ELECTROCHEMISTRY OXIDATION REDUCTION REACTIONS

OXIDATION REDUCTION Reactions and Electrochemical Cells Voltaic Cells: Using Spontaneous Reactions to Generate Electrical Energy Cell Potential: Output of a Voltaic Cell Free Energy and Electrical Work Electrolytic Cells: Using Electrical Energy to Drive Nonspontaneous Reactions

OXIDATION STATES The oxidation state (oxidation number) is an indicator of the degree of oxidation of an atom in a chemical compound. The formal oxidation state is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic.

ASSIGNING OXIDATION STATES (IUPAC) the oxidation state of a free element (uncombined element) is zero for a simple (monatomic) ion, the oxidation state is equal to the net charge on the ion hydrogen has an oxidation state of 1 and oxygen has an oxidation state of − 2 when they are present in most compounds. (Exceptions hydrides and peroxides) the algebraic sum of oxidation states of all atoms in a neutral molecule must be zero, while in ions the algebraic sum of the oxidation states of the constituent atoms must be equal to the charge on the ion

Key Points About Redox Reactions o Oxidation (electron loss) always accompanies reduction (electron gain). o The oxidizing agent is reduced, and the reducing agent is oxidized. o The number of electrons gained by the oxidizing agent always equals the number lost by the reducing agent.

REDOX reactions - transfer of electrons between species. All the redox reactions have two parts: OxidationReduction

The Loss of Electrons is Oxidation. An element that loses electrons is said to be oxidized. The species in which that element is present in a reaction is called the reducing agent. The Gain of Electrons is Reduction. An element that gains electrons is said to be reduced. The species in which that element is present in a reaction is called the oxidizing agent.

A summary of terminology for oxidation-reduction (redox) reactions.

A SUMMARY OF REDOX TERMINOLOGY Zn loses electrons. Zn is the reducing agent and becomes oxidized. The oxidation number of Zn increases from 0 to +2. REDUCTION One reactant loses electrons. Reducing agent is oxidized. Oxidation number increases. Hydrogen ion gains electrons. Hydrogen ion is the oxidizing agent and becomes reduced. The oxidation number of H decreases from +1 to 0. Other reactant gains electrons. Oxidizing agent is reduced. Oxidation number decreases. Zn(s) + 2H + (aq) → Zn 2+ (aq) + H 2 (g)

HALF-REACTION METHOD FOR BALANCING REDOX REACTIONS Summary: This method divides the overall redox reaction into oxidation and reduction half-reactions. o Each reaction is balanced for mass (atoms) and charge. o One or both are multiplied by some integer to make the number of electrons gained and lost equal. o The half-reactions are then recombined to give the balanced redox equation.

Half-Reaction Method for Balancing Redox Reactions Advantages: o The separation of half-reactions reflects actual physical separations in electrochemical cells. o The half-reactions are easier to balance especially if they involve acid or base. o It is usually not necessary to assign oxidation numbers to those species not undergoing change.

HALF-REACTION METHOD FOR BALANCING REDOX REACTIONS 1.Divide the equation into two half reactions. 2.Balance each half reaction - balance all elements other than H and O - balance the number of oxygen by adding H 2 O as needed - balance the number of hydrogen by adding H + as needed - balance the charges by adding e - as needed 3.Multiply each half-reactions by integers such as the number of electrons lost equals the number of electrons gained. 4.Add the reactions cancelling species that appear on both sides. 5.Check

Balancing Redox Reactions in Acidic Solution Mn +2 + BiO 3 - → Bi +3 + MnO 4 -

SAMPLE PROBLEM: Balance the following equation: NaCN + H 2 O + KMnO 4 → NaCNO + MnO 2 + KOH CN - + MnO 4 - → CNO - + MnO 2 (s) (basic solution)

Figure 21.3 General characteristics of voltaic and electrolytic cells.

Figure 21.5 A voltaic cell based on the zinc-copper reaction.

Notation for a Voltaic Cell Components of anode compartment (oxidation half-cell) Components of cathode compartment (reduction half-cell) Phase of lower oxidation state Phase of higher oxidation state Phase of lower oxidation state Phase boundary between half-cells Examples:Zn( s ) | Zn 2+ ( aq ) || Cu 2+ ( aq ) | Cu ( s ) Zn( s ) Zn 2+ ( aq ) + 2e - Cu 2+ ( aq ) + 2e - Cu( s ) graphite | I - ( aq ) | I 2 ( s ) || H + ( aq ), MnO 4 - ( aq ) | Mn 2+ ( aq ) | graphite inert electrodes

Figure 21.6 A voltaic cell using inactive electrodes.

SAMPLE PROBLEM The oxidation-reduction reaction, Cr 2 O 7 2- (aq) + 14H + (aq) + 6I - (aq) → 2Cr 3+ (aq) + 3I 2 (s) + 7H 2 O(l) is spontaneous. Draw a diagram of the cell and indicate the reactions occuring at the electrodes (anode, cathode), the direction of electron migration, the direction of ion migration, the signs of the electrodes. Write the notation for the cell.

