. For redox reactions in basic solutions, there is an eighth step. 8. Add one of these equations:

as needed to cancel out any H+. Here is an example of how to write a redox reaction equation for a reaction occurring in acidic conditions. Sample Exercise 2 Write a redox reaction equation for the reaction of permanganate ion, MnO4- , with tin(II) ion in acidic conditions to produce manganese(II) ion and tin(IV) ion.

Answer Step 1: Write the separate unbalanced half-reactions. Manganese is being reduced from +7 in MnO4- to +2 in Mn2+. Tin is being oxidized from +2 in Sn2+ to +4 in Sn4+. i)

ii) Step 2: Balance the elements except H and O. The elements, other than H and O, are already balanced. Step 3: Balance O in equation (i) by adding H2O. Water is the solvent for the reaction. Since there are four O's on the left of equation (i), add 4 H2O to the right side of the equation.

Equation (ii) does not require addition of H2O. Step 4: Balance H in equation (i) by adding H+. H+ are present in solution because the solution is acidic. There are eight hydrogen atoms on the right (4 H2O) of equation (i). To balance them, add 8 H+ to the left side of the equation.

Equation (ii) does not require addition of H. Step 5: Balance the charges by adding electrons. For equation (i), the net charge on the left side is 7+. The net charge on the right is 2+. Add five electrons to the left side to balance the charge.

For equation (ii), the net charge on the left is 2+ and the net charge on the right is 4+, so two electrons are added to the right side.

Step 6: Balance the electrons using LCM method. Since LCM of 2 and 5 is 10, multiply equations as follows: i)

ii)

Here is an example of how to write a redox reaction equation for a reaction occurring in basic conditions. Sample Exercise 3 Write a redox reaction equation for the reaction of chromate ion, CrO42- , with sulfide ion in basic conditions to produce chromium(III) hydroxide and solid sulfur.

Answer Step 1: Write separate unbalanced equations for oxidation and reduction. Chromium is being reduced from +6 in CrO42- to +3 in Cr3+. Sulfur is being oxidized from -2 in S2- to 0 in S. i)

ii)

Step 2: Balance all elements except H and O. The other elements are balanced. Step 3: Balance O by adding H2O. (Water is the solvent for the reaction.) Water is the solvent for the reaction. Since there are four O's on the left of equation (i), and three O's on the right, so add 1 H2O to the right side of the equation.

Equation (ii) does not require addition of H2O. Step 4: Balance H by adding H+. (The H+ comes from the acidic solution.) H+ are present in solution because the solution is acidic. There are five hydrogen atoms on the right (3 in Cr(OH)3 and 2 in H2O) of equation (i). To balance them, add 5 H+ to the left side of the equation.

Equation (ii) does not require addition of H. Step 5: Balance the charges by adding electrons. For equation (i), the net charge on the left side is 3+. The net charge on the right is 0. Add three electrons to the left side to balance the charge.

For equation (ii), the net charge on the left is 2- and the net charge on the right is 0, so two electrons are added to the right side.

The net charge on both sides of both equations is now 0. Step 6: Balance the electrons using LCM method. Since LCM of 3 and 2 is 6, multiply equations as follows:

i)

ii)

Step 8: Add 10 ( H2O –> H+ + OH - ) to cancel the 10 H+ ions.

Check the equation for net charge on both sides for for conservation of "atoms". Both sides have a net negative charge of 10-. Atoms are conserved: 2 Cr, 16 O, 16 H, and 3 S. Stoichiometry of a Redox Reaction Stoichiometry is a topic that you've encountered a number of times in chemistry so far. Most recently, you studied the stoichiometry associated with acid-base titration. A common misconception in chemistry is that titration involves acids and bases only. However, titration is a useful technique in redox chemistry as well because titration is simply the progressive addition of one reagent to another until chemical equivalents have reacted. Chemical reactions best suited to study using titration are spontaneous reactions - ones in which the mixing of reactants readily results in the formation of products. The following sample problem involves a typical spontaneous redox reaction.

Sample Exercise 4 Calculate the molar concentration of a 20.00 mL sample of aqueous sodium oxalate, Na2C2O4, if it was titrated to a light purple endpoint with 4.35 mL of 0.0100 M potassium permanganate, KMnO4.

