Electrochemistry Chapter 20

Redox Reactions Reduction – Oxidation, or redox, involves the transfer of electrons Reduction – gain of electrons Oxidation – loss of electrons

Redox Reactions LEO the lion goes GER Lose Electrons Oxidation Gain Electrons Reduction

Redox Reactions OIL RIG Oxidation Is Losing (OIL) Reduction Is Gaining (RIG)

Oxidation Numbers (States) Positive, negative or neutral values assigned to an atom to keep track of the number of electrons lost or gained. Charge

Oxidation Number Rules Elements alone = 0 Monatomic ion = charge Compound Σ atoms = 0 Polyatomic ion Σ atoms = charge

Common Oxidation Numbers Group 1 +1 Group 2  +2 Group 13  +3 Group 15  -3 Group 16  -2 Group 17  -1 Some exceptions to each above

Oxidation Number Exceptions H in metal hydrides H-1 when paired with a metal NaH, LiH O in peroxide O2-2 (O-1) H2O2 2 nonmetals More electronegative element will be negative OF2 F-1 O+2

Equations Net Ionic Half Reactions Shows only the ions involved in the redox reaction, not spectator ions Half Reactions Only shows one element and how many electrons are gained or lost (Regents)

Half Reactions Zn + CuSO4  Cu + ZnSO4 Zn + Cu+2  Cu + Zn+2 Net Ionic Zn  Zn+2 + 2e- Oxidation Cu2+ + 2e-  Cu Reduction

Electrochemical Cells Voltaic (Chemical) Spontaneous reaction based on potential difference between metals Electrolytic Non-spontaneous reaction using an outside power source

Electrodes Cathode – electrode where reduction takes place Red Cat Anode – electrode where oxidation takes place An Ox

Homework Read Chapter 20 Sections 1 and 2 (20.1-20.2)

Balancing Reactions The number of electrons lost must equal the number of electrons gained Example: 2Na + ZnCl2  Zn + 2NaCl Zn+2 + 2e-  Zn 2(Na Na + + e- )

Balancing Redox Reactions in Acidic solution Balance all elements except H, O Balance O by adding H2O Balance H by adding H+ Balance charge by adding e- Multiply half reactions to balance electrons lost/gained Combine & cancel like species

Example PbO2(s) + I-(aq) → Pb+2(aq) + I2(g) 2 I- → I2 + 2e- 2e- + Oxidation 2e- + 4H+ + PbO2 → Pb+2 + 2H2O Reduction 2e- + 4H+ + 2I- + PbO2 → Pb+2 + I2 + 2H2O + 2e- 4H+ + 2I- + PbO2 → Pb+2 + I2 + 2H2O

Balancing Redox Reactions in Basic solution Follow balancing for acidic solution Add OH- to both sides equally to balance out H+ Combine H+ and OH- to make H2O Combine and cancel like species

Example MnO4-(aq) + NO2-(aq) → MnO2(s) + NO3-(aq) 4OH- + 2( 3e- + Reduction 4OH- + 2( 3e- + 3e- + 4H2O + 4H+ + MnO4- → MnO2 + 2H2O + 4OH- ) 3( 2OH- + H2O + NO2- → NO3- + 2H2O + 2e- + 2H+ + 2e- + 2OH- ) Oxidation 6e- + 6OH- + 11H2O + 2MnO4- + 3NO2- → 2MnO2 + 3NO3- + 10H2O + 8OH- + 6e- H2O + 2MnO4- + 3NO2- → 2MnO2 + 3NO3- + 2OH-

Homework (Re)Read Chapter 20.2 Answer questions 20, 22

Electrochemical Cells Voltaic (Galvanic) Spontaneous reaction based on potential difference between metals Electrolytic Non-spontaneous reaction using an outside power source

Electrodes Cathode – electrode where reduction takes place Red Cat Anode – electrode where oxidation takes place An Ox

Spontaneous Reaction Some metals are more likely to ionize than another metals Lose electrons (oxidize) Activity Series from Regents When 2 metals are paired together, one will lose and one will gain electrons

Voltaic Cell

Potential Difference (V) Amount of work done per unit charge as a charged particle is moved between points From Physics Units V = J/C 𝑉= 𝑊 𝑞

Electromotive force (emf) The difference in the potential between the 2 metals is what drives the electrons to move From high potential to low potential

Half Reaction Potential Measured with a standard hydrogen electrode to determine the potential of each half reaction Reported as Reduction Potentials Table 20.1 Appendix E

