Dr. Namphol Sinkaset Chem 201: General Chemistry II Ch. 19: Electrochemistry Dr. Namphol Sinkaset Chem 201: General Chemistry II

I. Chapter Outline Introduction Balancing Redox Reactions Galvanic Cells Standard Reduction Potentials E°cell, ΔG°, and K Batteries Electrolysis

I. Introduction Reduction-oxidation (redox) reactions are a huge branch of chemistry. Basically, they are reactions in which electrons are in motion. Electrons in motion = electricity. Concepts in this chapter can be used to explain fuel cells, batteries, electroplating, etc.

I. Redox Reactions Recall that reduction and oxidation are coupled processes. Reduction is the gain of electrons. Oxidation is the loss of electrons. We identify what is oxidized and what is reduced by examining oxidation numbers.

I. Assigning Oxidation Numbers Atoms in elemental form have O.N. = 0. Charge on a monatomic ion equals its O.N. The sum of all O.N. must equal the total charge. For Group 1, O.N. = +1. For Group 2, O.N. = +2. For H, O.N. = +1 w/ nonmetals, -1 w/ metals and B. For F, O.N. = -1. For O, O.N. = -1 in peroxides and -2 in all others. For Group 17, typically O.N. = -1.

I. Sample Redox Reaction Ca(s) is oxidized; it is the reducing agent. H2O(l) is reduced; it is the oxidizing agent.

II. Balancing Redox Reactions Balancing a redox reaction is more involved because mass and charge must both be balanced. A common way to balance redox reactions is the “half-cell method.” In this method, we split the reaction into 2 half reactions: oxidation and reduction. Each half is balanced separately, and then the two are recombined.

II. The Half-Cell Method Identify what is being oxidized and reduced. Write each half-cell reaction. Balance each half-cell w/ respect to mass: Balance elements other than O and H. Balance O by adding H2O. Balance H by adding H+. Balance each half-cell w/ respect to charge by adding e-’s. Make the # of e-’s in each half-cell equal by multiplying by an appropriate factor. Add the two half-cells together. Add required # of OH- to each side and simplify if redox reaction takes place in basic solution.

II. Sample Problem Balance the following reaction in acidic solution: ClO-(aq) + Cr(OH)4-(aq)  CrO42-(aq) + Cl-(aq).

II. Sample Problem Balance the following reaction in basic solution: MnO4-(aq) + Br-(aq)  MnO2(s) + BrO3-(aq)

II. Inspection Method A different way to balance redox reactions when you are given all reactants and products already. e.g. Balance the following reaction. H2SO4 + HCl  H2S + Cl2 + H2O

II. The Inspection Method Assign oxidation states to all atoms. Identify species in which oxidation states change and write them as half reactions. Balance the atoms that are changing oxidation state in each half reaction. Identify the number of e-’s involved in each half reaction. Multiply the half reaction with the fewer number of e-’s so that the e-’s in each are equal. Without changing these coefficients, balance the rest of the reaction by inspection.

II. Sample Problem 19.1 Balance the following reactions. Cu(s) + HNO3(aq)  Cu(NO3)2(aq) + NO(g) + H2O(l) HNO3(aq) + H2S(aq)  NO(g)  + S(s)  +  H2O(l) ClO4- + I- + H+  I2 + Cl- + H2O H2SO4 + HCl  H2S + Cl2 + H2O

III. Generating Electricity As stated earlier, electrical current is simply charge in motion. Since an e- is charged, movement of electrons generates electrical current. Some redox reaction occur spontaneously. e.g. Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)

III. Zn/Cu2+ Redox Zn(s) is oxidized to Zn2+ which goes into solution. Cu2+ is reduced to Cu(s) which “plates out.” 2 e-’s are transferred from Zn to Cu2+.

III. Making e-’s Travel In this arrangement, we don’t really get electrons flowing, more of a transfer. If we separate the components, we can force electrons to travel externally, and even do work for us. This kind of setup is called an electrochemical cell.

III. Electrochemical Cell

III. Electrochemical Cells Voltaic (or galvanic) cells produce electrical current from a spontaneous redox reaction. Important aspects: Each part is called a half-cell. Oxidation occurs at the anode. Electrons flow away from the anode. Reduction occurs at the cathode. Electrons come to the cathode. A salt bridge is necessary to prevent charge build up and complete the circuit.

