Transcript ELECTROCHEMISTRY2(1)

ELECTROCHEMISTRY
Brief Review

OXIDATION—loss of electron(s) by a species;
increase in oxidation number; increase in oxygen.

REDUCTION—gain of electron(s); decrease in
oxidation number; decrease in oxygen; increase
in hydrogen.

OXIDIZING AGENT—electron acceptor; species
is reduced.

REDUCING AGENT—electron donor; species is
oxidized.
Electron Transfer
Reactions

Electron transfer reactions are oxidationreduction or redox reactions.

Results in the generation of an electric current
(electricity) or be caused by imposing an
electric current.

Therefore, this field of chemistry is often called
ELECTROCHEMISTRY.
OXIDATION-REDUCTION
REACTIONS
Direct Redox Reaction
Oxidizing and reducing agents in direct contact.
Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s)
OXIDATION-REDUCTION
REACTIONS
Indirect Redox Reaction
A battery functions by transferring electrons through an
external wire from the reducing agent to the oxidizing
agent.
CHEMICAL CHANGE ELECTRIC
CURRENT
Mg metal
With time, Cu plates out
onto Mg metal strip, and
Mg strip “disappears.”
Cu2+ ions
Mg
is oxidized and is the reducing agent
Mg(s) ---> Mg2+(aq) + 2e-
Cu2+
is reduced and is the oxidizing agent
Cu2+(aq) + 2e- ---> Cu(s)
Basic Concepts of Electrochemical Cells
wire
e le c t r o n s
Mg
Mg 2+ ions
Anode
Cu
salt
bridge
Cu
2+
ions
Cathode
CHEMICAL CHANGE ELECTRIC
CURRENT
To
obtain a useful current,
we separate the oxidizing
and reducing agents so that
electron transfer occurs thru
an external wire.
wire
elect rons
Mg
Cu
salt
bridge
Mg2+ ions
Cu2+ ions
This is accomplished in a GALVANIC or
VOLTAIC cell.
A group of such cells is called a battery.
Mg --> Mg2+ + 2e-
wire
Cu2+ + 2e- --> Cu
elect rons
Oxidation
Anode
Negative
Mg
salt
bridge
Cu
<--Anions
Cations-->
2+
Mg ions
Cu2+ ions
•Electrons travel thru external wire.
Salt bridge allows anions and cations to
move between electrode compartments.
Reduction
Cathode
Positive
wire
CELL POTENTIAL, (E )
0
elect rons
Mg

For Mg/Cu cell, potential is
2.71V at 25 ˚C
Cu
salt
bridge
Mg2+ ions
Cu2+ ions

This is the STANDARD CELL POTENTIAL, Eo

E0 a quantitative measure of the tendency of
reactants to proceed to products when all are in
their standard states at 25 ˚C.

If E0 is positive the reaction is spontaneous on
the otherwise the reaction is non- spontaneous.
Calculating Cell Voltage

Balanced half-reactions can be added together to
get overall, balanced equation.

If we know E0 for each half-reaction, we could get
E0 for net reaction.
Cu2+(aq) + 2e- ---> Cu(s)
Mg(s) ---> Mg2+(aq) + 2e-------------------------------------------Cu2+(aq) + Mg(s) ---> Mg2+(aq) + Cu(s)
Note that this is the
reduction potential. In
the reaction, Mg
underwent oxidation
such that the sign of
the E0 from -2.73.
reduction potential E0
of Cu is + 0.34
Mg/Cu Electrochemical Cell
wire
Anode,
negative,
source of
electrons
elect rons
Mg
Mg2+ ions
salt
bridge
Cu
Cathode,
positive, sink
for electrons
Cu2+ ions
Cu2+(aq) + 2e- ---> Cu(s)
Eo = +0.34 V
Mg(s) ---> Mg2+(aq) + 2eEo = - 2.37 V
--------------------------------------------------------------Cu2+(aq) + Zn(s)  Zn2+(aq) + Cu(s) Eo = +2.71 V
Spontaneous Reaction
Anode
Cathode
Oxidation occurs
Reduction occurs
Electrons produced
Electrons are consumed
Anions migrate toward
Cations migrate toward
Has negative sign
Has positive sign
+-
KIaq  K+1 + I -1
electrolysis
There are three possible
species:
K+1 + e- K (s)
I2 + 2e-  2IH2O + 2e- H2(g) + 2OH -
KI (aq)
Add: Phenolphthalein – base indicator
(pink)
Starch
- test for iodine
(blue –violet)
Which has higher SRP?
There are three possible species:
K+1 + e- K (s)
Eo = -2.925
I2 + 2e-  2I-
= + 0.535
H2O + 2e- H2(g) + 2OH -
= - 0.828
Cathode : H2O + 2e- H2(g) + 2OH Anode :
2I- I2 + 2e-
H2O + 2I-  I2 + H2(g) + 2OH -
E0 cell = E0 cat - E0 an
= (-0.828) - (+ 0.535)
= - 1.363 V
Eo = - 0.828
E0 = + 0.535
e-+
Cathode
+ IK
H2O
Anode
More About
Calculating Cell Voltage
Assume I- ion can reduce water.
2 H2O + 2e- ---> H2 + 2 OHCathode
2 I- ---> I2 + 2eAnode
------------------------------------------------2 I- + 2 H2O --> I2 + 2 OH- + H2
Assuming reaction occurs as written,
E˚ = E˚cat- E˚an= (-0.828 V) – (+0.535 V) = -1.363 V
Minus E˚ means rxn. occurs in opposite direction
(the connection is backwards or you are recharging the battery)
Eo for a Voltaic Cell
All ingredients are present.
Which way does reaction
proceed?
B
A
Cd  Cd2+ + 2e- Fe  Fe2+ + 2eFe2+ + 2e-  Fe Cd2+ + 2e-  Cd
Write the full equation and its net
reduction potential.
Eo for a Voltaic Cell
Cd2+ + 2e-  Cd
Fe  Fe2+ + 2e-
E0 = -0.40
E0 = -0.44
--------------------------------------------Fe + Cd2+  Fe+2 + Cd E0 = 0.04
From the table, you see
• Fe is a better reducing agent than Cd
• Cd2+ is a better oxidizing agent than
Fe2+