Redox and Electrochemistry

Redox and Electrochemistry Oxidation: Historically means the combination of a substance with oxygen 2Mg + O 2 2MgO Oxygen is called the oxidizing agent, and magnesium is the substance oxidized. Reduction: Historically, reduction is associated with the removal of oxygen from a substance. CuO + H 2 Cu + H 2 O Hydrogen is called the reducing agent, and copper (II) oxide is the substance that is reduced. Oxidation and reduction arises out of the competition for electrons in a chemical reaction. To know what really happens during these processes, we need a method of assigning electrically charges to atoms and ions. Oxidation Numbers (Oxidation States) An oxidation number is the charge that an atom or ion has- or appears to have- when certain 1

rules are applied. (appears to have because covalent compounds do not have charges; we just treat them as if they do) Rules: 1. The oxidation number of a free element is zero. Free element = any element that is not combined with another element. Therefore, Na, He, O 2, O 3 all have the oxidation number zero. 2. The oxidation number of a simple ion is its charge. For example, the oxidation number of Cl - is 1, the oxidation number of Al +3 is +3. 3. The metals in Groups 1 and 2 have oxidation numbers of +1 and +2 respectively 4. Hydrogen in combination usually has an oxidation number of +1. An exception to this rule are the metal hydrides (such as NaH), in which hydrogen has the oxidation number 1. 5. Oxygen in combination usually has an oxidation number of 2. Exceptions to this rule include peroxides (such as H 2 O 2 ), in which the oxidation number of oxygen is 1, and oxygen-fluorine compounds, in which the oxygen number is positive. 2

6. In a molecular or ionic compound, the sum of the oxidation number totals must add to zero since these compounds are electrically neutral. 7. In a polyatomic ion, the sum of the oxidation number totals must add to the charge on the ion. Questions Find the oxidation number in each of the following: a. S in H 2 SO 3 b. Cr in Na 2 CrO 7 c. Fe and Cl in FeCl 3 d. 3- P in PO 4 e. O in OF 2 Formal Definitions of Oxidation and Reduction 2Mg 0 + O 0 2 2Mg 2+ O 2- Magnesium increased oxidation number from zero to +2 Magnesium is oxidized Oxidation = a loss of electrons 2Mg 0 2Mg +2 + 4 e - (oxidation halfreaction) 3

Oxygen decreases oxidation number from zero to 2 Oxygen is reduced Reduction = a gain of electrons O 0 2 + 4e - 2O 2- Redox reactions = reactions involving oxidation and reduction LEO the lion says GER Oxidizing agent = the substance that is reduced (causes oxidation) Reducing agent = the substance that is oxidized (causes reduction) Redox Equations Is this (unbalanced) equation: NH 3 + O 2 NO + H 2 O a redox equation? To answer this question, we rewrite the equation and include the oxidation number for each element. Then, look for changes in the oxidation numbers: an increase means oxidation and a decrease means reduction. N 3- H + 3 + O 0 2 N 2+ O 2- + H + 2O 2- Nitrogen changes from 3 to +2; its oxidation number increases, therefore it has been oxidized. 4

Oxygen changes its oxidation number from zero to 2; its oxidation number decreases, and therefore it has been reduced. Half- Reactions Redox equations can be split into half reactions by identifying the elements that are oxidized and reduced. NH 3 + O 2 NO + H 2 O N 3- H + 3 + O 0 2 N 2+ O 2- + H + 2O 2- N 3- N 2+ + 5e - (oxidation half reaction) O 0 + 2 e - O 2- (reduction half reaction)* unbalanced* Question: Indicate whether each of the following equations is a redox equation. If it is a redox equation, write the half-reactions representing oxidation and reduction. a. Fe + 2HCl FeCl 2 + H 2 b. HNO 3 + I 2 HIO 3 + NO 2 + H 2 O (unbalanced) c. HCl + NaOH H 2 O + NaCl Balancing Redox Equations by the Half-Reaction Method Redox equations may be very difficult to balance simply by inspection. 5

