Transcript Unit 11: Electrochemical Cells

Unit 5:
Everything You Wanted to Know
About Electrochemical Cells, But
Were Too Afraid to Ask
By : Michael
“Chuy el Chulo” Bilow
And “H”Elliot Pinkus
Redox Reactions
A Redox Reaction features the transfer of
electrons between ions.
X + Y  Xn+ + YnOxidation Half-Reaction: X  Xn+ + eReduction Half-Reaction: Y + e-  Yn-
Redox Reactions
Determine Oxidation Numbers
Atoms in a pure element have oxidation number of zero
A monatomic ion has oxidation number equal to its
charge.
Sum of oxidation numbers equals overall charge of
compound.
Fluorine is always –1 with other elements
H is +1 and O is –2 in most compounds
Cl, Br, and I are –1 except with Oxygen or Fluorine.
Redox Reactions
H2SO4
– 1 H = +1
– 1 O = -2
– -8 + 2 + S = 0
– 1 S = +6
Cr2O72– 1 O = -2
– -14 + 2(Cr) = -2
– 1 Cr = +6
Balancing Redox Equations
Balance: Cr2O72- + Cl-  Cr3+ + Cl2
Split Equation into half-reactions
– Cr2O72-  2Cr3+
– Cl-  Cl2
Add H+, then H2O, then e- to balance.
– 6e- + 14H+ + Cr2O7  2Cr3+ + 7H2O
– 2Cl-  Cl2 + 2e-
Combine into overall reaction
– 6Cl- + 14H+ + Cr2O7  2Cr3+ + 7H2O + 3Cl2
Balancing Redox Equations
To balance in a BASIC solution:
Take final answer for Acidic Solution:
– 6Cl- + 14H+ + Cr2O72-  2Cr3+ + 7H2O + 3Cl2
Add OH- to cancel H+ and add H2O
– 6Cl- + 7H2O + Cr2O72-  2Cr3+ + 3Cl2 +14OH-
What are Electrochemical (EC) Cells?
An Electrochemical Cell converts chemical
energy into electrical energy by reducing
one substance and oxidizing another.
For example:
Cu+F2Cu2++2FThe copper is oxidized and the fluorine is
reduced because of a transfer of
electrons, thus creating a current.
What are EC Cells?
There are two types of EC cells:
Galvanic cells spontaneously produce
energy
Electrolytic cells must have work done on
them to go to completion, and are thus
nonspontaneous
Electrolytic and Galvanic Cells
In both electrolytic and galvanic cells,
oxidation takes place at the anode and
reduction takes place at the cathode
But, galvanic cells have positively charged
cathodes and negatively charged anodes
And electrolytic cells have negative
cathodes and positive anodes
WHY?
Because reduction is forced in electrolytic
cells, electrons collect there, giving a
negative charge.
And because the oxidation is not favored,
the anode develops a positive charge
How do You Make a Galvanic Cell?
Many EC cells are made with two metals
in a solution of one of their sulfate or
nitrate
The two metal bars are connected by a
salt bridge. The salt bridge allows anions
to pass through to the oxidized side to
restore charge
For example, take zinc and copper in
solutions of CuSO4 and ZnSO4.
How a Galvanic Cell is made
 In this reaction,
Zn(s) would be
oxidized to Zn2+(aq)
and Cu2+(aq) would
be reduced to Cu(s)
 The zinc-copper
galvanic cell would
look like this:
http://hyperphysics.phy-astr.gsu.edu/hbase/chemical/electrochem.html
How do Galvanic
Cells Produce Electricity?
The electron flow from cathode to anode
produces a current, and thus electricity.
Over time, the Zn anode will deteriorate as
it is oxidized to Zn2+, and Cu2+ ions will be
reduced to Cu and leave the solution,
plating the Cu cathode
How are Electrolytic Cells Made?
 There are many ways
to make electrolytic
cells, but all require
an outside source of
energy to force the
reaction towards the
products
 This shows the
electrolysis of NaCl(l)
to Na(l) and Cl2(g)
How Much Electricity?