Cell Potential, E Difference between potential of cathode and anode Cell emf 𝐸 𝑐𝑒𝑙𝑙 = 𝐸 𝐹𝑖𝑛𝑎𝑙 − 𝐸 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝐸 𝑐𝑒𝑙𝑙 = 𝐸 𝐶𝑎𝑡ℎ𝑜𝑑𝑒 − 𝐸 𝐴𝑛𝑜𝑑𝑒

Standard Cell Potential, E° Ecell when at standard conditions Concentration = 1M Standard emf 𝐸° 𝑐𝑒𝑙𝑙 = 𝐸° 𝐶𝑎𝑡ℎ𝑜𝑑𝑒 − 𝐸° 𝐴𝑛𝑜𝑑𝑒

Example Ecell Zn + Cu+2  Zn+2 + Cu Cu+2 + 2e-  Cu E = +0.34V Zn+2 + 2e-  Zn E = -0.76V Ecell = +0.34 - (-0.76) = 1.10V

Example Ecell 2Al + 3I2  2Al+3 + 6I- I2 + 2e-  2I- E = +0.54V Al+3 + 3e-  Al E = -1.66V Ecell = +0.54 - (-1.66) = 2.20V

Homework (Re)Read Chapter 20.3-20.4 Answer questions 6, 34, 36, 38

Spontaneous ΔG° = – RT∙ln(K) A reaction is spontaneous if: ΔG < 0

Spontaneous ΔG = – nFE ΔG° = – nFE° Relating E to ΔG n = number of electrons transferred in reaction

Faraday Constant Electron = 1.6x10-19C 1 mole of electrons = 96,485 C F = 96,485 C/mol e-

Example ΔG° = – nFE° ΔG° = – (2)(96485)(1.10) ΔG° = – 212 kJ Zn + Cu+2  Zn+2 + Cu E°cell = 1.10V ΔG° = – nFE° ΔG° = – (2)(96485)(1.10) ΔG° = – 212 kJ

Example ΔG° = – nFE° ΔG° = – (6)(96485)(2.20) ΔG° = – 1274 kJ 2Al + 3I2  2Al+3 + 6I- E°cell = 2.20V ΔG° = – nFE° ΔG° = – (6)(96485)(2.20) ΔG° = – 1274 kJ

Electricity Review What is current? Amount of charges flowing per unit time Unit : A = C/s 𝐼= 𝑞 𝑡

Example Suppose an electrochemical cell runs for 15 minutes with a current of 0.10A. For the following reaction, how much would the mass of the cathode increase by? Zn + Cu+2  Zn+2 + Cu 9.3𝑥 10 −4 𝑚𝑜𝑙 𝑒 − 2 𝑒 − =4.7x 10 −4 𝑚𝑜𝑙 𝑞=𝐼𝑡= 0.1𝐴 900𝑠 =90𝐶 (90𝐶) (96485𝐶/𝑚𝑜𝑙) =9.33𝑥 10 −4 𝑚𝑜𝑙 𝑒 − 4.7𝑥 10 −4 63.55 =0.030 𝑔

Homework (Re)Read Chapter 20.5 Answer questions #50, 52, 54, 56 Quiz Wed

Non-Standard Conditions ∆𝐺= ∆𝐺°+𝑅𝑇𝑙𝑛𝑄 −𝑛𝐹𝐸=−𝑛𝐹𝐸°+𝑅𝑇𝑙𝑛𝑄 𝐸=𝐸°− 𝑅𝑇 𝑛𝐹 𝑙𝑛𝑄

Non-Standard Conditions 𝐸=𝐸°− 𝑅𝑇 𝑛𝐹 𝑙𝑛𝑄 𝐸=𝐸°− 2.303𝑅𝑇 𝑛𝐹 𝑙𝑜𝑔𝑄 𝐸=𝐸°− 0.0592 𝑛 𝑙𝑜𝑔𝑄

Nernst Equation 𝐸=𝐸°− 0.0592 𝑛 𝑙𝑜𝑔𝑄

Non-Standard Conditions 𝐸=𝐸°− 0.0592 𝑛 𝑙𝑜𝑔 [𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠] [𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠]

Concentration Cells 𝐸=𝐸°− 0.0592 𝑛 𝑙𝑜𝑔 [𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠] [𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠] Potential difference can be provided by a difference in concentration 𝐸=𝐸°− 0.0592 𝑛 𝑙𝑜𝑔 [𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠] [𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠]

Concentration Cells E°cell = 0 [reactants]<[products]

Electrolysis Process of using electrical current to drive a non-spontaneous reaction Electrolytic Cell Oxidation at anode Reduction at cathode E (–) Electroplating Electropolishing

Electrolysis Can be used to separate compounds into their elements 2H2O  2H2 + O2