Cu + HNO 3 Cu(NO 3 ) 2 + NO + H 2 O Redox equations can be balanced, however, by identifying the elements that are oxidized and reduced and then writing their half reactions. Steps: 1. Rewrite the equation with its oxidation numbers Cu 0 + H + N 5+ O 3 2- Cu 2+ (N 5+ O 3 2- ) 2 + N 2+ O 2- + H 2 + O 2-2. Identify the elements that are oxidized and reduced. Cu is oxidized N is reduced 3. Write the oxidation and reduction half reactions Cu 0 Cu 2+ + 2e - N 5+ + 3e - N 2+ 4. Balance the half reactions. (This step is necessary because electric charge must be conserved.) Multiply each half reaction so that the number of electron lost by the oxidized element is equal to the number of electrons gained by the reduced element. 3 (Cu 0 Cu 2+ + 2e - ) = 3Cu 0 3Cu 2+ + 6e - 2 (N 5+ + 3e - N 2+ ) = 2N 5+ + 6e - 2N 2+ 6

5. Add the two balanced half reactions together, eliminating the electrons. The result is the balanced skeleton redox equation. 3Cu 0 + 2N 5+ 3Cu 2+ 2N 2+ 6. Insert the coefficients from the skeleton back into the original equation by matching each element and its oxidation number. There is one exception: Do not insert the coefficient of any item that appears in more than one place in the equation. 3Cu 0 + _H + N 5+ O 2-3 3Cu 2+ (N 5+ O 2-3) 2 + 2N 2+ O 2- + _H + 2O 2- since N 5+ appears twice, do not insert the coefficient for it. 3Cu + HNO 3 3Cu(NO 3 ) 2 + 2NO +H 2 O 7. Balance the rest of the equation (the nonredox part) by inspection. 3Cu + 8HNO 3 3Cu(NO 3 ) 2 + 2NO +4H 2 O Question: Balance the following redox equations by the half reaction method: a. Cu + HNO 3 Cu(NO 3 ) 2 + NO 2 + H 2 O b. HNO 3 + I 2 HIO 3 + NO 2 + H 2 O 7

Steps for Balancing Redox Equations in Acidic Solution Since the solution is acidic, H+ and H 2 O can be added either to the reactants or the products to balance hydrogen or oxygen 1. Divide the equation into two complete half reactions, one for oxidation and the other for reduction 2. Balance each half reaction. a.first, balance the elements other than H and O. b. Next, balance the O atoms by adding H 2 O c.then, balance the H atoms by adding H + d. Finally, balance the charge by adding e - to the side with the greater overall positive charge. 3. Multiply each half reaction by an integer so that the number of electrons lost in one half reaction equals the number gained in the other. 4. Add the two half reactions and simplify where possible by canceling species appearing on both sides of the equation 5. Check the equation to make sure that there are the same number of atoms of 8

each kind and the same total charge on both sides. Questions: Complete and balance the following oxidationreduction equations using the method of half reactions. Both reactions occur in acidic solution a. Cu(s) + NO 3 - (aq) Cu 2+ (aq) + NO 2 (g) b. Mn 2+ (aq) + NaBiO 3 (s) Bi 3+ (aq) + MnO 4 - (aq) Balancing Equations for Reactions Occurring in Basic Solution OH - and H 2 O are used instead of H + and H 2 O The half reactions can be balanced initially as if they occurred in acidic solution. The H + ions can then be neutralized by adding an equal number of OH - ions to both sides of the equation. Question: Complete and balance the following equations for oxidation reduction reactions that occur in basic solution: a. NO 2 - (aq) + Al(s) NH 3 (aq) + Al(OH) 4 (aq) 9

b. Cr(OH) 3 (s) + ClO - (aq) CrO 4 2- (aq) + Cl 2 (g) Reduction Potentials Every half reaction has a potential, or voltage, associated with it. A table of half reactions can be used to determine the tendency for a substance to be oxidized or reduced. Insert table of Reduction Potentials AP review The table provides reduction potentials. To determine oxidation potentials, reverse the reaction and change the sign on the voltage given. Reduction potential for Li + Li + + e - Li E o = -3.05 V not a favorable reaction 10

