Transcript The redox ppt without screencast

Redox Reactions &
Electrochemical Cells
Oxidation Reduction
 Type of reaction based on transfer of electrons between
species
 Used to be based on gain or loss of O
 Now: gain or loss of electrons
 Oxidation: loss of electrons
 Reduction: gain of electrons
 Numerous reactions/applications




Single displacement
Combustion and other organic reactions
Rust and corrosion
Batteries
Oxidation Numbers
Oxidation numbers are numbers assigned to atoms
that reflect the net charge an atom would have if
the electrons in the chemical bonds involving that
atom were assigned to the more electronegative
element
 Rules!
 Oxidation Number Rules
 The oxidation number for an atom in its elemental form is always zero.
 The oxidation number of a monoatomic ion = charge of the monatomic ion.
 Examples:
 Oxidation number of S-2 is -2.
 Oxidation number of Al+3 is +3.
 The oxidation number of all Group 1A metals = +1 (unless elemental).
 The oxidation number of all Group 2A metals = +2 (unless elemental).
 Hydrogen has two possible oxidation numbers:
 +1 when bonded to a nonmetal
 -1 when bonded to a metal
 Oxygen has two possible oxidation numbers:
 -1 in peroxides (O2-2)
 -2 in all other compounds...most common
 Fluorine always has an oxidation number of -1.
 The sum of the oxidation numbers of all atoms (or ions) in a neutral
compound = 0.
 The sum of the oxidation numbers of all atoms in a polyatomic ion =
charge on the polyatomic ion.
Examples
 NO2
O: 2 x (-2) = -4
N: must be +4 to make sum = 0
 K2CrO4
Separate into ions: K+1 and CrO4-2
K : +1
For CrO4-2 :
O: 4 x (-2) = -8
Cr: ion must add up to -2, so Cr = +6
Has an element been reduced or oxidized?
 To determine change in oxidation states, compare oxidation
numbers from one side of an equation to another
 CaCO3 → CaO + CO2
CaCO3
-Ca: +2
-CO3-2: O is -2, so C must be +4
CaO
-Ca+2 and O-2
CO2
- O2: O is 2 x (-2) = -4, so C is +4
No changes in oxidation numbers, so not a redox reaction!!!
Example #2
 Fe + CuSO4 → FeSO4 + Cu
 Fe: 0
Fe: +2
Lost electrons!
Oxidized!
 Cu: +2
Cu: 0
Gained electrons!
Reduced!
 S: +6
S: +6
Same oxidation
state
 O: -2
O: -2
Same oxidation
state
 All oxidation reactions involve a reduction occurring at
the same time. That is why they are called redox: one
cannot occur without the other
When one substance is oxidized, it is in turn reducing
another substance. Therefore it is called the “reducing
agent”. When a substance is reduced, another substance is
oxidized so the reduced substance is known as the
“oxidizing agent’
 Fe+Cu2+ → Fe2++Cu
Fe is oxidized so is the reducing agent
Cu is reduced so is the oxidizing
agent
Another Example:
Zn + 2HCl → Zn2+ + H2 +2Cl In this reaction Zn + 2H+ → Zn2+ + H2 (Cl does not
change oxidation state
Zn is oxidized to Zn2+ (loses 2 electrons)
 reducing agent (Reduces H+)
H+ is reduced to H2 (gains 2 electrons)
 oxidizing agent (Oxidizes Zn)
Strongest Oxidizing Agent
Weakest Reducing Agent
Ba 2+ (aq)
Ba (s)
Ca 2+ (aq)
Ca (s)
Mg 2+ (aq)
Mg (s)
Al 3+ (aq)
Al (s)
Zn 2+ (aq)
Zn (s)
Cr 3+ (aq)
Cr (s)
Fe 2+ (aq)
Fe (s)
Cd 2+ (aq)
Cd (s)
Tl + (aq)
Tl (s)
Co 2+ (aq)
Co (s)
Ni 2+ (aq)
Ni (s)
Sn 2+ (aq)
Sn (s)
Cu 2+ (aq)
Cu (s)