Describing a Voltaic Cell with Diagram and Notation Draw a diagram, show balanced equations, and write the notation for a voltaic cell that consists of one half-cell with a Cr bar in a Cr(NO 3 ) 3 solution, another half-cell with an Ag bar in an AgNO 3 solution, and a KNO 3 salt bridge. Measurement indicates that the Cr electrode is negative relative to the Ag electrode. Identify the oxidation and reduction reactions and write each half- reaction. Associate the (-)(Cr) pole with the anode (oxidation) and the (+) pole with the cathode (reduction).

Determining an unknown E o half-cell with the standard reference (hydrogen) electrode.

E o cell = E o cathode - E o anode.

Sample Problem 21.3Calculating an Unknown E o half-cell from E o cell A voltaic cell houses the reaction between aqueous bromine and zinc metal: Br 2 ( aq ) + Zn( s ) Zn 2+ ( aq ) + 2Br - ( aq ) E o cell = 1.83 V Calculate E o bromine, given E o zinc = V. The reaction is spontaneous as written since the E o cell is (+). Zinc is being oxidized and is the anode. Therefore, the E o bromine can be found using E o cell = E o cathode - E o anode.

Sample Problem 21.3Calculating an Unknown E o half-cell from E o cell PLAN: SOLUTION: The reaction is spontaneous as written since the E o cell is (+). Zinc is being oxidized and is the anode. Therefore, the E o bromine can be found using E o cell = E o cathode - E o anode. Anode: Zn( s ) Zn 2+ ( aq ) + 2e - E = V E o Zn as Zn 2+ ( aq ) + 2e - Zn( s ) is V E o cell = E o cathode - E o anode = 1.83 = E o bromine - (-0.76) E o bromine = (-0.76) = 1.07 V

Li+/Li Sn+4/Sn Br2/Br+1.07

Cd +2 /Cd Cu +2 /Cu Fe +2 /Fe Ni +2 /Ni Zn +2 /Zn -0.76

By convention, electrode potentials are written as reductions. When pairing two half-cells, you must reverse one reduction half-cell to produce an oxidation half-cell. Reverse the sign of the potential. The reduction half-cell potential and the oxidation half-cell potential are added to obtain the E o cell. When writing a spontaneous redox reaction, the left side (reactants) must contain the stronger oxidizing and reducing agents. Example:Zn( s ) + Cu 2+ ( aq ) Zn 2+ ( aq ) + Cu( s ) stronger reducing agent weaker oxidizing agent stronger oxidizing agent weaker reducing agent Writing Spontaneous Redox Reactions

Sample Problem 21.4 Writing Spontaneous Redox Reactions PROBLEM:(a) Combine the following three half-reactions into three balanced equations (A, B, and C) for spontaneous reactions, and calculate E o cell for each. E o = 0.96 V(1) NO 3 - ( aq ) + 4H + ( aq ) + 3e - NO( g ) + 2H 2 O( l ) E o = V(2) N 2 ( g ) + 5H + ( aq ) + 4e - N 2 H 5 + ( aq ) E o = 1.23 V(3) MnO 2 ( s ) + 4H + ( aq ) + 2e - Mn 2+ ( aq ) + 2H 2 O( l )

Relative Reactivities (Activities) of Metals 1. Metals that can displace H 2 from acid. 2. Metals that cannot displace H 2 from acid. 3. Metals that can displace H 2 from water. 4. Metals that can displace other metals from solution.

The activity series of the metals.

Free Energy and Electrical Work  G  -E cell -E cell = -w max charge charge = nF n = mol e - F = Faraday constant F = 1 V = 1 F =  G = w max = charge x (-E cell )  G = -nFE cell In the standard state -  G o = -nFE o cell  G o = - RT ln K E o cell = - ( ) ln K 96,485 C mol e x 10 4 J Vmol e - J C RT n Fn F

The interrelationship of  G 0, E 0 cell, and K. < 0 0 > 0 0 < 0 > 1 1 < 1  G o = -RT lnK

The Effect of Concentration on Cell Potential  G =  G o + RT ln Q -nF E cell = -nF E cell + RT ln Q E cell = E o cell -ln Q RT nF When Q [product], ln Q E o cell. When Q >1 and thus [reactant] 0, so E cell < E o cell. When Q = 1 and thus [reactant] = [product], ln Q = 0, so E cell = E o cell. E cell = E o cell - log Q n

Figure A concentration cell based on the Cu/Cu 2+ half-reaction.

Figure The laboratory measurement of pH.

Alkaline battery. Figure 21.14

Silver button battery. Figure 21.15

Lead-acid battery. Figure 21.16

Nickel-metal hydride battery. Figure 21.17

Lithium-ion battery. Figure 21.18

Figure Hydrogen fuel cell.

Figure The corrosion of iron.

Figure The effect of metal-metal contact on the corrosion of iron. Faster corrosion

Figure The use of sacrificial anodes to prevent iron corrosion.