Oxidation potential for Li Li Li + + e - E o = 3.05 V favorable reaction Li has a large oxidation potential, making it likely to lose electrons. Li is a very strong reducing agent. Reduction potentials can be used to determine the potential of a redox reaction. If the potential for the redox reaction is positive, the reaction is favored and will be spontaneous. Add the potential for the oxidation half reaction to the potential for the reduction half reaction Never multiply the potential for a half reaction by a coefficient E 0 cell = E oxidation + E reduction Is the following reaction a spontaneous reaction? Sn(s) + 2Ag + Sn 2+ + 2Ag(s) Oxidation: Sn Sn 2+ + 2e - E o = 0.14 V Reduction: Ag + + e - Ag E 0 = 0.80 V E 0 cell = E oxidation + E reduction E o cell = 0.14 + 0.80 = 0.94 V Notice, the coefficient for silver does not enter into the equation for the voltage for the cell. 11

Question: Calculate E 0 for the redox reaction: Fe(s) + Pb 2+ (aq) Fe 2+ (aq) + Pb(s) Is the reaction spontaneous? Galvanic Cells (Voltaic Cells) In every example discussed thus far, there has been a transfer of electrons between the substance that was oxidized and the substance that was reduced. If we could direct these electrons through an external wire, we would have the source of an electric current. An electrochemical cell uses a spontaneous redox reaction to provide a source of electrical energy. It is designed so that the oxidation and reduction half reactions occur in separate half-cells that are connected to one another. Insert cell diagram in Let s Review Chemistry 12

Oxidation occurs in the left half-cell. The electrons that are released travel through the external wire and enter the Cu(s). Reduction occurs in the right half-cell. The metal strips are called electrodes. The electrode that participates in oxidation is called the anode; the electrode that participates in reduction, the cathode. The direction of the electron flow in an electrochemical cell is always from the anode (where oxidation occurs) to the cathode (where reduction occurs). AN OX RED CAT As the reaction progresses, there would be an electrical imbalance in the half-cells that would stop the electrochemical cell from functioning. However, the porous barrier (salt bridge*) allows the migration of ions between the half-cells, keeping them electrically neutral. If we connect the external wire to an external circuit, we will have a usable source of electrical energy; we have created a battery! The anode of the cell is the negative terminal because electrons are flowing out of it; the cathode is the positive terminal because electrons are flowing into it. 13

After a period of time, the redox reaction of the cell reaches equilibrium and the cell no longer operates. The battery is dead! *Salt Bridges We can build an electrochemical cell whose half cells are completely separated, but we must provide a device known as a salt bridge to allow for the flow of ions between the halfcells. As shown in the accompanying diagram, the salt bridge contains an electrolyte, such as KCl, which is dispersed throughout a gel, such as agar. The gel provides firmness but allows ions to flow through it. Insert salt bridge diagram Let s Review Chemistry 14

Since Zn 2+ (aq) is produced in the left halfcell, negative ions are required to keep the solution electrically neutral. Therefore, Cl - ions flow into this half-cell. At the same time Cu 2+ (aq) is being used up in the right half-cell, which requires positive ions to keep the solution electrically neutral. Therefore, K + ions flow into this half-cell. Nerst Equation Under standard conditions (STP and all concentrations are 1 M) the voltage of the cell is the same as the total voltage of the redox reaction. Under nonstandard conditions, the cell voltage can be computed using the Nerst equation. E cell = E 0 RT cell ln Q nf E cell = cell potential under nonstandard conditions (V) E 0 cell = cell potential under standard conditions (V) R = the gas constant, 8.31 (volt-coulomb) / (mol-k) T = absolute temperature (K) n = the number of moles of electrons exchanged in the reaction (mol) 15