III. Current and Potential Two important aspects about electrons in motion are current and potential. Electrical current is measured in amperes (A), which has units of coulombs (measure of charge) per second, C/s. 1 A = 1 C/s. The current is driven by a potential energy difference called the potential difference. Potential difference is a measure of the difference in PE per unit of charge. The SI unit is the volt (V). 1 V = 1 J/C.

III. The River Analogy

III. Cell Potentials The potential difference is the force that drives the movement of e-’s, so it’s also called the electromotive force (emf). In a voltaic cell, the potential difference between the cathode and anode is called the cell potential (Ecell) or cell emf.

III. Standard Cell Potentials If the cell is under standard conditions, the cell potential is the standard cell potential (E°cell) or standard emf. For the Zn(s)/Cu2+(aq) cell, E°cell = 1.10 V. The cell potential is a measure of the tendency of the redox reaction to occur spontaneously.

III. Electrochemical Cell Notation Instead of drawing an electrochemical cell, line notation can be used. The Zn(s)/Cu2+(aq) can be represented as Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s) Oxidation is always written first, followed by the reduction. Double line represents the salt bridge. Different phases are separated by single lines. Multiple substances in solution are separated by commas.

III. Sample Cell Notation Fe(s)|Fe2+(aq)||MnO4-(aq), H+(aq), Mn2+(aq)|Pt(s)

III. Sample Problem 19.2 Draw and completely label the electrochemical cell represented by the line notation shown below. Additionally, write the overall balanced equation for the redox reaction. Sn(s)|Sn2+(aq)||NO(g)|NO3-(aq), H+(aq)|Pt(s)

IV. Calculating Cell Potentials In an electrochemical cell, we can think of each half-cell as having its own potential. Thus, E°cell is a sum of both half-cell potentials. The half-cell with the higher potential will occur in the forward direction, forcing the other half-cell to occur in the reverse direction.

IV. Half-Cell Potentials We can’t create a cell that has only reduction or only oxidation; thus, we can’t measure an absolute value of a half-cell. We have to assign a particular half-cell a value of 0.00 V and measure everything else relative to that.

IV. Standard Hydrogen Electrode The standard hydrogen electrode (SHE) is normally assigned a potential of 0.00 V. We attach different half-cells to SHE, watch what happens (red or ox), and whatever potential comes out is assigned to the half-cell.

IV. Measuring a Standard Reduction Potential

IV. Standard Reduction Potentials

IV. Important Points The reduction potential of SHE is 0.00 V. Half-cells that undergo reduction when attached to SHE have a positive reduction potential. Half-cells that undergo oxidation when attached to SHE have a negative reduction potential. Substances at the top have strong tendency to be reduced; they are strong oxidizing agents.

IV. Important Points Substances at the bottom have strong tendency to be oxidized; they are strong reducing agents. E°ox = -E°red For any cell, E°cell = E°ox + E°red. Never multiply half-cell potentials by coefficients used to balance redox reactions.

IV. Sample Problem 19.3 Calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 25 °C. 3Pb2+(aq) + 2Cr(s)  3Pb(s) + 2Cr3+(aq)

IV. Predicting Spontaneous Redox Reactions Positive cell potentials are spontaneous. We can identify the half-cells in a reaction, look up their potentials, and sum to find overall cell potentials. Thus spontaneity can be determined for any redox reaction.

IV. Sample Problem 19.4 Will the following redox reactions be spontaneous under standard conditions? Zn(s) + Ni2+(aq)  Zn2+(aq) + Ni(s) Zn(s) + Ca2+(aq)  Zn2+(aq) + Ca(s)

IV. Sample Problem 19.5 Determine whether the following metals dissolve in HCl(aq), HNO3(aq), both, or neither. Fe(s) Au(s) Ag(s)

V. Everything’s Related Positive E°cell means spontaneous. Negative ΔG° means spontaneous. K > 0 means spontaneous. Thus, all of these must be related somehow.

V. Potential and Work Potential difference can be expressed as a function of work. We will use this new view to derive the relationship between E°cell and ΔG°.

V. Eqn. Relating E°cell and ΔG°

V. Sample Problem 19.6 Using tabulated half-cell potentials, calculate ΔG° for the reaction 2Na(s) + 2H2O(l)  H2(g) + 2OH-(aq) + 2Na+(aq).

