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Chem 1C Final Review Fall 2013

Chem 1C Final Review Fall 2013. Rimjhim Hemnani Amin Mahmoodi Justin Nguyen. Electrochemistry - Overview. Branch of chemistry that deals with the interconversion of electrical energy to chemical energy What to know: Oxidation number rules Balancing Redox reactions

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Chem 1C Final Review Fall 2013

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  1. Chem 1C Final ReviewFall 2013 RimjhimHemnani Amin Mahmoodi Justin Nguyen

  2. Electrochemistry - Overview • Branch of chemistry that deals with the interconversion of electrical energy to chemical energy • What to know: • Oxidation number rules • Balancing Redox reactions • Galvanic Cells • Cell Diagram • Standard Reduction Potentials • Thermodynamics of Redox Reactions • Effect of Concentration on Cell EMF • Electrolysis

  3. Oxidation number rules Posted on Dr. A’s website: https://eee.uci.edu/12s/40200/Oxidation_Numbers.pdf In a nutshell… • Natural state element/compounds = 0 (ex. H2) • Oxidation # = monatomic ion charge (Na+ = +1) • Oxygen is usually -2, except in peroxides (H2O2, Na2O2) • Hydrogen is usually +1, except in metal hydrides (LiH, NaH) • Sum of oxidation numbers = 0 for electrically neutral compounds. For polyatomic ions, sum of oxidation numbers = charge of compound. (NH4+has total oxidation sum of 1)

  4. Electrochemistry – Redox Reactions • Electrons are transferred from one species to another • When a species loses electron = OXIDATION • Marked by oxidation number going up • A species that is oxidized is also called the reducing agent • When a species gains electron = REDUCTION • Marked by oxidation number going down • A species that is reduced is also called the oxidizing agent • “LEO the lion says GER”

  5. Balancing Redox Reactions • Problem: Write a balanced equation to represent the oxidation of iodide ion (I-) by permanganate ion (MnO4-) in basic solution to yield molecular iodine (I2) and manganese (IV) oxide (MnO2) Step 1: Write unbalanced equation: I- + MnO4- I2 + MnO2

  6. Balancing Redox Reactions • Problem: Write a balanced equation to represent the oxidation of iodide ion (I-) by permanganate ion (MnO4-) in basic solution to yield molecular iodine (I2) and manganese (IV) oxide (MnO2) Step 2: Assign oxidation numbers and divide into two half reactions (oxidation and reduction) I- + MnO4- I2 + MnO2 -1 +7 -2 0 +4 -2

  7. Balancing Redox Reactions • Problem: Write a balanced equation to represent the oxidation of iodide ion (I-) by permanganate ion (MnO4-) in basic solution to yield molecular iodine (I2) and manganese (IV) oxide (MnO2) Step 2: Assign oxidation numbers and divide into two half reactions (oxidation and reduction) I- + MnO4- I2 + MnO2 -1 +7 -2 0 +4 -2 Oxidation: I- I2 (I is going from -1 to 0)

  8. Balancing Redox Reactions • Problem: Write a balanced equation to represent the oxidation of iodide ion (I-) by permanganate ion (MnO4-) in basic solution to yield molecular iodine (I2) and manganese (IV) oxide (MnO2) Step 2: Assign oxidation numbers and divide into two half reactions (oxidation and reduction) I- + MnO4- I2 + MnO2 -1 +7 -2 0 +4 -2 Oxidation: I- I2 (I is going from -1 to 0) Reduction: MnO4- MnO2 (Mnis going from +7 to +4)

  9. Balancing Redox Reactions • Problem: Write a balanced equation to represent the oxidation of iodide ion (I-) by permanganate ion (MnO4-) in basic solution to yield molecular iodine (I2) and manganese (IV) oxide (MnO2) Step 3: Balance each half reaction for number and type of atoms and charges Oxidation: 2 I- I2 + 2e-

  10. Balancing Redox Reactions Step 3: Balance each half reaction for number and type of atoms and charges Reduction: MnO4- MnO2 First add O: MnO4- MnO2 + 2 OH- Next balance H: 2 H++ MnO4- MnO2 + 2 OH- *Neutralize H+: + 2 OH-+ 2 H+ + MnO4- MnO2 + 2 OH- + 2 OH- Combine H and OH: 2 H2O + MnO4- MnO2 + 4 OH- Balance charge: 2 H2O + MnO4- + 3e-  MnO2 + 4 OH- *Only do this for basic conditions.

