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Chem 1C Final Review Fall 2013 Rimjhim Hemnani Amin Mahmoodi Justin Nguyen.

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Presentation on theme: "Chem 1C Final Review Fall 2013 Rimjhim Hemnani Amin Mahmoodi Justin Nguyen."— Presentation transcript:

1 Chem 1C Final Review Fall 2013 Rimjhim Hemnani 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… 1.Natural state element/compounds = 0 (ex. H 2 ) 2.Oxidation # = monatomic ion charge (Na + = +1) 3.Oxygen is usually -2, except in peroxides (H 2 O 2, Na 2 O 2 ) 4.Hydrogen is usually +1, except in metal hydrides (LiH, NaH) 5.Sum of oxidation numbers = 0 for electrically neutral compounds. For polyatomic ions, sum of oxidation numbers = charge of compound. (NH 4 + 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 (MnO 4 - ) in basic solution to yield molecular iodine (I 2 ) and manganese (IV) oxide (MnO 2 ) Step 1: Write unbalanced equation: I - + MnO 4 -  I 2 + MnO 2

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

7 Balancing Redox Reactions Problem: Write a balanced equation to represent the oxidation of iodide ion (I - ) by permanganate ion (MnO 4 - ) in basic solution to yield molecular iodine (I 2 ) and manganese (IV) oxide (MnO 2 ) Step 2: Assign oxidation numbers and divide into two half reactions (oxidation and reduction) I - + MnO 4 -  I 2 + MnO Oxidation: I -  I 2 (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 (MnO 4 - ) in basic solution to yield molecular iodine (I 2 ) and manganese (IV) oxide (MnO 2 ) Step 2: Assign oxidation numbers and divide into two half reactions (oxidation and reduction) I - + MnO 4 -  I 2 + MnO Oxidation: I -  I 2 (I is going from -1 to 0) Reduction: MnO 4 -  MnO 2 (Mn is 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 (MnO 4 - ) in basic solution to yield molecular iodine (I 2 ) and manganese (IV) oxide (MnO 2 ) Step 3: Balance each half reaction for number and type of atoms and charges Oxidation: 2 I -  I 2 + 2e-

10 Balancing Redox Reactions Step 3: Balance each half reaction for number and type of atoms and charges Reduction: MnO 4 -  MnO 2 First add O: MnO 4 -  MnO OH - Next balance H:2 H + + MnO 4 -  MnO OH - *Neutralize H + : + 2 OH H + + MnO 4 -  MnO OH OH - Combine H and OH: 2 H 2 O + MnO 4 -  MnO OH - Balance charge: 2 H 2 O + MnO e-  MnO 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 (MnO 4 - ) in basic solution to yield molecular iodine (I 2 ) and manganese (IV) oxide (MnO 2 ) Step 4: Add half-reactions together to form overall reaction. Equalize number of electrons so that they cancel out later. Oxidation: 3[ 2 I -  I 2 + 2e- ] Reduction: 2[2 H 2 O + MnO e-  MnO OH - ] 6 I -  3 I 2 + 6e- 4 H 2 O + 2 MnO e-  2 MnO OH -

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

13 Balancing Redox Reactions Problem: Write a balanced equation to represent the oxidation of iodide ion (I - ) by permanganate ion (MnO 4 - ) in basic solution to yield molecular iodine (I 2 ) and manganese (IV) oxide (MnO 2 ) Step 6: Final check for balanced charges and atoms Answer: 6 I MnO H 2 O  3 I MnO 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, NH 4 NO 3 ). Ions prevent buildup of + charge on anode and (-) charge on cathode (maintains balance of charge) “An ox”, “red cat”

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: 1.Anode written first 2.| indicates a phase boundary (i.e. going from solid to aqueous) 3.A comma (,) indicates different components in the same phase (i.e. Fe 2+ to Fe 3+ ) 4.|| denotes a salt bridge, thus separating cathode and anode 5.Pt (s), aka platinum, is used when there are no metals present for electrons to travel, (i.e. Fe 2+ to Fe 3+, or H + (aq) to H 2 (gas)) Example #1: Cu 2+ (aq) + Zn (s)  Cu (s) + Zn 2+ (aq) Cell notation: Zn (s) | Zn 2+ (1 M) || Cu 2+ (1 M)| Cu (s) Oxi = anode Red. = cath