Hg 2+ (aq)
Hg (s)
Ag 2+ (aq)
Ag (s)
Pt 2+ (aq)
Pt (s)
Au 1+ (aq)
Au (s)
Weakest Oxidizing Agent
Strongest Reducing Agent
Half reactions
 Half reactions are a way to separate the
substance that is oxidizing from the substance
that is being reduced. It is useful when
balancing an oxidation reduction reaction
 For the reaction in the previous slide (Zn +
2HCl → Zn2+ + H2 +2Cl-) the half reactions
would be:
 Zn →Zn2+ + 2e- (oxidation)
 2H+ + 2e- →H2 (reduction: notice that
the equation is balanced)
 Another example:
Balancing more complex reactions using the
half reaction method
When balancing a more complex reaction it is useful to first split
the reaction into two half reactions, balance the half reactions and
then put them back together to get a net equation. The following
steps will be useful:
1. Write the half reaction for the oxidation and reduction reactions
2. Balance elements other than H & O
3. Balance O with H2O
4. Balance H with H+
5. Balance charge by adding electrons as needed
6. Multiply each half reaction if necessary to make the number of
electrons in each half reaction equal
7. Add the half reactions & cancel out like terms where possible
Balancing Redox Reactions (continued)
Cr2O72- + C2H6O
→ Cr3+
+
C2H4O
What is oxidized? Reduced? Write the half reactions
for each
Oxidized: C2H6O → C2H4O
Reduced: Cr2O72- → Cr 3+
Balance all atoms but H & O
C2H6O → C2H4O
Cr2O72- → 2Cr3+
Continue next slide…..
Balance O using H2O
C2H6O → C2H4O
Cr2O72- → 2 Cr3+ + 7 H2O
Balance H using H+ (this is assuming an acid
solution..we will look at basic solutions next)
C2H6O → C2H4O + 2 H+
14 H+ + Cr2O72- → 2 Cr3+ + 7 H2O
Continued on next slide….
Balance charge by adding the appropriate number of electrons
to either side (pay attention to which is oxidation and which is
reduction!) The net charge does not have to equal zero, they
just have to equal each other.
This is done by adding electrons. To determine the number of
electrons required, find the net charge of each side the
equation.
C2H6O → C2H4O + 2 H+
0
+2
14 H+ + Cr2O72- → 2 Cr3+ + 7 H2O
+12
+6
2e- must be added to the first equation on product side and 6eto the reactant side of the second equation (Check! Does this
make sense???)
Almost Done!
Multiply each reaction (if necessary) to make the number of electrons lost
equal to the number of electrons gained
6 e- + 14 H+ + Cr2O72- → 2 Cr 3+ + 7 H2O
3 x (C2H6O → C2H4O + 2 H+ + 2 e)
The final step involves adding the two half reactions and reducing to the
smallest whole number by cancelling species which are on both sides of
the arrow.
6 e- + 14 H+ + Cr2O72- + 3 C2H6O → 2 Cr3+ + 7 H2O + 3 C2H4O + 6
H+ + 6 eThe equation can be further simplified by subtracting out 6 e- and 6 H+ ions
from both sides of the equation to give the final equation.
8 H+ + Cr2O72- + 3 C2H6O → 2 Cr 3+ + 7 H2O + 3 C2H4O
 Note: the equation is completely balanced in terms of having an equal
number of atoms as well as charges.
Balancing Redox Reactions : another example!
Balance following reaction:
CuS (s) + NO3 - (aq) → Cu2+(aq) + SO42- (aq) + NO (g)
1. Split into 2 Half-Reactions (Which is oxidized? Reduced?)
CuS → Cu2+ + SO42-
NO3 -1 → NO
Balance all atoms but H &O
Already balanced!