  11. Balancing Redox Reactions • Problem: Write a balanced equation to represent the oxidation of iodide ion (I-) by permanganate ion (MnO4-) in basic solution to yield molecular iodine (I2) and manganese (IV) oxide (MnO2) Step 4: Add half-reactions together to form overall reaction. Equalize number of electrons so that they cancel out later. Oxidation: 3[2 I- I2 + 2e- ] Reduction: 2[2 H2O + MnO4- + 3e-  MnO2 + 4 OH- ] 6 I- 3 I2 + 6e- 4 H2O + 2 MnO4- + 6e-  2 MnO2 + 8 OH-

  12. Balancing Redox Reactions • Problem: Write a balanced equation to represent the oxidation of iodide ion (I-) by permanganate ion (MnO4-) in basic solution to yield molecular iodine (I2) and manganese (IV) oxide (MnO2) Step 4: Add half-reactions together to form overall reaction. Equalize number of electrons so that they cancel out later. Oxidation: 3[2 I- I2 + 2e- ] Reduction: 2[2 H2O + MnO4- + 3e-  MnO2 + 4 OH- ] 6 I- 3 I2 + 6e- 4 H2O + 2 MnO4- + 6e-  2 MnO2 + 8 OH-

  13. Balancing Redox Reactions • Problem: Write a balanced equation to represent the oxidation of iodide ion (I-) by permanganate ion (MnO4-) in basic solution to yield molecular iodine (I2) and manganese (IV) oxide (MnO2) Step 6: Final check for balanced charges and atoms Answer: 6I- + 2 MnO4- + 4 H2O  3 I2 + 2 MnO2 + 8 OH-

  14. Galvanic (aka voltaic) Cells • Experimental apparatus for generating electricity through the use of a spontaneous reaction

  15. Galvanic (aka voltaic) Cells • Cathode: where REDUCTION occurs • Electrons flow to the cathode • Anode: where OXIDATION occurs • Electrons flow away from the anode • Salt bridge: an inverted U tube containing an inert electrolyte solution (i.e. KCl, NH4NO3). • Ions prevent buildup of + charge on anode and (-) charge on cathode (maintains balance of charge) • “Anox”, “redcat”

  16. Galvanic (aka voltaic) Cells • Cell voltage – difference in electrical potential between the anode and cathode • Also called cell potential, or electromotive force (EMF) • Cell voltage is dependent on: • Nature of electrode and ions • Concentrations of ions • Temperature

  17. Cell Diagram Rules: • Anode written first • | indicates a phase boundary (i.e. going from solid to aqueous) • A comma (,) indicates different components in the same phase (i.e. Fe2+ to Fe3+) • || denotes a salt bridge, thus separating cathode and anode • Pt (s), aka platinum, is used when there are no metals present for electrons to travel, (i.e. Fe2+ to Fe3+, or H+ (aq) to H2(gas)) Example #1: Cu2+ (aq) + Zn (s)  Cu (s) + Zn2+ (aq) Cell notation: Zn (s) | Zn2+ (1 M) || Cu2+ (1 M)| Cu (s) Oxi = anode Red. = cath

  18. Cell Diagram Rules: • Anode written first • | indicates a phase boundary (i.e. going from solid to aqueous) • A comma (,) indicates different components in the same phase (i.e. Fe2+ to Fe3+) • || denotes a salt bridge, thus separating cathode and anode • Pt (s), aka platinum, is used when there are no metals present for electrons to travel, (i.e. Fe2+ to Fe3+, or H+ (aq) to H2(gas)) • Example #2: Zn (s) + 2 H+  Zn2+ + H2 • Cell notation: Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (g) | Pt (s) Oxi = anode Red. = cath

  19. Standard Reduction Potentials • The voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm . • The reduction of H+ to H2 is arbitarily set at 0.00 V to determine relative potentials of other substances. • The more positive E° is, the greater the tendency for the substance to be reduced (aka the strongest oxidizing agent) • E°cell = E°cathode - E°anode • A positive E°cellmeans the reaction is favored • Changing the stoichiometric coefficients of a half-cell reaction will NOT affect the value of E°. • When a reaction is reversed, the sign of E° changes but not its magnitude.