Answer Step 1: Identify wanted and given information. Potassium permanganate dissociates in to potassium, K+, and permanganate ions in a 1:1 ratio, so [KMnO4] = [MnO4-].

Step 2: Find moles of MnO4-.

Step 3:Use mole ratio to find moles of oxalate ion.

Step 4: Calculate molar concentration of oxalate ion.

Sodium oxalate, Na2C2O4, dissociates to produce oxalate ions in a 1:1 ratio, so the molar concentration of the oxalate ion equals that of the sodium oxalate solution. The molar concentration of the sodium oxalate solution is 0.00544 M.

Electrochemical Cells Electron transfer is what redox is all about. In a redox reaction, one species loses electrons (the reducing agent) and another species gains electrons (the oxidizing agent). If the electron transfer takes place along a path like a wire, electricity is produced. Electrochemistry is the study of converting chemical energy to electrical energy. A spontaneous redox reaction is one that does not require the input of energy. Certain redox reactions are spontaneous; that is, electrons are readily transferred from one species to another. In order to produce an electron flow, the reducing agent has to be separated from the oxidizing agent. This is achieved using separate containers called half-cells. The half-cells are connected by two paths: one that allows for flow of electrons (like a wire) and another that allows for flow of ions (like a porous material). This arrangement of half-cells is called an electrochemical cell, which is also known as a galvanic cell or sometimes a voltaic cell. The Typical Electrochemical Cell When a zinc strip is placed in a solution of copper(II) sulfate, this redox reaction takes place spontaneously.

The spontaneous reaction releases heat to the surroundings. This reaction can also be carried out in half-cells. This is well illustrated in most chemistry textbooks including pages 758-759 of the MHR text. Each half-cell contains an electrode and electrolytes. Consider the zinc-copper(II) redox reaction again. One half-cell contains a strip of zinc metal in a solution of zinc sulfate. Zinc metal is the anode. Zinc sulfate solution contains the electrolytes (aqueous ions). In this cell, electrons leave zinc atoms resulting in the formation of zinc ions. In other words, oxidation occurs at the anode.

The other half-cell contains a strip of copper in a solution of copper(II) sulfate. The copper strip

is the cathode. In this cell, electrons combine with aqueous copper(II) ions to form copper atoms. Reduction occurs at the cathode.

The two cells are separated from each other by a porous material that allows ions to flow from one cell to the other but prevents copper ions from contacting the zinc metal. Ions migrate across the porous material to maintain the electrical balance of the solutions. Cations migrate to the cathode; anions migrate to the anode. In this case, zinc ions migrate towards the copper half-cell and sulfate ions migrate to the zinc half cell. The electrons lost by zinc cannot travel through the electrolyte - they must pass through an external medium like a wire. The build-up of electrons in the zinc strip resulting from the formation of zinc ions gives the anode a negative charge. The copper strip is more positive in relation to the zinc strip, so electrons pass along the wire from the anode to the cathode. For each zinc ion formed as zinc atoms loses electrons, a copper atom forms as a copper(II) ion gains the electrons. In some electrochemical cells, the porous membrane is replaced with a salt bridge. The salt bridge is filled with an electrolyte that does not interfere with the redox reaction. The bridge is also designed to regulate flow of ions from one half-cell to the other. Summary of Electrochemical Cell Components 1. An electrochemical cell consists of two half-cells. 2. One half-cell contains the anode and its electrolyte. 3. Another half-cell contains the cathode and its electrolyte. 4. A porous material or a salt bridge allows ions to flow from one half-cell to another. 5. Cations migrate to the cathode; anions migrate to the anode. 6. Electrons flow from the anode to the cathode through an external circuit. 7. Together, the flow of electrons and the flow of ions produces an electrical circuit. Cell Notation Drawing electrochemical cell diagrams can be a time consuming not to mention artistic venture. Chemists have developed an internationally recognized convention for representing an electrochemical cell. The general form of the cell notation is: anode | electrolyte || electrolyte | cathode - The anode and its half cell solution are located on the left of a double line, and the cathode and its half-cell solution are located on the right. - The single lines represent the boundary of the phases that are in direct contact. - The double line represents the porous material or the salt bridge that separates the halfcells.

