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TRANSCRIPT
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Essentially dealing with OXIDATION-REDUCTION REACTIONS
Electron transfer reaction
SPONTANEOUS REACTIONS: Examples: voltaic cells, batteries.
NON-SPONTANEOUS REACTIONS:
Examples: electrolysis, electrolytic cells. QUANTITATE REACTIONS
How much current flows 6(,7&8%'98"&:;<%'=(6(
Electrochemistry involves the relationship between electrical energy and chemical energy.
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Ionic equation: Net ionic equation:( What is oxidized? What is reduced? What is the oxidizing agent? What is the reducing agent?
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Zn(s) + CuSO4(aq) R ZnSO4(aq) + Cu(s)
Look at this oxidation reduction reaction
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VA9C(%"&('I7&<(E9'(G&%&':;A;AM(9F;G#H9A(4%#%&<(1. Oxidation state of atom in elemental form is zero.
e.g. Cl2 O2 P4 C(s) S8 2. The oxidation number of a monatomic ion equals its charge. 3. Some elements have “common” oxidation numbers that can be used as reference in
determining the oxidation numbers of other atoms in the compound.(Alkali metals +1 Alkaline earth metals +2 Fluorine –1 H usually +1 (Hydrides: metal-H compounds (–1)) O usually –2 (peroxides (-1) & superoxides possible) Cl, Br, I almost always –1
4. Sum of oxidation numbers is equal to overall charge of molecule or ion: • For a neutral compound the sum of oxidation numbers equals zero. • For a polyatomic ion, the sum of the oxidation numbers is equal to the charge on the
ion. 5. Shared electrons are assigned to the more electronegative atom of the pair:
• more electronegative atom will have a negative oxidation number.
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For main group elements (s and p block) The highest possible positive oxidation state is
equal to the group number!
(We won’t worry about the
transition metals right now)
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Driving force: atoms tend to lose or gain electrons to achieve:
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It helps to know the Periodic trends in oxidation states
Can we explain these common oxidation states?
H -1 +1
Li +1
Be +2
F -1
Na +1
Mg +2
Al +3
P +3 +5
Cl -1
+1,3,5,7
K +1
Ca +2
Ga +3
Ge +2 +4
As +3 +5
Br -1
+1,3,5,7
Rb +1
Sr +2
In +1 +3
Sn +2 +4
Sb +3 +5
I -1
+1,3,5,7
Ba +2
Tl +1 +3
Pb +2 +4
Bi +3 +5
O -2,-1
S -2
+2,4,6 Se -2
+2,4,6 Te -2
+2,4,6
Blue: most common oxidation states in Group 5 Z(,7&8%'98"&:;<%'=(6(2>(?9@#A(
1. Assign Oxidation Numbers.
2. Write incomplete half-reactions.
3. Balance each half-reaction separately. a. Balance atoms undergoing redox. b. Balance remaining atoms.
i. Add H2O to balance oxygens. ii. Add H+ to balance hydrogens.
4. Balance charges by adding electrons. 5. Multiply each half-reaction so that the same number of electrons
are involved in the reduction and the oxidation.
6. Add the half-reactions.
7. In basic solutions, add OH! to neutralize H+
This is the method for BALANCING REDOX REACTIONS
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Example: Balancing Redox Reactions in acid(
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Example: Balancing Redox Reactions in base(
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Example: Balancing Redox Reactions (
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SPONTANEOUS ELECTROCHEMICAL REACTIONS can do work.
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Voltaic Cells Consist of: (OJ ,7&8%'9G&<+(((
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Which electrode will increase in mass? Which will decrease?
Which direction do the electrons flow?
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Cell voltage (EMF or Ecell) is the measure of reaction spontaneity (
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Cell voltage depends on:
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2)
3)
The more spontaneous a reaction, ! ! !