  20. Standard Reduction Potentials Chart offered by Dr. A: https://eee.uci.edu/12s/40200/Reduction_Potentials.pdf More extensive chart: http://highered.mcgraw-hill.com/sites/dl/free/0023654666/650262/Standard_Reduction_Potential_19_01.jpg

  21. Standard Reduction Potentials • Example: Can Sn reduce Zn2+ (aq) under standard-state conditions? • Sn2+ has a greater tendency to be reduced (more positive E value), while Zn2+ has a greater tendency to be oxidized. • Answer: No.

  22. Standard Reduction Potentials • A galavanic cell consists of a Mg electrode in a 1.0 M Mg(NO3)2 solution and a Ag electrode in a 1.0 M AgNO3 solution. Calculate the standard emf • Ag+ has a greater tendency to be reduced (aka cathode) • Mg is therefore oxidized Half Reactions: Reduction: Ag+ + e- Ag (s) Oxidation: Mg (s)  Mg2+ (aq) + 2 e- Calculation: E°cell = E°cathode - E°anode = 0.80 – (-2.37) = 3.17 V

  23. Thermodynamics of Redox Reactions • E°cellis related to ΔG° and K. • 1 Joule = 1 Coulomb * 1 Volt • Joule = energy • Coulomb = electrical charge • Volt = voltage • Faraday’s constant = 9.65 * 104 C/mol e- (or 96,500) • total charge of one mole of electrons • Total charge can now be expressed as nF (n = # of moles of e- transferred in reaction)

  24. Thermodynamics of Redox Reactions • Useful equations: • ΔG = -nFEcell( a negative ΔG = spontaneous) • ΔG = -RT ln K • Ecell = RT ln k • nF R = 8.314 J/K*mol F = 98,500 J/V*mol T = in KELVINS! n = moles of e- from balanced redox reaction

  25. Thermodynamics of Redox Reactions • Example: Calculate standard free-energy change for the following reaction at 25°C: 2 Au (s) + 3 Ca2+ (1.0 M)  2 Au3+ (1.0 M) + 3 Ca (s) -------------------------------------------------------------------------- Step 1: Determine what equation to use: In this case, you are looking for ΔG . Equation: ΔG = -nFEcell

  26. Thermodynamics of Redox Reactions • Example: Calculate standard free-energy change for the following reaction at 25°C: 2 Au (s) + 3 Ca2+ (1.0 M)  2 Au3+ (1.0 M) + 3 Ca (s) -------------------------------------------------------------------------- Step 2: Determine what is being reduced and oxidized. Then find Ecelland n. Half Reactions: Reduction: 3 Ca2+ + 6e- 3 Ca (s) (-2.87 V) Oxidation: 2 Au (s)  2 Au3+ (1.0 M) + 6e- (1.50 V) Ecell = -2.87 – (1.50) = -4.37 V n = 6

  27. Thermodynamics of Redox Reactions • Example: Calculate standard free-energy change for the following reaction at 25°C: 2 Au (s) + 3 Ca2+ (1.0 M)  2 Au3+ (1.0 M) + 3 Ca (s) -------------------------------------------------------------------------- Step 3: Plug and chug ΔG = -nFEcell = -(6)(96500)(-4.37) = 2.53*106 J/mol Or 2.53*103 kJ/mol

  28. Thermodynamics of Redox Reactions • Example: Calculate standard free-energy change for the following reaction at 25°C: 2 Au (s) + 3 Ca2+ (1.0 M)  2 Au3+ (1.0 M) + 3 Ca (s) -------------------------------------------------------------------------- Notes: the negative Ecell value that you calculated indicates an unfavorable reaction, which explains why ΔG turned out to be positive (unspontaneous)

  29. The effect of concentration on Cell EMF • Until now, we’ve dealt with STANDARD conditions (1 M) • But what if the concentrations of species aren’t 1 M? • Use Nernst equation: • E = E° - RTln Q nF • E = new EMF that you want to solve • E° = standard EMF • Q = reaction quotient (product / reactant)

  30. The effect of concentration on Cell EMF • Example: Calculate the emf for the following reaction and determine if it will occur spontaneously, given that [Co2+] = 0.15 M and [Fe2+] = 0.68 M: Co (s) + Fe2+ (aq)  Co2+ + Fe (s) ------------------------------------------------------------------------------------ Step 1: Determine what is being reduced and oxidized to find E° and n. Half Reactions: Reduction: Fe2+ + 2 e- Fe (s) (-0.44) Oxidation: Co (s)  Co2++ 2e- (-0.28) E° = (-0.44) – (-0.28) = -0.16 V n = 2