ELECTROLYTIC CELL

ELECTROLYSIS OF MOLTEN SALTS

Predicting the Electrolysis Products of Aqueous Ionic Solutions What products form during electrolysis of aqueous solutions of the following salts: (a) KBr; Compare the potentials of the reacting ions with those of water, remembering to consider the 0.4 to 0.6 V overvoltage. The reduction half-reaction with the less negative potential, and the oxidation half- reaction with the less positive potential will occur at their respective electrodes. E o = V(a) K + ( aq ) + e - K( s ) E o = V2H 2 O( l ) + 2e - H 2 ( g ) + 2OH - ( aq ) The overvoltage would make the water reduction -0.8 to -1.0 V, but the reduction of K + is still a higher potential so H 2 ( g ) is produced at the cathode. The overvoltage would give the water half-cell more potential than Br -, so the Br - will be oxidized. Br 2 ( l ) forms at the anode. E o = 1.07 V2Br - ( aq ) Br 2 ( l ) + 2e - 2H 2 O( l ) O 2 ( g ) + 4H + ( aq ) + 4e - E o = 0.82 V

A summary diagram for the stoichiometry of electrolysis. Faraday’s Law of Electrolysis: The amount of substances that undergoes oxidation or reduction at each electrode during electrolysis is proportional to the amount of electricity that passes through the cell 1.0 Faraday = 96,500 C 1 ampere = 1 C/s

CORROSION PROTECTION 1.Plating the metal with a thin layer of less easily oxidized 2.Connecting a metal to a sacrificial anode (a more active metal – preferentially oxidized 3.Metal oxides to form naturally on the surface 4.Galvanizing (coating with zinc) 5.Protective coating, such as paint

Latimer Diagram Frost Diagram DIAGRAMMATIC REPRESENTATIONS OF REDUCTION POTENTIAL DATA Pourbaix Diagram Table of redox potentials are organized by the voltage of the process rather than by the chemical species involved. Difficult to use if one wants to understand the complex redox chemistry of the elements

Reminder: Thermodynamic data will predict which reactions ought to occur, but cannot determine whether they happen at an observable rate or not. Most of the redox reactions of inorganic compounds are rapid reactions, but there are many times when thermodynamics predicts more than one possible product, and where the actual product is selected by the rate of reaction.

Written with the most oxidized species on the left and the most reduced species on the right. Oxidation number decrease from left to right and the E 0 values are written above the line joining the species involved in the couple. LATIMER DIAGRAMS

Look at the Latimer diagram of nitrogen in acidic solution a bc de fgh

FROST DIAGRAM N 2

FROST DIAGRAMS - INTERPRETATION The lowest point represent the most thermodynamically stable form of the element A move downward the plot represent a thermodynamically favorable process A species at the to right of the diagram is oxidizing. Strengths relative to their position. The slope of the line drawn between two points divided by the number of electrons transferred is the E o for the half reaction. A positive slope means the reduction process is positive. Any state represented on a ‘convex’ point is thermodynamically unstable with respect to disproportionation Any state represented on a ‘concave’ point is thermodynamically stable with respect to disproportionation,

the lowest lying species corresponds to the most stable oxidation state of the element What do we really get from the Frost diagram?

The oxidizing agent - couple with more positive slope - more positive E The reducing agent - couple with less positive slope If the line has –ive slope- higher lying species – reducing agent If the line has +ive slope – higher lying species – oxidizing agent Oxidizing agent? Reducing agent?

Identify all the species that are unstable with respect to disproportionation

DISPROPORTIONATION…. ANOTHER EXAMPLE

Comproportionation reaction

Comproportionation is spontaneous if the intermediate species lies below the straight line joining the two reactant species.

NE 0 Identify species that are unstable to disproportionation and comproportionation

Disproportionation

Comproportionation In acidic solution… Mn and MnO2 Mn 2+ Rate of the reaction hindered insolubility? In basic solution… MnO2 and Mn(OH)2 Mn2O3

Consider the Latimer Diagrams for Pb and Tin in acid solution. A)Calculate values of nE for each oxidation state of lead and tin, and construct a comparative Frost diagram for these two elements in acid solution. B) Which of the six species you have plotted is the strongest oxidizing agent? Justify your answer. C) Which of the six species you have plotted is the strongest reducing agent? Justify your answer. D) What product(s) would form if PbO 2 was mixed with Sn 2+ in acidic solution? Write a balanced equation.

POURBAIX DIAGRAMS Graphical representations of thermodynamic and electrochemical equilibria between metal and water, indicating thermodynamically stable phases as a function of electrode potential and pH. -predicts the spontaneous direction of reactions. -estimates the composition of corrosion products. -predicts environmental changes that will prevent or reduce corrosion attack.

SAMPLE PROBLEM: Consider the Pourbaix diagram for manganese. (a)What form(s) will manganese take in lake and stream water, where the pH = 6-8 and E o = V?