Balance O using H2O
CuS + 4H2O → Cu 2+ + SO42NO3-1 → NO + 2H2O
Balance H using H+
CuS + 4H2O → Cu2+ + SO42- + 8H+
NO3-1 + 4H+ → NO + 2H2O
Balance Charge
CuS + 4H2O → Cu 2+ + SO42- + 8H+ + 8eNO3-1 + 4H+ + 3e- → NO + 2H2O
Multiply to get charges equal for both equations
3 x (CuS + 4H2O → Cu 2+ + SO42- + 8H+ + 8e-)
8 x (NO3-1 + 4H+ + 3e- → NO + 2H2O)
Add and cancel like terms
3CuS + 12H2O + 8NO3-1 + 32H+ + 24e- → 3Cu2+ + 3SO42- + 24H+ + 8NO
+16H2O + 24e8H+
4H2O
Final equation:
3CuS(s) + 8 NO3-1(aq) + 8H+
(aq)
→ 3Cu2+ (aq) + 3 SO42-(aq) + 8NO(g) +4H2O(l)
Check mass balance and charge balance in equation
Left
Right
3 x Cu
3 x S
8 x N
24 x O
8 x H
3 x Cu
3 x S
8 x N
24 x O
8 x H
(8 x 1-) + (8 x H+) = 0
(3 x 2+ )+(3 x 2- ) = 0
What if the solution was basic?
In these examples we have assumed the solution was
acidic - we added H+ to balance the equation.
To make a basic solution, we will add enough OH- ions
(to both sides of the equation) to neutralize the H+
added in the reaction.
Balance the equation in acid, then add enough OH- to
turn the H+ into H2O. You must add OH- to both
sides, so that one side then ends up with excess OH-,
and the solution is now a base
Look at our previously balanced equation:
3CuS(s) + 8 NO3-1(aq)+ 8H+(aq) → 3Cu2+(aq) + 3 SO42-(aq) + 8NO(g) +4H2 O(l)
If we add 8OH- to both sides, this is what happens:
3CuS(s) + 8 NO3-1(aq)+ 8H+(aq) →3Cu2+(aq) +3 SO42-(aq) + 8NO(g)+4H2 O(l) + 8OH8OH
-
8H2O
4
Final equation:
3CuS(s) + 8 NO3-1(aq)+ 4H2Oaq) →3Cu2+(aq) +3 SO42-(aq) + 8NO(g)+ 8OH-
Notice! After adding OH-, we have 8 waters on the left and 4
on the right. If we simplify, the waters on the right cancel and
there are 4 waters remaining on the left. The right side now
has an excess of 8 OH-, so the solution is a base
Balancing Redox Equations Practice
Balance in acid: H2C2O4 + MnO4-1 → Mn2+ + CO2
Balance in base: CN- + MnO4-1 → CNO-1 + MnO2
Redox Reactions in Electrochemistry
Redox reactions may be used to produce electricity through
chemical work. These reactions are what we use to make
batteries
Two Types of Electrochemical Cells:
1. Galvanic
2. Electrolytic
Galvanic Cell - Converts chemical potential energy into an
electrical potential to perform work
Electrolytic Cell- Uses electrical energy to force a chemical
reaction to happen that would not otherwise occur
Galvanic (Voltaic) Cells
What happens when a strip of zinc is put into a solution
of copper sulfate?
CuSO4 + Zn → Cu + ZnSO4
The half reactions would be:
Cu2+ + 2 e− → Cu
Zn → Zn2+ + 2 e−
Zinc is added to a
blue solution of
copper(II) sulfate
The blue color
disappears…the zinc
metal reacts, and
solid copper metal
precipitates on the
zinc strip
The zinc is oxidized
and the copper ions
are reduced
The galvanic cell
If the half reactions in a
redox reaction are separated
into two “cells” and
connected by a wire, the
reaction will still occur.
This is a simple form of a
battery and is known as a
galvanic or voltaic cell
The metals are known as
electrodes
ANODE - Where
OXIDATION takes place
CATHODE - Where
REDUCTION takes
place
Think of them alphabetically:
Anode, Cathode. Oxidation,
Reduction
Generally: When drawing cell,
anode is on the left, cathode is
on the right but this is not
always true!
Remember the activity series??
 This is what the activity series compares: the ability for an
element to be reduced or oxidized
 What metals have the greatest reducing power? (are able to
give electrons most readily?). These are at the top of the list
 All are compared to hydrogen as the standard (although H2 is
not a metal, it is readily “displaced” from compounds by
many metals).
 If we separate two 1/2 cells physically, then provide a conduit
through which the electrons travel from one cell to the other,
the electrons flow and a current is produced
 Over time the reaction will stop. Why?