  31. The effect of concentration on Cell EMF • Example: Calculate the emf for the following reaction and determine if it will occur spontaneously, given that [Co2+] = 0.15 M and [Fe2+] = 0.68 M: Co (s) + Fe2+ (aq)  Co2+ + Fe (s) ------------------------------------------------------------------------------------ Step 2: Determine Q Q = [product]/[reactant] = [0.015]/[0.68] = .221

  32. The effect of concentration on Cell EMF • Example: Calculate the emf for the following reaction and determine if it will occur spontaneously, given that [Co2+] = 0.15 M and [Fe2+] = 0.68 M: Co (s) + Fe2+ (aq)  Co2+ + Fe (s) ------------------------------------------------------------------------------------ Step 3: Plug and chug • E = E° - (RT/nFlnQ) E° R T Q n F -0.16 – (8.314)(298)(ln(.221))/(2*96500) = -0.14 Emf = negative = NOT spontaneous

  33. Electrolysis • The mass of product formed (or reactant consumed) at an electrode is proportional to both the amount of electricity transferred at the electrode and the molar mass of the substance in question. • Basically a lot of conversions to get from current to mass (or liters) of substance. • Current * time  Coulombs • Coulombs / Faraday’s constant  number of moles of e- • Use mole ratios to find moles of substance being oxidized/reduced • Moles * molar mass = mass of substance

  34. Electrolysis • Example: A current of 1.26 A is passed through an electrolytic cell containing a dilute sulfuric acid solution for 7.44 h. Write the half-cell reactions and calculate the volume of gases (H2 and O2) generated at STP. ----------------------------------------------------------------------------------- Step 1: Determine half-cell reactions. • You know that sulfuric acid will give you H+ ions in water. • You also know that sulfuric acid is electrolytic and will conduct electricity Reactions: Oxidation: 2 H2O (l)  O2 (g) + 4 H+ (aq) + 4e- Reduction: 2 H+ + 2 e-  1 H2 (g) We see that the gases we need to calculate for are O2 and H2

  35. Electrolysis • Example: A current of 1.26 A is passed through an electrolytic cell containing a dilute sulfuric acid solution for 7.44 h. Write the half-cell reactions and calculate the volume of gases (H2 and O2) generated at STP. ----------------------------------------------------------------------------------- Step 2: Find the volume of O2 gas generated. convert current to charge 1.26 A * 7.44 hr * (3600 s / 1 hr) = 3.37*104 C Step 3: convert charge to moles of electrons 3.37*104C * ( 1 mol e- / 96,500 C) = 0.349 mole e-

  36. Electrolysis • Example: A current of 1.26 A is passed through an electrolytic cell containing a dilute sulfuric acid solution for 7.44 h. Write the half-cell reactions and calculate the volume of gases (H2 and O2) generated at STP. ----------------------------------------------------------------------------------- Step 4: Use mole ratio of oxygen and electrons. 2 H2O (l)  O2 (g) + 4 H+ (aq) + 4e- For every mole of oxygen formed, 4 moles of e- are transferred. 0.349 mol e- * (1 mol O2 / 4 mol e-) = 0.0873 mol O2

  37. Electrolysis • Example: A current of 1.26 A is passed through an electrolytic cell containing a dilute sulfuric acid solution for 7.44 h. Write the half-cell reactions and calculate the volume of gases (H2 and O2) generated at STP. ----------------------------------------------------------------------------------- Step 5: PV = nRT! V = nRT/P = (0.0873 mol)(0.0821)(273 K)/(1atm) = 1.96 L

  38. Electrolysis • Example: A current of 1.26 A is passed through an electrolytic cell containing a dilute sulfuric acid solution for 7.44 h. Write the half-cell reactions and calculate the volume of gases (H2 and O2) generated at STP. ----------------------------------------------------------------------------------- Now for H2 gas, use same methods. 2 H+ + 2 e-  1 H2 (g) 3.37*104 C * (1 mole e- / 96500 C) * (1 mol H2 / 2 mol e-) = 0.175 mol H2 V = nRT/P = (.175 mol)(.0821)(273 K)/(1 atm) = 3.92 L

  39. Thank you and good luck!! • TIPS ON STUDYING: • DO NOT PROCRASTINATE • PRACTICE PROBLEMS! • WHEN IN DOUBT, LOOK AT UNITS! • SLEEP WELL • EAT WELL • AND….DON’T FORGET EVALUATIONS PLZ 

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