18 Cell Diagram Rules: 1.Anode written first 2.| indicates a phase boundary (i.e. going from solid to aqueous) 3.A comma (,) indicates different components in the same phase (i.e. Fe 2+ to Fe 3+ ) 4.|| denotes a salt bridge, thus separating cathode and anode 5.Pt (s), aka platinum, is used when there are no metals present for electrons to travel, (i.e. Fe 2+ to Fe 3+, or H + (aq) to H 2 (gas)) Example #2: Zn (s) + 2 H +  Zn 2+ + H 2 Cell notation: Zn (s) | Zn 2+ (1 M) || H + (1 M) | H 2 (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 H 2 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° cell means 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: hill.com/sites/dl/free/ /650262/Standard_Reduction_Potenti al_19_01.jpg

21 Standard Reduction Potentials Example: Can Sn reduce Zn 2+ (aq) under standard-state conditions? Sn 2+ has a greater tendency to be reduced (more positive E value), while Zn 2+ 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(NO 3 ) 2 solution and a Ag electrode in a 1.0 M AgNO 3 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)  Mg 2+ (aq) + 2 e - Calculation: E° cell = E° cathode - E° anode = 0.80 – (-2.37) = 3.17 V

23 Thermodynamics of Redox Reactions E° cell is related to ΔG° and K. 1 Joule = 1 Coulomb * 1 Volt Joule = energy Coulomb = electrical charge Volt = voltage Faraday’s constant = 9.65 * 10 4 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 = -nFE cell ( a negative ΔG = spontaneous) -ΔG = -RT ln K -E cell = RT ln k - nF R = 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 Ca 2+ (1.0 M)  2 Au 3+ (1.0 M) + 3 Ca (s) Step 1: Determine what equation to use: In this case, you are looking for ΔG. Equation: ΔG = -nFE cell

26 Thermodynamics of Redox Reactions Example: Calculate standard free-energy change for the following reaction at 25 °C: 2 Au (s) + 3 Ca 2+ (1.0 M)  2 Au 3+ (1.0 M) + 3 Ca (s) Step 2: Determine what is being reduced and oxidized. Then find E cell and n. Half Reactions: Reduction: 3 Ca e -  3 Ca (s) (-2.87 V) Oxidation: 2 Au (s)  2 Au 3+ (1.0 M) + 6e - (1.50 V) E cell = – (1.50) = 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 Ca 2+ (1.0 M)  2 Au 3+ (1.0 M) + 3 Ca (s) Step 3: Plug and chug ΔG = -nFE cell = -(6)(96500)(-4.37) = 2.53*10 6 J/mol Or 2.53*10 3 kJ/mol

28 Thermodynamics of Redox Reactions Example: Calculate standard free-energy change for the following reaction at 25 °C: 2 Au (s) + 3 Ca 2+ (1.0 M)  2 Au 3+ (1.0 M) + 3 Ca (s) Notes: the negative E cell 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° - RT ln 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 [Co 2+ ] = 0.15 M and [Fe 2+ ] = 0.68 M: Co (s) + Fe 2+ (aq)  Co 2+ + Fe (s) Step 1: Determine what is being reduced and oxidized to find E° and n. Half Reactions: Reduction: Fe e -  Fe (s) (-0.44) Oxidation: Co (s)  Co e - (-0.28) E° = (-0.44) – (-0.28) = 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 [Co 2+ ] = 0.15 M and [Fe 2+ ] = 0.68 M: Co (s) + Fe 2+ (aq)  Co 2+ + 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 [Co 2+ ] = 0.15 M and [Fe 2+ ] = 0.68 M: Co (s) + Fe 2+ (aq)  Co 2+ + Fe (s) Step 3: Plug and chug E = E° - (RT/nF ln Q) E° R T Q n F – (8.314)(298)(ln(.221))/(2*96500) = 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 (H 2 and O 2 ) 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 H 2 O (l)  O 2 (g) + 4 H + (aq) + 4e - Reduction: 2 H e -  1 H 2 (g) We see that the gases we need to calculate for are O 2 and H 2

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 (H 2 and O 2 ) generated at STP Step 2: Find the volume of O 2 gas generated. convert current to charge 1.26 A * 7.44 hr * (3600 s / 1 hr) = 3.37*10 4 C Step 3: convert charge to moles of electrons 3.37*10 4 C * ( 1 mol e - / 96,500 C) = 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 (H 2 and O 2 ) generated at STP Step 4: Use mole ratio of oxygen and electrons. 2 H 2 O (l)  O 2 (g) + 4 H + (aq) + 4e - For every mole of oxygen formed, 4 moles of e - are transferred mol e - * (1 mol O 2 / 4 mol e - ) = mol O 2

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 (H 2 and O 2 ) generated at STP Step 5: PV = nRT! V = nRT/P = ( 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 (H 2 and O 2 ) generated at STP Now for H 2 gas, use same methods. 2 H e -  1 H 2 (g) 3.37*10 4 C * (1 mole e - / C) * (1 mol H 2 / 2 mol e - ) = mol H 2 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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