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:
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

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
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
Oxidation: I-  I2 (I is going from -1 to 0)
Reduction: MnO4-  MnO (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 (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-  I e-

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-  MnO OH-
Next balance H: 2 H+ + MnO4-  MnO OH-
*Neutralize H+: OH- + 2 H+ + MnO4-  MnO OH OH-
Combine H and OH: H2O + MnO4-  MnO OH-
Balance charge: 2 H2O + 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 (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-  I e- ]
Reduction: 2[2 H2O + MnO e-  MnO OH- ]
6 I-  3 I e-
4 H2O + 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 (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-  I e- ]
Reduction: 2[2 H2O + MnO e-  MnO OH- ]
6 I-  3 I e-
4 H2O + 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 (MnO4-) in basic solution to yield molecular iodine (I2) and manganese (IV) oxide (MnO2)
Step 6: Final check for balanced charges and atoms
Answer:
6 I- + 2 MnO 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)
“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: 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°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:
More extensive chart:

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°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 * 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 = 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 Ecell and n.
Half Reactions:
Reduction: 3 Ca e-  3 Ca (s) (-2.87 V)
Oxidation: 2 Au (s)  2 Au3+ (1.0 M) + 6e (1.50 V)
Ecell = – (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 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° - 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 [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: Fe e-  Fe (s) (-0.44)
Oxidation: Co (s)  Co2+ + 2e (-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 [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/nF ln Q)
E° R T Q n F
-0.16 – (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 (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*104 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 (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-) = 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 = ( 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- / C) * (1 mol H2 / 2 mol e-)
= 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 