For example the zinc-copper redox reaction can be represented using this cell notation: Zn(s) | ZnSO4(aq) || CuSO4(aq) | Cu Sample Exercise 1 The half-cells for this redox reaction:

contain cadmium chloride and nickel(II) chloride. a. Write the half-reaction equations. b. Sketch the electrochemical cell and label the anode and cathode. c. Assume a salt bridge is used to connect the cells. Suggest an electrolyte for the bridge. d. Write the cell notation. Answer

a. The half-reactions. Cadmium metal loses electrons to become cadmium ions as nickel(II) ions gain electrons to become nickel metal. Oxidation:

Reduction: b. The Electrochemical Cell. Oxidation occurs at the anode. Reduction occurs at the cathode.

c. Chose an electrolyte for the salt bridge option. A suitable electrolyte for the salt bridge is potassium chloride. Potassium will not be displaced by nickel or cadmium.

d. The Cell Notation

Cell Notations Involving Inert Electrodes Platinum and graphite (carbon) electrodes are often used in half-cells in which the half-reaction does not involve a solid species. Examples include:  reduction of iron(III) ions to iron(II) ions  reduction of manganese in permanganate ions to manganese(II) ions  oxidation of hydrogen gas to hydrogen ions  oxidation of chromium(II) ions to chromium(III) ions. In these cases, the transfer of electrons occurs on an inert medium that can conduct electricity. Inert means generally not reactive. Sample Exercise 2 Write the cell notation for the electrochemical cell in which this redox reaction takes place:

The anode and cathode are made of graphite, C(s). The salt bridge contains potassium nitrate. Answer Step 1: Identify the half-reactions. Oxidation:

Reduction: Step 2: Identify the reaction locations. The oxidation of iodide ions occurs at the anode, so iodine will build up there. The reduction of manganese occurs at the cathode. Step 3: Write the cell notation. The oxidation half cell contains a graphite rod, solid iodine, and iodide ions. The reduction half cell contains a graphite rod and several different aqueous ions.

Standard Reduction Potentials In physics, potential energy is often illustrated by placing an object some distance above a surface. The higher the object, the greater its potential energy. Potential energy can be thought of as the "potential" to become kinetic energy. In chemistry, the extent to which particles are separated is communicated in terms of potential energy. The more separated particles are, the greater their potential energy. This energy is released when the separated particles bond again. In electrochemistry, potential can refer to the relationship between an atom and its valence electrons. You can think of potential as a charge separation. Consider placement of a piece of zinc metal in water. Some of the zinc atoms will shed their valence electrons to become Zn2+ ions which mix with water near the metal.

The valence electrons lost by the dissolved zinc atoms remain in the metal and the metal acquires a net negative charge as a result. Meanwhile, the solution around the metal becomes positive due to the presence of the metallic cations. The metal and the surrounding solution make up a halfcell. There is a charge difference between the metal and the solution - this is the half-cell potential. A different metallic element will dissolve to a different extent in water resulting in a different charge build-up in the metal. For example, copper atoms in a metal strip dissolve to a lesser extent in water than zinc atoms do.

The charge build-up in the copper strip is less negative than in the zinc strip. The potential of these metals in water is different.

The difference between the positiveness of the solution and negativeness of the metal can be expressed in volts. It is the half-cell voltage. This potential cannot be measured directly. Why not? The answer is because in order to measure voltage, there has to be a flow of electrons - an electrical circuit must exist. There is no circuit in a half-cell. The Standard Half-Cell A circuit can be created by joining two half-cells. You have already seen several examples of this. The electrical potential (E) or cell potential of the electrochemical cell is a function of the half-cell combination. The cell potential for a zinc/copper cell will be different than that of a magnesium/zinc cell. The problem of determining the half-cell voltage can be solved using the standard hydrogen halfcell. The standard hydrogen electrode consists of hydrogen gas at SATP and aqueous hydrogen ions at a concentration of 1.0 mol/L. An inert metal like platinum serves to conduct electrons. At standard conditions, this equilibrium is established: Hydrogen ions are reduced at the same rate that hydrogen gas molecules are oxidized. The hydrogen gas is oxidized on a platinum electrode; in other words, the platinum receives the electrons lost by the hydrogen gas molecules. By international agreement, the half-cell potential, E, of the standard hydrogen electrode is set at 0 volts. Since hydrogen ions are reduced in this halfcell, we say that the reduction potential of H+ is 0 V. You might be wondering something like "why is this important?" The answer is that connecting any other half-cell to the standard hydrogen electrode will produce a cell potential that is equal to its half-cell potential. Determining Standard Reduction Potentials