 Charge build-up in the cells will stop the electron flow.
 Salt bridge: allows charge neutrality in each cell to be
maintained.
Salt bridge/porous disk:
allows for ion migration so
that the solutions will remain
neutral.
Example: Zinc as the anode:
A piece of Zn metal is immersed in ZnSO4 (aq)
Atoms of the zinc metal lose electrons and become zinc ions
in solution.
The zinc electrode loses mass
Half-reaction : Zn → Zn+2 + 2e-
(oxidation half-reaction)
Oxidation always occurs at the anode.
Copper is the cathode:
A piece of Cu metal is immersed in CuSO4 (aq)
Copper ions in solution gain electrons to add copper atoms
to the copper metal.
The half-reaction: Cu+2 + 2e- → Cu
The copper electrode gains mass
This is a reduction half-reaction.
Reduction always occurs at the cathode.
The electrons move from the anode to the cathode
through the external circuit.
Zn+2 go into solution.
SO4-2 ions in solution move from the cathode to
anode through the salt bridge.
The cell continues to operate as long as there is a
potential energy difference between the half-cells.
Cell Notation
A shorthand method of representing a galvanic cell:
Zn(s) | Zn 2+ (aq),1M; SO42-(1M) || Cu2+(aq), 1M; SO42-(1M)| Cu(s)
 Begin at the solid zinc anode. A single vertical line
indicates phase change from solid zinc anode to aqueous
zinc sulfate solution. The double vertical line indicates the
salt bridge between the anode and the cathode. A single
vertical line indicates a phase change from aqueous copper
(II) solution in the cathode to the solid copper cathode.
REVIEW!!!
Voltmeter
e–
Anode Salt bridge
+
Zn (–) Na
SO 2–
(+) Cu
4
Zn2+
Oxidation half-reaction
Zn(s)
Zn2+(aq) + 2e–
Cu2+
e–
2e– lost
per Zn atom
oxidized
Zn
Zn2+
Voltmeter
e–
Anode Salt bridge
+
Zn (–) Na
SO 2–
(+) Cu
4
Zn2+
Oxidation half-reaction
Zn(s)
Zn2+(aq) + 2e–
Cu2+
e–
2e– lost
per Zn atom
oxidized
Zn
Zn2+
Voltmeter
e–
e–
Anode Salt bridgeCathode
+
Zn (–) Na
SO 2– (+) Cu
4
Zn2+
Cu2+
Oxidation half-reaction
Zn(s)
Zn2+(aq) + 2e–
Reduction half-reaction
Cu2+(aq) + 2e–
Cu(s)
e–
2e– gained
per Cu2+ ion
reduced
2e– lost
per Zn atom
oxidized
Zn
Cu2+
Zn2+
–
Cu e
Voltmeter
e–
e–
Anode Salt bridgeCathode
+
Zn (–) Na
SO 2– (+) Cu
4
Zn2+
Cu2+
Oxidation half-reaction
Zn(s)
Zn2+(aq) + 2e–
Reduction half-reaction
Cu2+(aq) + 2e–
Cu(s)
e–
2e– gained
per Cu2+ ion
reduced
2e– lost
per Zn atom
oxidized
Zn
Cu2+
Zn2+
–
Cu e
Voltmeter
e–
e–
Anode Salt bridgeCathode
+
Zn (–) Na
SO 2– (+) Cu
4
Zn2+
Cu2+
Oxidation half-reaction
Zn(s)
Zn2+(aq) + 2e–
Reduction half-reaction
Cu2+(aq) + 2e–
Cu(s)
Overall (cell) reaction
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
Galvanic cell animation
 Galvanic cell
 http://www.mhhe.com/physsci/chemistry/essentialche
mistry/flash/galvan5.swf
Standard Hydrogen Electrode (SHE CELL)
 Calculates the potential for an electrochemical cell : A
standard electrode is needed to provide a reference point.
Standards:
Concentration 1 M
Partial pressure of gases 1
atm
Temperature 25˚C (298K).
