electrochemistry - lamar university · two parts to electrochemistry a chemical reaction either 1....
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Electrochemistry
Chapter 20
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Chemical vs Electro-chemical Reactions
• What is the difference between an “ordinary” Chemical Reaction and an Electro-chemical Reaction ?
“Ordinary” Reactions produce Heat
Electrochemical ones produce Voltage
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Electrochemistry
20.1 Oxidation-Recuction (Redox) Reactions20.1 Balancing Redox Reactions20.3 Voltaic Cells (Batteries)
20.4 EMF 20.5 Spontaneity of Redox Reactions
20.9 Electrolysis
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Two Parts to Electrochemistry
A CHEMICAL REACTION EITHER
1. GENERATES VoltageTherefore A BATTERY
or2. USES Voltage
WHICH IS ELECTROLYSIS
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Two Parts to Electrochemistry
1. The Energy Released By A SPONTANEOUS Chemical Reaction
Is Converted to Electricity
Or In Which
2. Elecrical Energy is USED To CAUSE A NONSPONTANEOUS Chemical Reaction To Occur
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Electrochemical Cells
• Electrodes: are where Oxidation Reduction reactions occur.
• Anode: is the electrode where oxidation takes place.
• Cathode: is the electrode where reduction takes place.
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Review
Chapter 4Page 128 - 129
Oxidation & Reduction ReactionsOxidation Numbers
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OXIDATION – REDUCTION(redox) REACTIONS
electron-TRANSFER REACTIONS
TWO STEP PROCESSOR
TWO HALF-REACTIONS INVOLVED
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1. OXIDATION – LOSS of electrons
2. REDUCTION – GAIN of electrons
• AN OXIDIZING AGENT “CAUSES”OXIDATION
• A REDUCING AGENT “CAUSES”REDUCTION
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OXIDATION REDUCTION (Redox) REACTIONS
Examples
HgO(s) ���� Hg(liq) + O2 (gas)
Fe(s) + O2(gas) � Fe2O3(s)
Zn(s) + HCl(aq) � H2(g) + ZnCl2(aq)
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Rules for Assigning Oxidation Numbers
MUST KNOW OXIDATION “RULES”
See Page 128 – 129 Text
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OXIDATION NUMBERS
1. EACH ATOM in a PURE ELEMENT Has An OXIDATION NUMBER OF ZERO
. 2. For IONS Consisting of a SINGLE ATOM, the Ox Num IS EQUAL TO THE CHARGE ON THE ION
3. F ALWAYS Has an Ox Num of –1 IN ALL OF ITS COMPOUNDS
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OXIDATION NUMBERS
4. Cl, Br, and I are – 1 EXCEPT when combined with O or F
5. H is + 1 and O is – 2 EXCEPT in
6. Hydrides (CaH2) & Peroxides (H2O2)
7. The ALGEBRAIC SUM of the Oxidation Numbers in a NEUTRAL Compound MUST be ZERO
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Prob 4.40 Determine the oxidation number for
1. Ti in TiO2
2. Sn in SnCl2
3. C in C2O42-
4. N in (NH4)2SO4
5. N in HNO3
6. Cr in Cr2O72-
+4
+2
+3
-3
+5
+6
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Give OXIDATION NUMBER of Underlined Atom
1 H Cl O4
2. Cl O3-
3. Cl F
4. Cl 2
+7
+5
+1
0
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Redox Reactions
• Redox reaction are those involving the oxidation and reduction of species.
• OIL – Oxidation Is Loss of electrons.
• RIG – Reduction Is Gain of electrons.
• Oxidation and reduction must occur
together.
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OXIDATION – REDUCTION(redox) REACTIONS
Zn(s) + HCl(aq) � H2(g) + ZnCl2(aq)
1. How do you know that this is an Oxidation Reduction (Redox) Reaction?
2. Does this reaction occur spontaneously?
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Oxidation Reduction
• What is The OXIDATION NUMBER For EACH Element Present ?
• What is OXIDIZED ?
• What is REDUCED ?
• What is The OXIDIZING AGENT ?
• What is The REDUCING AGENT ?
