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Electrochemistry

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Electrochemistry. The electricity produced by chemical reactions or . Electrochemistry Cell Types. Electrolytic Cells External source of electricity drives ... – PowerPoint PPT presentation

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Title: Electrochemistry


1
Electrochemistry
The electricity produced by chemical reactions or
.. The chemical changes brought about by
electricity.
Electrochemical reactions Oxidation-Reduction
Reactions
Cell System where chemical reactions occur
Electrode means of adding/removing electric
current to/from system.
Cathode electrode at which reduction occurs.
(usually -) Anode electrode at which
oxidation occurs. (usually )
Conductivity flow of electrons many solid
metals conduct electricity electrolyte
solutions (acids, bases, or soluble ionic cpds)
2
Electrochemistry Cell Types
Electrolytic Cells External source of
electricity drives a nonspontaneous
(DG gt 0) reaction. Electrolysis Used to
convert water into H2 and O2 gas. Used for
electroplating put a thin layer of one metal
unto
another metal at cathode.
Voltaic Cells A spontaneous (DG lt 0) reaction
is used to generate an electric
current. e.g. batteries
3
Voltaic Cell
Zn

Cu
ZnZn2(1M)Cu2(1M)Cu
4
Voltaic Cell
CuCu2(1M)Ag(1M)Ag
5
SHE
Standard Hydrogen electrode
ZnZn2(1M)H(1M)H2(1atm)Pt
Zn
6
Voltaic Cell
PtH2(1atm)H(1M)Cu2(1M)Cu
H2

Cu
7
Reduction ½ rx Standard Reduction
Potential E? volts
Zn2 2e- ? Zn(s)
-0.763 V
Fe2 2e- ? Fe(s)
-0.44 V
Ni2 2e- ? Ni(s)
-0.25 V
Pb2 2e- ? Pb(s)
-0.126 V
2H 2e- ? H2(g)
0.000 V
Cu2 2e- ? Cu(s)
0.337 V
Hg2 2e- ? Hg(s)
0.789 V
Ag2 e- ? Ag(s)
0.799 V
Cl2 2e- ? 2Cl-
1.360 V
The higher up a ½ rx is on the table the more
readily that element/substance is oxidized.
(Reverse ½ reaction E)
8
Calculating The Standard Electrical Potential for
any cell from the tabulated Standard Reduction
Potentials.
1. Choose the appropriate ½ rxs from table
2. Write the ½ rx for the more () or less (-)
substance
  • Write the ½ rx for the less () or more (-)
    substance
  • as an oxidation reaction. (reverse the
    sign on E?)
  • Write the net balanced reaction (the electrons
    must be balanced
  • but do not multiply E? by the balancing
    coefficient!)
  • Sum the values of E? for each ½ rx to get the
    Cells
  • standard electrical potential.

9
Reduction ½ rx Standard Reduction
Potential E? volts
Zn2 2e- ? Zn(s)
-0.763 V
Fe2 2e- ? Fe(s)
-0.44 V
Ni2 2e- ? Ni(s)
-0.25 V
Pb2 2e- ? Pb(s)
-0.126 V
2H 2e- ? H2(g)
0.000 V
Cu2 2e- ? Cu(s)
0.337 V
Hg2 2e- ? Hg(s)
0.789 V
Ag2 e- ? Ag(s)
0.799 V
Cl2 2e- ? 2Cl-
1.360 V
The higher up a ½ rx is on the table the more
readily that element/substance is oxidized.
(Reverse ½ reaction E)
10
Reaction Spontaneity and DG
DG lt 0 ? reaction is spontaneous proceeds as
written
DG 0 ? reaction is at equilibrium
DG gt 0 ? reaction proceeds in the reverse
direction unless enough energy
provided to drive reaction forward.
DG DG? RT ln Q
DG? - RT ln K
  • Represents standard conditions where P 1atm
  • and the s of all reagents are 1M. Q 1

11
Reaction Spontaneity and E
E gt 0 ? reaction is spontaneous proceeds as
written with voltage output
E 0 ? reaction is at equilibrium no current
flow
E lt 0 ? reaction proceeds in the reverse
direction unless enough
current provided to drive reaction
forward. (e.g. electrolysis)
The Nernst Equation E E? - 2.303RT log
Q nF
DG DG? RT ln Q
12
Voltaic Cell
Zn

Cu
ZnZn2(1M)Cu2(1M)Cu
E -0.763 - 0.0592 log(1/1.2)
n
13
The Nernst Equation
E E? - 2.303 RT log Q lump constants
together. nF
E E? - 0.0592 log Q at 25? C
n
E? 0.0592 log K RT ln K at 25? C
n nF
ZnZn2(1M)Cu2(1M)Cu E 1.10
V
ZnZn2(1.2M)Cu2(0.8M)Cu E ??? V
14
Reaction Spontaneity and the Nernst Equation
E E? - 0.0592 log Q at 25? C
n
E? 0.0592 log K RT ln K at 25? C
n nF
DG -RT ln K
DG -nFE or DG -nFE
Calculate DG or DG and K from cell potentials,
E/E.
15
Electrolysis of Water
2H2O ? 2H2(g) O2(g)
Reduction (2H 2e- ? H2(g)) 2
Oxidation 2H2O ? 4H O2 4e-
16
Graphite or Platinum are common inert
electrodes

Molten NaCl (801 C)
Electrolysis of molten NaCl
17
Electron Stoichiometry
A Coulomb (C) is the SI unit of charge 1 e-
1.602 x 10-19 C or 1.602 x 10-19 C per e- 1
mole of e- 96,485 C 1 Faraday (F)
Current charge per time C s-1 Ampere (A)
How many grams of H2 gas can be produced from
water through which 1.35 x 106 C have been
passed?
Reduction 2H 2e- ? H2(g)
18
3.4 amps for 15 seconds How much Cu is
electroplated?

Cu2 SO42-
Electroplating copper
3.4C x 15s x 1 mol e- x 1 mol Cu x
63.546g Cu 16.8 mg 1s
96,485C 2 mol e- 1 mol Cu
19
Anode (oxidation) Zn ? Zn2 2e-
Cathode (reduction) 2NH4 2e- ? 2NH3 H2
2MnO2 H2 ? 2MnO(OH) (removes H2 gas)
Zn2 4NH3 ? Zn (NH3)42 (removes NH3)
20
Anode (oxidation) Zn 2OH- ? Zn(OH)2 2e-

Cathode (reduction) 2MnO2 2H2O 2e- ?
2MnO(OH) 2OH-
Alkaline cells have longer shelf lives
21
Rechargeable Batteries Nickel-Cadmium
Anode (oxidation) Cd 2OH- ? Cd(OH)2(s)
2e-
Cathode (reduction) NiO2 2H2O 2e- ?
Ni(OH)2(s) 2OH-
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