Applications of Redox Your last chapter! I know, …… kinda sad. Electrochemistry Applications of Redox Your last chapter! I know, …… kinda sad.

Redox Review Oxidation reduction reactions involve a transfer of electrons. OIL- RIG Oxidation Involves Loss Reduction Involves Gain LEO-GER Lose Electrons Oxidation Gain Electrons Reduction

Applications Moving electrons is electric current. 8H++MnO4-+ 5Fe+2 +5e-  Mn+2 + 5Fe+3 +4H2O Helps to break the reactions into half reactions. 8H++MnO4-+5e-  Mn+2 +4H2O 5(Fe+2  Fe+3 + e- ) In the same mixture it happens without doing useful work. If separate, current can flow.

Connected this way the reaction starts Stops immediately because charge builds up. Fe2+ MnO4- H+

Galvanic Cell Salt Bridge allows current to flow H+ MnO4- Fe+2

Electricity travels in a complete circuit Instead of a salt bridge e- H+ MnO4- Fe+2

Porous Disk H+ MnO4- Fe+2

e- e- e- e- Anode Cathode e- e- Oxidation - supplies electrons Reduction - uses electrons

17.2 Cell Potential Oxidizing agent pulls the electron. Reducing agent pushes the electron. The push or pull (“driving force”) is called the cell potential Ecell Also called the electromotive force (emf) Unit is the volt(V) = 1 joule of work/coulomb of charge Measured with a voltmeter (BDVD)half-reactions

0.76 H2 in Cathode Anode H+ Cl- Zn+2 SO4-2 1 M ZnSO4 1 M HCl Pt

Standard Hydrogen Electrode This is the reference all other oxidations are compared to Eº = 0 º indicates standard states of 25ºC, 1 atm, 1 M solutions. H2 in H+ Cl- 1 M HCl

Cell Potential Zn(s) + Cu+2 (aq)  Zn+2(aq) + Cu(s) The total cell potential is the sum of the potential at each electrode. Eº cell = EºZn Zn+2 + Eº Cu+2  Cu We can look up reduction potentials in a table. One of the reactions must be reversed, so change its sign. Overall cell must be + Potentials are not multiplies by the coefficients in the balanced equation. (BDVD) Galvanic cells

Cell Potential Determine the cell potential for a galvanic cell based on the redox reaction. Cu(s) + Fe+3(aq)  Cu+2(aq) + Fe+2(aq) Fe+3(aq) + e-  Fe+2(aq) Eº = 0.77 V Cu+2(aq)+2e-  Cu(s) Eº = 0.34 V Cu(s)  Cu+2(aq)+2e- Eº = -0.34 V 2Fe+3(aq) + 2e-  2Fe+2(aq) Eº = 0.77 V

Cell Potential A galvanic cell based on the reaction: Al3+(aq) + Mg(s) Al(s) + Mg2+(aq) Give the balanced cell reaction and calculate E° for the cell. Ex.17.1 MnO4-(aq) + H+(aq) + ClO3-(aq) ClO4-(aq) + Mn2+(aq) + H2O (l) (ClO4- + 2H+ + 2e- --> ClO3- + H2O = 1.19V)

Line Notation solidAqueousAqueoussolid Anode on the leftCathode on the right Single line different phases. Double line porous disk or salt bridge. If all the substances on one side are aqueous, a platinum electrode is indicated. For the last reaction Cu(s)Cu+2(aq)Fe+2(aq),Fe+3(aq)Pt(s)

Galvanic Cell Description The reaction always runs spontaneously in the direction that produced a positive cell potential. Four things for a complete description. Cell Potential. Direction of flow. Designation of anode and cathode. Nature of all the components- electrodes and ions.

Practice Completely describe (sketch) the galvanic cell based on the following half-reactions under standard conditions. Ex.17.2 Ag+(aq) + e-  Ag(s) Fe+3(aq) + e-  Fe2+(aq) Describe the cell with this reaction: IO3-(aq) + Fe+2(aq) Fe3+(aq) + I2(aq) #26a

Practice Completely describe (sketch) the galvanic cell based on the following half-reactions under standard conditions. #30 H2O2 + 2H+ + 2e- 2H2O 1.78 V O2 + 2H+ + 2e- H2O2 0.68 V Give the standard line notation for the previous problems.