Zn2+ Ion

Consider an electrochemical cell that consists of a zinc half-cell connected to the standard hydrogen electrode. Zinc metal loses electrons more easily than hydrogen gas does. You could say that zinc metal is oxidized more readily than hydrogen gas. Conversely, hydrogen ions have a greater tendency to be reduced than zinc ions. Any way you look at it, the build-up of electrons on the zinc metal will be greater than on the platinum metal. Electrons will flow from the zinc electrode to the platinum electrode. Placing a voltmeter between the zinc and platinum allows measurement of the cell potential (voltage potential). At standard conditions, the cell potential is 0.76 V. Since the hydrogen half-cell potential is set at 0 V, the zinc half-cell potential is must be 0.76 V. Copper Half-Cell

Now consider a cell consisting of a copper half-cell connected to the standard hydrogen electrode.

Copper does not lose its electrons as easily as hydrogen does. In terms of oxidation, hydrogen gas loses electrons more readily than copper metal. Consequently, there is a greater build-up of electrons on the platinum electrode than there is on the copper electrode. Electrons flow from platinum to copper. Copper ions are reduced more readily than hydrogen ions. At standard conditions, the cell potential is 0.34 V. Again, since the half-cell potential of the standard hydrogen electrode is set to 0 V, the cell potential (0.34 V) is equal to the half-cell potential of the copper cell.

Determining the Sign of a Reduction Potential

Compare the flow of electrons in each example above. What do you notice? How can the differences be communicated? The answer is simple, attach a sign to a reduction potential.  For a species that is reduced more easily than hydrogen, the cell potential is assigned a positive sign.  For a species that is not reduced more easily than hydrogen, the cell potential is assigned a negative sign. Applying this standard:

Cell Notation Revisited In Lesson 1, you were informed that in cell notation, the anode is written on the left and the cathode is written on the right. These are the cell notations for zinc and copper combined with the hydrogen half cell.  Zn|Zn2+||H+|H2|Pt  Pt|H2|H+||Cu2+|Cu Building a Table of Standard Reduction Potentials The Standard Reduction Potentials of Half-Cells Table was constructed by building electrochemical cells using the standard hydrogen electrode and measuring cell potentials. The species are listed based on their standard reduction potentials, . The E values represent cell potentials for species at SATP conditions. The electrolytes in each half-cell have a molar concentration of 1.0 mol/L.  A positive E for a half-cell indicates that electrons flow from the hydrogen half-cell.  A negative E for a half-cell indicates that electron flow from it to the hydrogen half-cell. At this point you may have another question. Why is a table of standard reduction potentials useful? Calculating Standard Cell Potentials By knowing the potentials of half-cells, you can calculate the electrical potential of a cell instead of having to build it first. Your MHR textbook describes two methods for calculating standard cell potentials. In order to calculate the standard cell potential: 1. Identify the anode and the cathode and write the half-reactions. 2. Look-up the standard reduction potentials of the anode and cathode. 3. Substitute the values into this formula:

If the standard cell potential (E) is a positive value, the redox reaction is said to be spontaneous. One last important point: a cell potential is a function of the species involved, not their mole amounts. If you are given a balanced redox equation to work with, ignore the coefficients when processing the E values. Sample Exercise 1 Calculate the standard cell potential for an electrochemical cell in which this reaction takes place:

Answer Step 1: Write the half reactions. 

Zinc is oxidized:

; zinc metal is the anode. ; cadmium is the cathode.

Step 2: Look-up the standard reduction potentials. Ezinc = -0.76 V Ecadmium = -0.40 V Step 3: Subtract the reduction potential of the anode from the reduction potential of the cathode.