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Zn(s) + HCl(aq) � H2(g) + ZnCl2(aq)
Zn 0 � Zn2+ OXIDATION
Cl -� Cl - NO CHANGE
H + � H2O REDUCTION
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Back to Electrochemistry
Electrochemical Cells
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Construct An Electrochemical Cell
1. Two Beakers
2. One containing a solution of Zinc Nitrate with a strip of Zinc Metal in it
3. The other beaker containing a solution of Copper II Nitrate with a strip of Cu Metal
4. A liquid connection between beakers
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Electrochemical Cells
Electrodes: are usually metals strips/wires
Salt Bridge: is a U–shaped tube that contains a salt. Used to connect solutions
Anode: electrode where oxidation takes place
Cathode: electrode where reduction takes place
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Electrochemical Cells
ElectrodeOXIDATION REDUCTION REACTIONS
OXIDATION Zn(s) � Zn2+ (aq) + 2 e -
REDUCTION 2 e - + Cu2+ (aq) � Cu(s)Net Cell Reaction
Zn(s) + Cu2+ (aq) � Zn2+ (aq) + Cu(s)
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Electrochemical cell Convention
Zn(s) | Zn2+ (aq) || Cu2+ (aq) | Cu(s)
Anode is placed on left by convention
| Used to indicate a change of phase
|| Used to indicate a salt bridge
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Overall: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
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Electrochemical cell
A O
C R
�������� �������������� ���������
� � � ��� �� ��� ��� �� � ��� �� � �����
���� ���������������� �������
� ���� �������������� ��������� �
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Electrochemical Cells
Given the following cell notation,
Pt(s) | Sn2+(aq), Sn4+(aq) || Ag+(aq) | Ag(s)
write the anode and cathode reactions
Sn2+(aq) → Sn4+(aq) + 2 e-
2 e- +Ag1+(aq) → Ag(s)
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Write oxidation reduction reactions forFe(s) + Sn2+(aq) → Fe2+(aq) + Sn(s)
Anode Fe(s) → Fe2+(aq) + 2e oxidation
Cathode 2e + Sn2+(aq) → Sn(s) reduction
Write the cell shorthand notation
Fe(s) | Fe2+(aq) || Sn2+(aq) | Sn(s)
Anode saltbridge Cathode
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E°Cell = E°oxidation + E°reduction
The standard potential of any galvanic cell is
the sum of the standard half–cell potentials
for the oxidation and reduction half–cells.
Standard half–cell potentials are always
quoted as a reduction process. The sign
must be changed for the oxidation process.
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Standard Hydrogen Electrode (SHE)
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standard half–cell potentials
• The reference point is called the standard hydrogen electrode (S.H.E.) and consists of a platinum electrode in contact with H2 gas (1 atm) and aqueous H+ ions (1M).
• The standard hydrogen electrode is assigned an arbitrary value of exactly 0.00V.
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The SHE consists of a Pt electrode in a tube placed in H+(aq). H2 is bubbled over electrode
E° = zero by definition
2H+(aq) + 2e- → H2(gas) E° = zero
H2(gas) → 2H+(aq) + 2e- E° = zero
Standard Hydrogen Electrode (SHE)
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Zn2+(aq) + 2e- → Zn(s), E°red = -0.76 V.
Changing the stoichiometric coefficient does
not affect E°
2 Zn2+(aq) + 4e- → 2 Zn(s), E°red = -0.76 V
Reactions with E < 0 are nonspontaneous
reduction potentials
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Only the sign is changed !
Zn(s) → Zn2+(aq) + 2e- E°red = +0.76 V.
Changing the stoichiometric coefficient does not affect E°
2Zn(s) → 2Zn2+(aq) + 4e- E°red = +0.76 V.