F = Faraday s constant, 96,500 coulombs / mole Q = the reaction quotient (same as the equilibrium expression, but with initial concentrations instead of equilibrium concentrations) At 25 o C, the Nerst equation reduces to: E cell = E 0 cell 0.0592 log Q n As the concentration of the products of a redox reaction increases, the voltage decreases; and as the concentration of the reactants in a redox reaction increases, the voltage increases. Electrolytic Cells and Electrolysis Calculate E 0 for the redox reaction involving molten NaCl (l) 2NaCl 2Na(s) + Cl 2 (g) Half reactions: 2Na + + 2e - 2Na(s) (Na + is reduced) 2Cl - Cl 2 (g) + 2e - (Cl - is oxidized) Then, E 0 = E Cl + E Na = -1.36V + -2.71V = -4.07V 16

The negative sign indicates the reaction is not spontaneous It would be useful if we could force this reaction to occur: we could generate two elements that do not occur freely in nature. To accomplish this purpose, we construct a special type of cell. Electrolytic cells are used to force nonspontaneous redox reactions to occur. Unlike galvanic or voltaic cells, they do not generate electrical energy they use it. In practice, an electric current is passed through the substance inside the cell. The process is called electrolysis. Insert operation of an electrolytic cell diagram Let s Review Chemistry Electroplating In this process an electric current is used to deposit a layer of metal, such as silver, on the 17

object to be plated. The accompanying diagram illustrates how electroplating is accomplished. Insert Electroplating diagram Let s Review Chemistry The plating solution is a salt of the metal. A bar of metal is made the anode (positive electrode) and provides positive ions needed for plating. The material to be plated is made the cathode (negative electrode) and receives the positive ions which it reduces to the metallic layer of electroplate. 18

Additional Application of Redox and Electrochemistry Reduction of Metals Most metals do not occur freely in nature, but occur in oxidized state (i.e. as positive ions). A metal that belongs to Group 1 or 2 is recovered by electrolysis of its fused salt. Other metals are recovered by reduction of their ores; the method used depends on the activity of the metal and the nature of the ore. In the production of chromium metal, chromium (III) oxide is fairly stable and is treated with aluminum metal, which is a relatively strong reducing agent: 2Al + Cr 2 O 3 Al 2 O 3 + 2Cr Metals such as zinc and iron are extracted by the reduction of their oxides. Carbon, in the form of coke, or carbon monoxide is used as the reducing agent. ZnO + C + Heat energy Zn + CO Fe 2 O 3 + 3CO + Heat energy 2Fe + 3CO 2 19

Preventing the Corrosion of Metals Corrosion occurs when a metal is attacked slowly by elements in its environment. In many cases, the metal ceases to be useful. Corrosion is a redox reaction, and agents such as moisture may contribute to the process. Metals such as aluminum and zinc oxidize readily, but the oxide adheres to the metal s surface tightly producing a protective coating. Some metals, such as iron, form oxides that flake off the metal face allowing oxidation of the remaining metal surface to occur. Iron can be protected by plating it with a corrosionresistant metal such as chromium or nickel or coating it with paint, oil, or porcelain. Another protection technique is to coat the iron with zinc-a process known as galvanizing. If the coating is broken, exposing fresh iron and zinc, the more active zinc will be oxidized first and will produce a protective oxide coating. Commercial Batteries Lead Acid Battery Car and boat batteries The positive electrode (cathode) is PbO 2, the negative electrode (anode), Pb. A solution of sulfuric acid is the electrolyte for the redox 20

reaction- a reaction that lead changes its oxidation state. Pb + PbO 2 + 2H 2 SO 4 CHARGE DISCHARGE 2PbSO 4 + 2H 2 O Since the redox reaction is reversible, the battery may be recharged for further use. Nickel Oxide-Cadmium Battery NiO 2 is the positive electrode, Cd is the negative electrode. A solution of KOH is the electrolyte, and its concentration does not change. NiO 2 + Cd + 2H 2 O charge discharge Ni(OH) 2 + Cd(OH) 2 21