17.3 Potential, Work and G emf = potential (V) = work (J)/Charge(C) E = work done by system / charge E = -w/q Charge is measured in coulombs. -w = qE (- sign means work is leaving the system) Faraday (F) = 96,485 C/mol e- q = nF = moles of e- x charge/mole e- w = -qE = -nFE = G

Potential, Work and G Gº = -nFE º if E º < 0, then Gº > 0 nonspontaneous if E º > 0, then Gº < 0 spontaneous Calculate Gº for the following reaction: Cu+2(aq)+ Fe(s)  Cu(s) + Fe+2(aq) Fe+2(aq) + 2e- Fe Eº = -0.44 V Cu+2(aq)+2e-  Cu(s) Eº = 0.34 V Ex.17.3

Practice The amount of manganese in steel is determined by changing it to permanganate ion. The steel is first dissolved in nitric acid, producing Mn2+ ions. These ions are then oxidized to the deeply colored MnO4- ions by periodate ion (IO4-) in acid solution. #38 Complete and balance an equation describing each of the above reactions. Calculate E° and G° at 25°C for each reaction.

Practice The G° formula can also be used for half-reactions. Use standard reduction potentials to estimate G°f for Fe2+(aq) and Fe3+(aq). (G°f for e- = 0.) #42 Using data from Table 17.1 place the following in order of increasing strength as reducing agents. #44 Cr3+, H2, Zn, Li, F-, Fe2+

17.4 Cell Potential and Concentration Qualitatively - We can predict direction of change in E from LeChâtelier. Shift to the left reduces potential. To the right increases potential. 2Al(s) + 3Mn+2(aq)  2Al+3(aq) + 3Mn(s) Predict if Ecell will be greater or less than Eºcell if [Al+3] = 2.0 M and [Mn+2] = 1.0 M Ex. 17.5 if [Al+3] = 1.0 M and [Mn+2] = 3.0M if [Al+3] = 1.5 M and [Mn+2] = 1.5 M

Concentration Cells A potential can be generated with the same components but in different concentrations. The reaction proceeds in the direction that will equalize the ion concentration in the components. These voltages are typically small. By using the Nernst Equation, we can calculate the potential of a cell in which some or all of the components are not in their standard states.

Concentration Cells This cell has a driving force that transfers electrons from left to right. Silver will be deposited on the right electrode, thus lowering the concentration of Ag+ in the right compartment. In the left, the silver electrode dissolves (producing Ag+ ions) to raise the concentration of Ag+ in solution.

Practice Consider this cell. Predict the direction of electron flow, and designate the anode and cathode. If the Fe2+ concentration in the right compartment is changed from 0.1 M to 1 x 10-7 M Fe2+, what would change. #52

The Nernst Equation G = Gº + RTln(Q) -nFE = -nFEº + RTln(Q) E = Eº - RTln(Q) nF At 25°C, We use Always have to figure out n by balancing. If concentration can give voltage, then from the voltage we can tell the concentration.

Example 2Al(s) + 3Mn+2(aq)  2Al+3(aq) + 3Mn(s) Eº = 0.48 V under standard conditions. Consider a cell [Mn2+] = 0.50 M and [Al3+] = 1.50 M Q= [Al3+]2 = (1.50)2 = 18 [Mn2+]3 (0.50)3 n = 6 from the balanced equation = 6e- Using Nernst Ecell =0.48 - 0.0591 log(18) 6 = 0.48 - 0.01 = 0.47V (p.844)

The Nernst Equation As reactions proceed concentrations of products increase and reactants decrease. Reach equilibrium where Q = K and Ecell = 0, this is a “dead” cell. At equilibrium, the components in the two cell compartments have the same free energy and G = 0 0 = Eº - 0.0591 log(K) n log(K) = nEº 0.0591