Assuming standard conditions, the cell potential is 0.36 V. Determining Spontaneity A table of reduction potentials can be used to determine whether an equation for a redox reaction represents a spontaneous reaction. Sample Exercise 2 Determine whether this equation, as written, represents a spontaneous redox reaction.

Answer Step 1: Write the half reactions. As written, the equation shows that zinc is oxidized and chromium is reduced. Zinc is the anode and chromium is the cathode.

Step 2: Look-up the standard reduction potentials. Ezinc = -0.76 V Echromium = -0.91 V Step 3: Calculate the cell potential.

The sign of the cell potential is negative. Therefore the reaction as written is not spontaneous. It requires the input of electricity. The phrase "as written" is critical to the classification of redox reactions as spontaneous or nonspontaneous. All electrochemical reactions are spontaneous; however, when communicated in the form of a redox reaction equation, some redox reactions are not spontaneous - they do not proceed as written. The reverse redox reaction is the spontaneous reaction in these cases. The classic example is the zinc/copper cell. The oxidation of zinc is spontaneous, but the oxidation of copper is not.

Electrolytic Cells When two different half-cells are joined together, a galvanic cell is created: electrons flow from the anode to the cathode, anions flow from the reduction half-cell to the oxidation half-cell, and cations flow in the opposite direction (i.e. towards the cathode). The redox reaction in a galvanic cell is spontaneous - it proceeds as soon as the circuit is completed with wire or a salt bridge.

Oxidation:

Reduction:

Redox: The reverse redox reaction for a galvanic cell is not spontaneous - it does not proceed as written. Can a non-spontaneous redox reaction be made to proceed? If so, how? What can cause the electrons to flow in the opposite direction? Roll your mouse over the diagram above. Any redox reaction occurring in electrochemical cell can be reversed by adding a power source to the circuit. The voltage supplied by the power source must exceed the cell voltage of the electrochemical cell. This idea is known as overvoltage. A cell that requires the input of electrical energy to drive a non-spontaneous redox reaction is called an electrolytic cell. How much voltage should be supplied to drive the non-spontaneous redox reaction? Let's compare the galvanic cell with the electrolytic cell?

Do you notice any similarities? What are the differences? The key differences between a galvanic cell and an electrolytic cell are as follows.  The charges of the anodes and cathodes differ. In a galvanic cell, the anode is negative relative to the cathode. In an electrolytic cell, the anode is positive relative to the cathode.  There is an external power source in the circuit for an electrolytic cell.  The voltage supplied to the electrolytic cell is greater than the voltage potential of the corresponding galvanic cell. The use of electricity to split a chemical species is called electrolysis. Finding the oxidation and reduction half-reactions for the electrolytic cell involves the same skill used to find them for a galvanic cell. In the electrolytic cell, nickel is oxidized and zinc is reduced. Oxidation:

Reduction: The sum of the half-reactions is the redox reaction equation.

Applications of Electrolysis Unlike the galvanic cell, an electrolytic cell does not have to contain a salt bridge. In many cases, the redox reaction that occurs in a cell can only occur if external voltage is supplied, so the halfreactions need not be separated. A good example of this is the refining of pure copper from copper blister Electrolysis now has many significant applications including:  isolation of metals like aluminum,  purification of metals like nickel and copper,  plating of metals like chromium, silver, and gold. Sir Humphrey Davy was the pioneer of electrolysis. Using this technique, he discovered sodium, potassium, calcium and barium, and was the first person to isolate a number of other elements including chlorine. Davy's discovery of sodium involved melting sodium chloride and passing electricity through the liquid. Sodium metal formed at the cathode and chlorine gas formed at the anode. This process is now carried out industrially in Down's Cell.