Cell Potentials
To find how much electricity is produced or
needed, you must use the oxidation and
reduction potentials of each of the half-reactions
that take place in the system.
Reduction Potentials show how much energy is
either released or needed to cause a reduction
half-reaction to occur
Since oxidation is the opposite of reduction,
reduction potentials are the opposite of oxidation
potentials.
Cell Potentials
To find a cell’s potential difference
(voltage), first find its standard oxidation
and reduction potentials of its halfreactions, usually listed as Eo.
Then, subtract the standard reduction
potential for the oxidized species from the
standard reduction potential of the
reduced species to get:
Eocell=Eored-Eoox
Cell Potentials
Let’s go back to the zinc-copper cell
Make two half-reactions: Zn(s)  Zn2+(aq)
+2eAnd Cu2+(aq)+2e- Cu(s)
The reduction potential for the copper (II)
ion to copper metal is +0.34 V
The reduction potential for the zinc (II) ion
to zinc metal is -0.76 V
Cell Potentials
So, Eocell=EoCu -EoZn
Or Eocell= .34 V- (-.76)V
Therefore Eocell=1.10 V
Remember that oxidation and reduction
potentials change, and most are only listed
for 1M concentrations of electrolytes at
25oC and 1 atm of pressure. Changes to
this will result in changes in potentials.
2+
Cell Potentials
Determine the spontaneous cell reaction
and the cell potential of a cell that has
these two half reactions
Al3++3e-Al(s) EoAl =-1.66V
Cu2++2e-Cu(s) EoCu =0.34V
First determine which species is to be
oxidized and which to be reduced
The oxidized substance in a spontaneous
cell will always have the lesser potential
3+
2+
Cell Potentials
Remember to reverse the equation of the
oxidized species and balance the total
ionic equations so that no electrons are
left over.
3Cu2+ +2Al2Al3++3Cu
Now, find the Eocell
Eocell=Eored-Eoox
Eocell=EoCu -EoAl
2+
3+
Cell Potentials
Eocell=.34-(-1.66)
Eocell= 2.00 V
Note that the reduction potentials are not
multiplied by the coefficients in the
equation.
What else?
The SI unit of electric current is the
ampere (A) and the SI unit of charge is the
coulomb (C).
1A= 1 coulomb per second
It has been determined that the charge of
one mole of electrons is 9.65x104 C, which
is referred to as Faraday’s constant and
symbolized F
Faraday’s Constant
From this, we can determine how much
anode material is used up or how much is
produced at the cathode
For example, how many grams of copper
will be deposited on the cathode of an
electrolytic cell if a current of 4.00 A is run
through a solution of CuSO4 for 10.0 min?
Faraday’s Constant
First, convert the minutes to seconds to
coulombs
10.0 min*60.0sec*min-1*4.00A=2.40x103 C
Then coulombs to moles of electrons
2.40x103 C*1mol e-/9.65x104 C=.0249 mol eTo grams of copper. Remember that it takes 2
mol e- to reduce 1 mol Cu2+
.0249 mol e-*63.55 g Cu/2 mol e-=.791 g Cu
Gibbs Free Energy
The maximum amount of work that can be
done is the opposite of ΔG, the change in
Gibbs Free Energy
Since 1 Volt= 1 joule/1coulomb, and the
joule is the SI unit of work, we get
ΔG=-nFEcell
Where n is the moles of electrons
transferred and F is Faraday’s constant
Equilibrium Constants
and Cell Potentials
To find the equilibrium constant of an
equation from its Eocell, the equation is:
Eocell=RTln(Kc)/nF
Where R=8.314 J/molK, T is temperature
in Kelvin, ln(Kc) is the natural logarithm
(log base e) of the equilibrium constant, n
is the number of moles of electrons
transferred, and F is faraday’s constant.
The Nernst Equation
The Nernst Equation relates the calculated
potential of a cell to its potential at a certain
time.
Ecell=Eocell-RT/nFln(Q)
Where R,T, n, and F are the same as above
and Q is the mass-action constant of the
equation, which equals the concentrations
of products that can change concentration
to their coefficient’s power, divided by
reactants that act similarly.
The End