SHE continued
The reaction show reduction potential compared to hydrogen
H2(g) → 2H+(aq) + 2e-
A platinum electrode provides a surface on which the hydrogen
gas can be in contact with the hydrogen ions (aq)
The hydrogen electrode is always placed as the negative
electrode of a cell and is written as:
The potential of this cell is assigned the value of 0 V. The
potentials for other cells are then relative to this value.
If E is positive, the metal is less reactive than hydrogen (More
easily reduced)
If E is negative, the metal is more reactive than hydrogen (More
easily oxidized )
The potential difference of the zinc half-cell may be found by
comparison.
Zinc has a greater tendency to ionize than does hydrogen gas so
when the circuit is closed, the electrons will flow down the
potential gradient from the zinc strip to the platinum electrode.
This difference in potential is what causes the electrons to flow.
This is known as the electromotive force (emf).
Electromotive force
The potential difference that drives the flow of electrons in a
galvanic cell
May be measured with a voltmeter and found by adding the
potentials of the oxidation and reduction half-cells
Going back to the zinc and copper half cells:
Use the standard reduction potential table to find the potentials
for the half-cells and calculate the voltage for the zinc-copper cell
above. All values given are for reductions:
Calculations
Zn → Zn+2 + 2eCu+2 + 2e- → Cu
+0.76 Volts (oxidized, so cell potential reversed in sign)
+0.34 Volts
Total = +1.10 Volts
Calculating cell potentials
 Example: An electrochemical cell is created using
gold and magnesium half-cells. Determine what is
the anode, cathode and overall voltage of the cell
 Look at chart. Which metal is more easily reduced?
 The more positive, the more readily reduced:
Au+3 + 3e- → Au E = +1.50
 The more negative, the more readily oxidized:
Mg → Mg+2 + 2e- E = +2.37 (reverse to show oxidation, the sign
is reversed)
 Add the two for total voltage: +1.50 + +2.37 = +3.87
 A positive voltage means the reaction is spontaneous!
Electrolysis
 Non spontaneous reactions forced to occur through an outside
energy source
 Anode positive (connected to positive end of battery) and
Cathode negative (connected to negative end of battery)
 Molten NaCl: Na+ accepts e-‘s at cathode and Cl- release e-‘s at
anode
 Movement of ions maintains current in cell.
 Half reactions:

Na + + e- → Na(s)
-2. 71V
NOT Spontaneous!

Cl- → ½ Cl2 + e-1.36V
 Overall: Na+ + Cl- → Na + ½ Cl2
 Forces a chemical system into non-equilibrium states
 For NaCl in solution:
H2O more easily reduced than Na+ so cathode reaction:
2H2O + 2e- → H2(g) +2OH-
 electrolysis animation
 http://www.sepuplhs.org/high/hydrogen/electrolysis_si
m.html
Faraday’s Laws:
 Found chemical changes related to the quantity of the
electricity and the nature of the substances in the reaction
 Mass formed or consumed proportional to charge or
amount of energy consumed
 mass formed or consumed proportional to atomic or
molecular weight
 mass inversely proportional to the number of electrons
/mole needed to change oxidation state
 96,487 Coulombs of charge is generated by 1 mole of
electrons (1 Faraday)
 Electricity is measured in amperes
 Charge per time or 1 Coulomb/second
 Used to determine the amount of substance formed at
each electrode due to an applied electric current.
Examples
 How much product will be produced when a current of
0.452 amperes is passed through a cell of molten calcium
chloride for 1.50 hours?
 What are the half reactions?

Ca+2 + 2e- → Ca (l)

2Cl- → Cl2 + 2e Quantities formed depends on the number of electrons
that pass (charge)
 C = 1.50h x 60min/h x 60sec/min x 1A/sec x .452A = 2.44 x
103 Coulombs
 Mass calcium: 2.44 x 103 Coulombs x 1mole/96,500C x
1mole Ca/2 mol e- x 40.1gCa/1mol Ca = .507 g Ca
 Mass chlorine: 2.44x 103 C x 1mole e-/96,500C x 1 mol
Cl2/2mol e- x 70.9g Cl2/mol = .896gCl2
 Practice:
 How many grams of barium are produced from molten
barium chloride by applying 0.500A for 30 minutes?