V. Potential and Equilibrium Constants Using our new equation that relates standard cell potential to standard free energy, we can derive an equation between E°cell and K. We start with the equation that relates ΔG° to K.

V. Eqn. Relating E°cell and K

V. Sample Problem 19.7 Use tabulated half-cell potentials to calculate the equilibrium constant for the two electron oxidation of copper metal by H+.

V. Relating ΔG°, K, and E°cell With the last equation derived, we summarize how we can convert between ΔG°, K, and E°cell.

V. Nonstandard Potentials Just like any other system, electrochemical cells may not be under standard conditions. Why is the potential higher?*

V. Calculating Nonstandard Cell Potentials Although we can qualitatively predict whether Ecell is higher or lower than E°cell, we’d like to calculate an exact value. We can derive an equation starting with the nonstandard ΔG equation.

V. Deriving the Nernst Equation

V. Nernst Eqn. Generalizations Under standard conditions, log Q = log 1 = 0, so Ecell will equal E°cell. If Q < 1, there are more reactants than products, so redox reaction shifts right; Ecell will be > than E°cell. If Q > 1, there are more products than reactants, so redox reaction shifts left; Ecell will be < than E°cell. If Q = K, then Ecell = 0.

V. Sample Problem 19.8 Calculate the cell potential for the electrochemical cell represented by Ni(s)|Ni2+(aq, 2.0 M)||VO2+(aq, 0.010 M), H+(aq, 1.0 M), VO2+(aq, 2.0 M)|Pt(s).

V. Concentration Cells

V. Concentration Cells If the two half-cells are the same, there is no reason to reduce something on one side and oxidize the same thing on the other side. However, if there are [ ] differences, there is a push to get to equilibrium. Electrons flow in order to increase the [ ] of the dilute cell and decrease the [ ] of the concentrated cell.

VI. Using Electrochemistry If constructed correctly, electrochemical cells can be used to store and deliver electricity. We briefly look at dry-cell batteries, lithium ion batteries, and fuel cells.

VI. Dry-cell Batteries Called dry-cell because there’s very little water. Voltage derived from the oxidation of Zn(s) and the reduction of MnO2(s). In acidic battery, MnO2(s) reduced to Mn2O3(s). In alkaline battery, MnO2(s) reduced to MnO(OH)(s).

VI. Lithium Ion Batteries This rechargeable battery works a bit differently. Motion of Li+ from anode to cathode causes e-’s to flow externally and reduce the transition metal. Battery is recharged by oxidizing the transition metal in the cathode.

VI. Fuel Cells

VII. Nonspontaneous Redox Reactions Nonspontaneous reactions can be forced to occur by inputting energy. Nonspontaneous redox reactions can be forced to occur by inputting energy in the form of electrical current. This type of electrochemical cell is called an electrolytic cell.

VII. Voltaic vs. Electrolytic Cells

VII. Predicting Products In an aqueous electrolytic cell, it’s possible that H2O can be reduced or oxidized. To predict products, must consider potentials of all processes that could occur. Processes that are easiest (least negative or most positive half-cell potential) will occur.

VII. Overvoltage There is one problem in predicting electrolysis products. Some half-cell reactions don’t occur at their expected voltage potentials! e.g. 2H2O(l)  O2(g) + 4H+(aq) + 4e- has Eox = -0.82 V when [H+] = 1 x 10-7 M. However, kinetic factors require a voltage of 1.40 V for this to occur.

VII. Example Electrolysis What happens when a solution of NaI undergoes electrolysis? Oxidation 2I-(aq)  I2(aq) + 2e- E°ox = -0.54 V 2H2O(l)  O2(g) + 4H+(aq) + 4e- E°ox = -1.4 V Reduction Na+(aq) + e-  Na(s) E°red = -2.71 V 2H2O(l) + 2e-  H2(g) + 2OH-(aq) E°red = -0.41 V

VII. Electrolysis of NaI Solution

VII. Electrolysis Stoichiometry The # of electrons in any redox reaction can be used as a stoichiometric ratio. If we know how long an electrolysis takes place, and the magnitude of current that flowed, we can do stoichiometric calculations. Important relationships: 1 A = 1 C/s F = 96,485 C/mole e-

VII. Sample Problem 19.9 Copper can be plated out of a solution containing Cu2+ according to the half-reaction: Cu2+(aq) + 2e-  Cu(s). How long (in minutes) will it take to plate 10.0 g of copper using a current of 2.0 A?