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:
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Oxidation number rules
Balancing Redox reactions
Galvanic Cells
Cell Diagram
Standard Reduction Potentials
Thermodynamics of Redox Reactions
Effect of Concentration on Cell EMF
Electrolysis
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. H2)
2.
Oxidation # = monatomic ion charge (Na+ = +1)
3.
Oxygen is usually -2, except in peroxides (H2O2, Na2O2)
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. (NH4+ has total oxidation sum of 1)
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”
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
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
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)
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 (Mn is going from +7 to +4)
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-
Balancing Redox Reactions
Step 3: Balance each half reaction for number and type of
atoms and charges
Reduction: MnO4-  MnO2
First add O:
Next balance H:
*Neutralize H+:
MnO4-  MnO2 + 2 OH2 H+ + MnO4-  MnO2 + 2 OH+ 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.
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 + 6e4 H2O + 2 MnO4- + 6e-  2 MnO2 + 8 OH-
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 + 6e4 H2O + 2 MnO4- + 6e-  2 MnO2 + 8 OH-
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 MnO4- + 4 H2O  3 I2 + 2 MnO2 + 8 OH-
Galvanic (aka voltaic) Cells
• Experimental apparatus for generating electricity through the
use of a spontaneous reaction
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”
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
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.
Fe2+ to Fe3+)
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. Fe2+ to Fe3+, or H+ (aq) to H2 (gas))
Oxi = anode
Example #1:
Cu2+ (aq) + Zn (s)  Cu (s) + Zn2+ (aq)
Red. = cath
Cell notation:
Zn (s) | Zn2+ (1 M) || Cu2+ (1 M)| Cu (s)
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.
Fe2+ to Fe3+)
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. Fe2+ to Fe3+, or H+ (aq) to H2 (gas))
• Example #2:
Oxi = anode
Zn (s) + 2 H+  Zn2+ + H2
Red. = cath
• Cell notation:
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (g) | Pt (s)
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.
Standard Reduction Potentials
Chart offered by Dr. A:
https://eee.uci.edu/12s/40200/Reduction_Potentials.pdf
More extensive chart:
http://highered.mcgrawhill.com/sites/dl/free/0023654666/650262/Standard_Reduction_Potenti
al_19_01.jpg
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.
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:
Oxidation:
Ag+ + e-  Ag (s)
Mg (s)  Mg2+ (aq) + 2 e-
Calculation: E°cell = E°cathode - E°anode
= 0.80 – (-2.37) = 3.17 V
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 etransferred in reaction)
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
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
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 Ca2+ + 6e-  3 Ca (s)
Oxidation: 2 Au (s)  2 Au3+ (1.0 M) + 6e-
Ecell = -2.87 – (1.50) = -4.37 V
n=6
(-2.87 V)
(1.50 V)
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
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)
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)
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)
Oxidation: Co (s)  Co2+ + 2e-
E° = (-0.44) – (-0.28) = -0.16 V
n=2
(-0.44)
(-0.28)
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
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) = -0.14
Emf = negative = NOT spontaneous
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
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) + 4eReduction: 2 H+ + 2 e-  1 H2 (g)
We see that the gases we need to calculate for are O2 and H2
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) = 0.349 mole e-
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) + 4eFor 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
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
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
Thank you and good luck!!
• TIPS ON STUDYING:
•
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DO NOT PROCRASTINATE
PRACTICE PROBLEMS!
WHEN IN DOUBT, LOOK AT UNITS!
SLEEP WELL
EAT WELL
• AND….DON’T FORGET EVALUATIONS PLZ 