Faraday's Law Michael Faraday was the preeminent figure in the study of electrochemistry. He began his science career as an assistant to Sir Humphrey Davy and went on to make numerous scientific contributions of his own including discovery of the laws of electrochemistry and invention of the electric motor. Faraday proposed this idea: the amount of a substance consumed or produced in an electrolysis reaction is directly proportional to the quantity of electricity that flows through the circuit. This idea is now known as Faraday's Law. This law may be more easily illustrated using a half-reaction equation:

If one mole of aluminum is oxidized in a redox reaction, three moles of electrons will pass through a circuit. In this lesson, you will apply your stoichiometry skills; however, instead of dealing only with familiar quantities like mass and moles, you will be required to use or determine quantities of electrical charge too. Therefore it is important that you have some knowledge of key terms concerning electricity. Some Key Terms Concerning Electricity A good starting point for this lesson is a review of (or for some of you an introduction to) key terms about electricity.  Electric current is the flow of electrons through an external circuit.  The unit of electric current is the Ampere (A).  Electric charge is a function of electric current and time.  The unit of electric charge is the Coulomb (C).  One ampere is the electric charge that passes through a circuit in one second. The formula that expresses this relationship is: where I represents a current in Amperes, Q represents a charge in Coulombs, and t represents time in seconds. Electron Stoichiometry Since electrons are particles of matter, they can be counted. In chemistry, atoms, ions, molecules and of course, subatomic particles are so small, that counting them one at a time is impractical, not to mention impossible. In chemistry, amounts are "counted" in moles. Like anything in a balanced chemical equation, electrons can be counted in moles. For example, in this half reaction:

three moles of electrons are lost as one mole of aluminum atoms are oxidized. "Electron" stoichiometry involves conversion of electric charge to or from moles. The conversion factor is Faraday's constant (F), 96,500 C/mol, or 96,500 C of electric charge per mole of electrons. Faraday's constant is to electrons what molar volume is gases. Electric Charge to Mass Stoichiometry Think back to your first experience with mass to mass stoichiometry problems. These problems involved writing a balanced chemical equation, converting mass to moles, applying a mole ratio, and converting moles to mass. The steps in electrolysis stoichiometry are basically the same, except that instead of converting mass to moles, electric charge is converted to moles. Sample Exercise 1 Zinc metal is plated onto an iron nail by passing electricity through an external circuit. A current of 3.75 A is applied for a period of 2.0 hours. Calculate the mass of zinc that can be plated during this time. Answer Step 1: Write the reduction half-reaction equation.

Step 2: Convert time to seconds.

Step 3: Calculate the charge.

Step 4: Convert charge to moles.

Step 5: Apply mole ratio.

Step 6: Convert moles to mass.

The mass of zinc that can be plated onto the iron nail is 9.1 g. There are two other things to solve for in these stoichiometry problems. You may be asked to solve for the current or the time needed to carry out a certain electrolysis reaction. Sample Exercise 2 Zinc metal is plated onto an iron nail by passing electricity through an external circuit. Calculate the current required to plate 25 g of zinc onto an iron spike during a 1.5 hour period. Answer Step 1: Write the reduction half-reaction equation.

Step 2: Convert time to seconds.

Step 3: Convert mass to moles.

Step 4: Apply the mole ratio.

Step 5: Calculate the charge.

Step 6: Calculate the current.

The current required to plate 25 g of zinc in a 1.5 hour period is 14 A. Sample Exercise 3 Zinc metal is plated onto an iron nail by passing electricity through an external circuit. Calculate the time in seconds that a 6.0 A current should be applied to plate 25 g of zinc onto an iron spike.

Answer Step 1: Write the reduction half-reaction equation.

Step 2: Convert mass to moles.

Step 3: Apply the mole ratio.

Step 4: Calculate the charge.

Step 5: Calculate the time.

The time required to plate 25 g of zinc using a current of 6.0 A is 13,000 s or 3 hours and 37 minutes.