Practice For the redox reaction: log(K) = nEº 0.0591 S4O62-(aq) + Cr2+(aq)  Cr3+(aq) + S2O32-(aq) Half reactions are: S4O62- + 2e- S2O32- E°= 0.17 V Cr3+ + e-  Cr2+ E°= -0.50 V Balance the redox reaction, and calculate E° and K at 25°C Ex. 17.8 log(K) = nEº 0.0591

Practice The overall reaction in a lead storage battery is Pb(s) + PbO2(s) + 2H+(aq) + 2HSO4-(aq)  2PbSO4(s) + 2H2O(l) Calculate E at 25°C for this battery when [H2SO4] = 4.5 M. At 25°C, E° = 2.04V for the lead storage battery. #56 Al|Al3+(1.00 M)||Pb2+(1.00M)|Pb In the above cell, calculate the cell potential after the reaction has operated long enough for the [Al3+] to have changed by 0.60 mol/L. #58

17.5 Batteries are Galvanic Cells A battery is a galvanic cell or, more commonly, a group of galvanic cells connected in series. Car batteries are lead storage batteries. Pb + PbO2 + H2SO4 PbSO4(s) + H2O Dry Cell Zn + NH4+ + MnO2  Zn+2 + NH3 + H2O Alkaline Zn + MnO2  ZnO + Mn2O3 (in base) NiCad NiO2 + Cd + 2H2O  Cd(OH)2 + Ni(OH)2

17.6 Corrosion Corrosion (rusting) - spontaneous oxidation. Many metals develop a metal oxide coating that prevents further oxidation. Most structural metals have reduction potentials that are less positive than O2 . Fe  Fe+2 +2e- Eº= 0.44 V O2 + 2H2O + 4e-  4OH- Eº= 0.40 V Fe+2 + O2 + H2O  Fe2 O3 + H+ Reaction happens in two places.

Salt speeds up process by increasing conductivity Water Rust Anodic area Cathodic area e- Iron Dissolves - Fe  Fe+2

Preventing Corrosion Coating to keep out air and water. Galvanizing - Putting on a zinc coat. Has a lower reduction potential, so it is more easily oxidized. Alloying with metals that form oxide coats. Cathodic Protection - Attaching large pieces of an active metal like magnesium that get oxidized instead.

17.7 Electrolysis Forcing a current through a cell to produce a chemical change for which the cell potential is negative. Running a galvanic cell backwards. Put a voltage bigger than the potential and reverse the direction of the redox reaction. Used for electroplating.

1.10 e- e- Zn Cu 1.0 M Zn+2 1.0 M Cu+2 Anode Cathode

A battery >1.10V e- e- Zn Cu 1.0 M Zn+2 1.0 M Cu+2 Cathode Anode

Stoichiometry of Electrolysis How much chemical change occurs with the flow of a given current for a specified time? Measure current I (in amperes) 1 amp = 1 coulomb of charge per second 1 Faraday (F) = 96,485 C/mol e- Problems are essentially a unit conversion. current and time  quantity of charge moles of electrons  moles of substance grams of substance How long must 5.00 amp current be applied to produce 10.5 g of Ag from Ag+? Ex.17.9

Practice How long would it take to produce 10.0 g of Bi by the electrolysis of a BiO+ solution using a current of 25.0A? #74 Aluminum is produced commercially by the electrolysis of Al2O3 in the presence of a molten salt. If a plant has a continuous capacity of 1.00 million amp, what mass of aluminum can be produced in 2.00 hours? #76 Electrolysis of an alkaline earth metal chloride using a current of 5.00A for 748 seconds deposit 0.471 g of metal at the cathode. What is the identity of the alkaline earth metal? #78

Other uses Electrolysis of water. Separating mixtures of ions. More positive reduction potential means the reaction proceeds forward. A solution of Ag+, Cu2+, and Zn2+ were introduced to a low voltage that was slowly turned up. The most positive reduction potential is easiest to plate out of solution. Therefore Ag+> Cu2+ >Zn2+