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The standard potential for an electrochemical cell is the potential (voltage) generated when reactants and
products of a redox reaction are in their standard states.(
It is convenient to tabulate redox reactions as half reactions. Standard half-cell potentials When all substances are in standard state:
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Standard State defined so potentials can be tabulated: T = 25°C. Gases, p = 1 atm. [Solutions] = 1M
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Reduction Half-reaction Eo1/2(V)
F2 (g) + 2e– ! 2F– (aq) +2.87 H2O2 (aq) + 2H+(aq) + 2e– !2H2O(!) +1.776
Au+(aq) + e– ! Au(s) +1.69 PbO2(s) + 4H+(aq) + SO4
2– (aq) + 2e– ! PbSO4(s) + 2H2O(!) +1.685
Cl2 (g) + 2e– ! 2Cl–(aq) +1.359 O2 (g) + 4H+(aq) + 4e– ! 2H2O(!) +1.23
ClO4– (aq) + 2H+(aq) + 2e– ! ClO3
– (aq) + H2O(!) +1.23 Pt2+(aq) + 2e– ! Pt(s) +1.20 2IO3
–(aq) + 12H+(aq) + 10e– ! I2(s) + 6H2O(!) +1.195 Br2 (!) + 2e– ! 2Br–(aq) +1.065
NO3– (aq) + 4H+(aq) + 3e– ! NO(s) + 2H2O(!) +0.96
Ag+(aq) + e– ! Ag(s) +0.799 Fe3+(aq) + e– ! Fe2+(aq) +0.771 O2 (g) + 2H+(aq) + 2e– ! H2O2(aq) +0.68 I2 (s) + 2e– ! 2I–(aq) +0.536 I3
– (aq) + 2e– ! 3I–(aq) +0.53
Cu+(aq) + e– ! Cu(s) +0.521 Cu2+(aq) + 2e– ! Cu(s) +0.337 Sn4+(aq) + 2e–! Sn2+(aq) +0.154 Cu2+(aq) + e– ! Cu+(aq) +0.153 2H+ (aq) + 2e– ! H2(g) +0.00 Pb2+(aq) + 2e– ! Pb(s) –0.126 Sn2+(aq) + 2e– ! Sn(s) –0.136 Co+2(aq) + 2e–! Co(s) "0.277 Ni2+(aq) + 2e– ! Ni(s) –0.28 Cd2+(aq) + 2e– ! Cd(s) –0.403 Fe2+(aq) + 2e– ! Fe(s) –0.440 Cr3+(aq) + 3e– ! Cr(s) –0.74 Zn2+(aq) + 2e– ! Zn(s) –0.763 2H2O (!) + 2e– ! H2(g) + 2OH– (aq) –0.83 SO4
2– (aq) + H2O(!) + 2e– ! SO32– (aq) + 2OH– (aq) –0.93
Mn2+(aq) + 2e– ! Mn(s) –1.18 Al3+(aq) + 3e– ! Al(s) –1.66 Mg2+(aq) + 2e– ! Mg(s) –2.37 Na+(aq) + e– ! Na(s) –2.71 Ca2+(aq) + 2e– ! Ca(s) –2.87 Li+(aq) + e– ! Li(s) –3.05
The 1O-_]!,--(b0.,a.3O-(is the potential associated with the half-reaction.(
Rules for half-cell potentials: 1. The sum of two half-cells potentials in a cell equals the overall cell potential:
E°cell = E°1/2(oxid) + E°1/2(reduc)
2. For any half-reaction: E°1/2(oxid) = ! E°1/2(reduc)
3. Standard half-cell is a hydrogen electrode:((
H2(g,1atm) " 2H+ (aq, 1M) + 2e!
E°1/2(oxid) = E°1/2(reduc) = 0 V
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E0cell = E°1/2(reduc) + E°1/2(oxid) =
What is E°1/2(reduc) ? What is E°1/2(oxid)?
What is E0cell for Cu/Zn cell?
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E0cell = E°1/2(reduc) + E°1/2(oxid) =
What is E°1/2(reduc) ? What is E°1/2 (oxid)?
How are half cell potentials measured? A definition of zero potential is needed.
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E0cell = E°1/2(reduc) + E°1/2(oxid) =
What is E°1/2(reduc) ? What is E°1/2 (oxid)?
What is E°cell for the following reaction?
Al(s) + Cu+2(aq) R Al+3(aq) + Cu(s)
Use the table of half cell potentials to find the CELL POTENTIAL for any redox reaction.
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Br2 O2 Fe2+ Na+
Which one of these is the best oxidizing agent? Strategy: write out half cell potentials for REDUCTION
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Which is the best reducing agent of the following? Cl!(aq) Fe(s) Fe+3(aq) Fe+2(aq) Cl2(g)
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Which is the best oxidizing agent of the following? Cl!(aq) Fe(s) Fe+3(aq) Fe+2(aq) Cl2(g)
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What is the value of "G° for the reaction?
The standard cell potential (E°cell) for the reaction below is 0.89V.
2Cr(s) + 3 Sn4+ (aq) R 2 Cr3+ (aq) + 3 Sn2+ (aq)
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Corresponds to Standard conditions 1M solution or 1 atm gas pressure T = 298K
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Replace "G with !nF((((E and "G° with !nF((((E°
cAF(((,((t(cAF(((,q((^(/.(+)(x((((((
For electrochemical cell at equilibrium: "G = 0, E cell = 0
Nernst equation gives the concentration dependence of cell potential Ecell.(
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!
E = Eo "RTn#
lnQ
!
E 12
= E 12
0 " RTn#
ln C[ ]c D[ ]d
A[ ]a B[ ]b
!
E 12
= E 12
o " 2.303 RT
n#log
C[ ]c D[ ]d
A[ ]a B[ ]b
!
E 12
= E 12
o " 0.059
nlog
C[ ]c D[ ]d
A[ ] a B[ ]b at 298K
So for a half reaction: aA + bB + n e! R cC + dD
Sample problem using the Nernst equation for a half cell potential.(
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What is the half cell potential of the Ag/Ag+ redox couple (E0 = +0.799 V) in a 1 M NaCl solution that contains solid
AgCl (Ksp = 1.1 x 10!10)?
For an overall cell reaction: lL + mM R pP + qQ
!
Ecell = E cell0 "
RTn#
ln P[ ]p Q[ ]q
L[ ]l M[ ]m
!
Ecell = E cell0 "
2.303 RTn#
logP[ ] p Q[ ]q
L[ ]l M[ ]m
!
Ecell = E cell0 "
0.0592n
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Nernst equation gives the concentration dependence of cell potential Ecell.(
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