Transcript Unit 5: Electrochemistry

Unit 5: Electrochemistry
An AWESOME presentation by Dana and Brendan
Part 1: Oxidation and Reduction
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Determining oxidation numbers
Oxidizing and Reducing agents
Oxidation and Reduction half-reactions
Balance redox reactions in acidic and basic
solutions
Determining Oxidation Numbers
 Used to determine whether any elements are
changing oxidation states.
 Look at the charge of the element, use your
knowledge about the charge of the anion to
determine what the charge of the cation must be to
balance the charge of the anion and produce the
final, overall charge.
Ex. 1 MnO4- We know O has a charge of 2X+4(-2)=-1
X=+7
Ex. 2
Zn(s) + 2H+(aq)
0
1+
Zn2+(aq) + H2(g)
2+
0
The oxidation number of Zn changes from 0 to 2+
and that of H changes from 1+ to 0 so this is an
oxidation-reduction reaction.
Oxidizing and Reducing Agents
 LEO the lion goes GER
 OIL RIG
 The substance that loses its electrons is
oxidized and the one that gains electrons is
reduced.
From Ex. 1, Zn went from 0 to 2+ so it loses electrons
and is oxidized. H goes from 1+ to 0 so it gains
electrons and is reduced.
Oxidation and Reduction HalfReactions
 When there is an equation, separate the
half-reactions.
 Then balance the elements other than H and
O.
 Balance Oxygen by adding H20
 Balance the H atoms by adding H+
 Last, balance the charge by adding e-
Ex. 3
MnO4–(aq)+ C2O42–(aq)
Mn2+(aq) + CO2(g)
1. Separate the reactions
MnO4-(aq)
Mn2+(aq)
C2O42-(aq)
CO2(g)
2. Balance the half-reactions
5e-+8H++MnO4-(aq)
C2O42-(aq)
Mn2+(aq)+4H2O(l)
2CO2(g)+2e-
Ex. 3 continued
3. Multiply the equations to make the amount of
electrons of each equal.
[5e-+8H++MnO4-(aq)
[C2O42-(aq)
Mn2+(aq)+4H2O(l)]2
2CO2(g)+2e-]5
4. Add the equations
10e-+16H++2MnO4-(aq)
5C2O42-(aq)
16H++2MnO4-(aq)+5C2O42-(aq)
2Mn2+(aq)+8H2O(l)
10CO2(g)+10e2Mn2+(aq)+10CO2(g)+8H20(l)
In Acidic and Basic Solutions
 The previous example is how you would balance
reactions in acidic solutions, but there are extra
steps to balance them in basic solutions.
 You must balance the half-reactions as you would
for acidic half-reactions then add OH- to both
sides to cancel out the H+
Ex. 4
NO2-(aq) + Al(s)
NH3(aq) + Al(OH)4-(aq)
1. Solve as you would in acidic solution
6e-+7H++NO2-(aq)
4H2O+Al(s)
NH3(aq)+2H2O
Al(OH)4-(aq)+4H++3e-
2. Add OH- to both sides
7OH-+6e-+7H++NO2-(aq)
4OH-+4H2O+Al(s)
NH3(aq)+2H2O+7OHAl(OH)4-(aq)+4H++3e-+4OH-
H+ and OH- combine to form H2O
Ex. 4 continued
7H2O+6e-+NO2-(aq)
[4OH-+4H2O+Al(s)
NH3(aq)+2H2O+7OHAl(OH)4-(aq)+3e-+4H2O]2
3. Add the equations
7H2O+6e-+NO2-(aq)
NH3(aq)+2H2O+7OH-
8OH-+8H2O+2Al(s)
2Al(OH)4-(aq)+6e-+8H2O
OH-+5H2O+2Al(s)+NO2-(aq)
NH3(aq)+2Al(OH)4-(aq)
Part 2: Electrochemical cells
1. Half Cells (Voltaic Cell or Galvanic cell)
Points to Remember:
 Anions flow toward the anode, cations flow
toward the cathode.
 Electrons flow from the anode to the
cathode.
 The oxidation (losing electrons) reaction
takes place at the anode.
 The reduction (gaining electrons) reaction
takes place at the cathode.
Anode (oxidation half-reaction): Zn(s)
Zn2+(aq)+2eCathode (reduction half-reaction): Cu2+(aq)+2eCu(s)
2. Using Standard Reduction Potentials to predict Spontaneous Reactions
The larger the Potential (in V), the more
readily reduced the substance is. The
smaller the Potential, the more readily
oxidized it is.
So, to choose which is oxidized and
which is reduced, look to the table and
the one with the higher value is reduced.
Eocell = Eored(cathode) - Eored(anode)
Find this
experimentally
One will be given
If Ecell is positive, the reaction is spontaneous
3. More about standard reduction potential
and Cell Potential
 Cell potential, also called electromotive force, or emf, is measured in
volts and sometimes referred to as the cell voltage.
 For Eocell, the circle denotes that this is the cell potential at standard
conditions.
 Standard conditions are 1 atm, 25 degrees Celsius, and one mole.
 This measures the difference in potential energy between the cathode
and the anode. If there is a greater amount of potential energy in the
anode then the cathode, then the electrons will flow from the anode to
the cathode spontaneously.
Eocell = Eored(cathode) - Eored(anode)
4. Cell potential at non-standard
conditions
 Nernst Equation:
Temperature(K)
8.314 J/molK
E = Eo - (RT/nF)lnQ
Already
Solved
for
Number of
electrons transferred
Reaction Quotient:
[Reactants]/[Products]
96485 C/mol
5. Gibbs Free Energy
 A measure of the spontaneity of the process.
If ∆G is negative, the reaction is
spontaneous.
∆Go = -nFEo
n: number of electrons transferred
F: Faraday’s constant
Eo: Cell potential
6. Equilibrium Constant
 If the reaction is spontaneous, K will be a large,
positive number.
∆Go = -RTlnK
R: 8.314 J/molK
T: Temperature (K)
7. Relationship between electrode mass,
current, and time
Given type of metal, current, amount of days, and cell
potential, solve for the amount of grams of element lost.
Ex. 5 Metal: Zinc, Current: 50 mA, Days: 12, Cell
Potential:1.30 V
Note: Normally this information will be provided in a word problem
Zn
Zn2++2e-
12 days x 86400 s x .050 C x 1 mole e- x 1 mol Zn x 65.4 g = 17.57 grams
1
1 day
1s
96485 C
2 mol e- 1 mol Zn