Energy Production and Electrochemical Cells Batteries are all over the place -- in our cars, our PCs, laptops, portable MP3 players and cell phones. A battery is essentially a can full of chemicals that produce electrons. Chemical reactions that produce electrons are called electrochemical reactions. In this article, you'll learn all about batteries -- from the basic concept at work to the actual chemistry going on inside a battery to what the future holds for batteries and possible power sources that could replace them! Battery Basics If you look at any battery, you'll notice that it has two terminals. One terminal is marked (+), or positive, while the other is marked (-), or negative. In an AA, C or D cell (normal flashlight batteries), the ends of the battery are the terminals. In a large car battery, there are two heavy lead posts that act as the terminals. Electrons collect on the negative terminal of the battery. If you connect a wire between the negative and positive terminals, the electrons will flow from the negative to the positive terminal as fast as they can (and wear out the battery very quickly -- this also tends to be dangerous, especially with large batteries, so it is not something you want to be doing). Normally, you connect some type of load to the battery using the wire. The load might be something like a light bulb, a motor or an electronic circuit like a radio. Inside the battery itself, a chemical reaction produces the electrons. The speed of electron production by this chemical reaction (the battery's internal resistance) controls how many electrons can flow between the terminals. Electrons flow from the battery into a wire, and must travel from the negative to the positive terminal for the chemical reaction to take place. That is why a battery can sit on a shelf for a year and still have plenty of power -- unless electrons are flowing from the negative to the positive terminal, the chemical reaction does not take place. Once you connect a wire, the reaction starts. Battery Chemistry: Voltaic Pile The first battery was created by Alessandro Volta in 1800. To create his battery, he made a stack by alternating layers of zinc, blotting paper soaked in salt water, and silver, like this:

This arrangement was known as a voltaic pile. The top and bottom layers of the pile must be different metals, as shown. If you attach a wire to the top and bottom of the pile, you can measure a voltage and a current from the pile. The pile can be stacked as high as you like, and each layer will increase the voltage by a fixed amount. Battery Chemistry: Daniell Cell In the 1800s, before the invention of the electrical generator (the generator was not invented and perfected until the 1870s), the Daniell cell (which is also known by three other names -- the "Crowfoot cell" because of the typical shape of the zinc electrode, the "gravity cell" because gravity keeps the two sulfates separated, and a "wet cell," as opposed to the modern "dry cell," because it uses liquids for the electrolytes), was extremely common for operating telegraphs and doorbells. The Daniell cell is a wet cell consisting of copper and zinc plates and copper and zinc sulphates. To make the Daniell cell, the copper plate is placed at the bottom of a glass jar. Copper sulfate solution is poured over the plate to half-fill the jar. Then a zinc plate is hung in the jar as shown and a zinc sulfate solution poured very carefully into the jar. Copper sulfate is denser than zinc sulfate, so the zinc sulfate "floats" on top of the copper sulfate. Obviously, this arrangement does not work very well in a flashlight, but it works fine for stationary applications. If you have access to zinc sulfate and copper sulfate, you can try making your own Daniell cell. Battery Reactions Probably the simplest battery you can create is called a zinc/carbon battery. By understanding the chemical reaction going on inside this battery, you can understand how batteries work in general. Imagine that you have a jar of sulfuric acid (H2SO4). Stick a zinc rod in it, and the acid will immediately start to eat away at the zinc. You will see hydrogen gas bubbles forming on the zinc, and the rod and acid will start to heat up. Here's what is happening: The acid molecules break up into three ions: two H+ ions and one SO4-- ion. The zinc atoms on the surface of the zinc rod lose two electrons (2e-) to become Zn++ ions. The Zn++ ions combine with the SO4-- ion to create ZnSO4, which dissolves in the acid. The electrons from the zinc atoms combine with the hydrogen ions in the acid to create H2 molecules (hydrogen gas). We see the hydrogen gas as bubbles forming on the zinc rod. If you now stick a carbon rod in the acid, the acid does nothing to it. But if you connect a wire between the zinc rod and the carbon rod, two things change: The electrons flow through the wire and combine with hydrogen on the carbon rod, so hydrogen gas begins bubbling off the carbon rod. There is less heat. You can power a light bulb or similar load using the electrons flowing through the wire, and you can measure a voltage and current in the wire. Some of the heat energy is turned into electron motion. The electrons go to the trouble to move to the carbon rod because they find it easier to combine with hydrogen there. There is a characteristic voltage in the cell of 0.76 volts. Eventually, the zinc rod dissolves completely or the hydrogen ions in the acid get used up and the battery "dies."

Silver-zinc battery - This is used in aeronautical applications because the power-to-weight ratio is good. Metal-chloride battery - This is used in electric vehicles. Have you ever looked inside a normal 9-volt battery? It contains six, very small batteries producing 1.5 volts each in a serial arrangement!