Reactions with E > 0 are spontaneous
oxidation potentials
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Cell Voltage
Zn & H2
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Zn(s) | Zn2+ (aq) || H+ (aq) | H2(gas)
Ecell measured relative to the SHE
E°cell = E°red(cathode) - E°red(anode) = +0.76
Zn(s) → Zn2+(aq) + 2e- E° = +0.76 V
2H+(aq) + 2e- → H2(gas) E° = +0.00 V
Zn(s) + HCl(aq) � H2(g) + ZnCl2(aq)
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For Copper
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H2(gas) | H+ (aq) || Cu2+ (aq) | Cu(s)
Ecell measured relative to the SHE
E°cell = E°red(cathode) - E°red(anode) = +0.34
Cu2+(aq) + 2e- → Cu(s) E° = +0.34 V
H2(gas) → 2H+(aq) + 2e- E° = +0.00 V
H2(g) + CuCl2(aq) � Cu(s) + HCl(aq)
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Zn(s) | Zn2+ (aq) || Cu2+ (aq) | Cu(s)
E°cell measured = + 1.10 V
Zn(s) → Zn2+(aq) + 2e- E° = +0.76 V
Cu2+(aq) + 2e- → Cu(s) E° = +0.34 V
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) E°cell = 1.10
Reactions with E > 0 are spontaneous
Reactions with E < 0 are nonspontaneous
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When selecting two half cell reactions the more negative value will form the
oxidation half–cell.Consider the reaction between zinc and silver:
Ag+(aq) + e– → Ag(s) E° = 0.80V
Zn2+(aq) + 2e– → Zn(s) E° = – 0.76V
•Therefore zinc forms the oxidation half–cell: Zn(s) →→→→ Zn2+(aq) + 2e– E° = – (–0.76V)
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Electrochemical Cells
Can Fe 2+(aq) oxidize Al(s) ?(a) yes (b) no (c) I do not know
Write ReactionFe 2+(aq) + Al(s) � Fe (s) + Al+3(aq) 2e + Fe 2+(aq) � Fe (s) Ered = -0.44
Al(s) � Al+3(aq) + 3e Eoxid = + 1.66Ecell = Eoxid + Ered = + therefore YES!
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Electrochemical Cells
• Can Pb2+(aq) oxidize Cu(s) ?
• Calculate E°cell formed with lead and copper
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Batteries
They produce POSITIVE voltage
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Batteries 101
Batteries are the most important practical application of galvanic cells.
Single cell batteries consist of one galvanic cell.
Multicell batteries consist of several galvanic cells linked in series to obtain the desired voltage
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Lead Storage Battery:
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The overall electrochemical reaction is PbO2(s) + Pb(s) + 2SO4
2-(aq) + 4H+(aq) → 2PbSO4(s) + 2H2O(l)
for whichE°cell = E°red(cathode) - E°red(anode)
= (+1.685 V) - (-0.356 V)= +2.041 V.
Wood or glass-fiber spacers are used to prevent the electrodes form touching
Lead-Acid Battery
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Lead Storage Battery:
A typical “12 volt” battery consists of six individual cells connected in series.
Cell Potential: 1.924V {each cell
Electrolyte: 38% by mass Sulfuric Acid.
Anode: Lead grid packed with spongy lead.
Cathode: Lead grid packed with lead IV oxide.
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Lead Storage Battery:
Anode: leadPb(s) + HSO4
–(aq) →PbSO4(s) + H+(aq) + 2e–
Cathode: lead IV oxidePbO2(s) + 3H+(aq) + HSO4
–(aq) + 2e– →PbSO4(s) + 2H2O(l)
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Lead Storage Battery:
• Pb is oxidized to PbSO4 at the anode, and
• PbO2 is reduced to PbSO4 at the cathode
• PbSO4 adheres to the electrodes causing the battery to “run–down.”
• Recharging the battery reverses this reaction to reform Pb and PbO2.
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Fuel Cell:Uses externally fed CH4 or H2, which react to
form water. Most common is H2.Anode: Porous carbon containing metallic catalysts
2H2(s) + 4OH–(aq) → 4H2O(l) + 4e–
Cathode: Porous carbon containing metallic catalysts O2(s) + 2H2O(l) + 4e– → 4OH–(aq)
Electrolyte: Hot aqueous KOH solution.
Cell Potential: 1.23V
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Fuel Cell:
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Fuel Cell:
• Fuel cells are not batteries because they are not self contained.
• Fuel cells are typically have about 40% conversion to electricity, the remainder is lost as heat.
• Excess heat can be used to drive turbine generators.
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Electrolysis
It is the OPPOSITE of batteriesIt does not occur UNLESS
you make it happen
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Electrolysis 101
• Electrolysis: is the process in which electrical energy is used to drive a
nonspontaneous chemical reaction.
• Processes in an electrolytic cell are the
reverse of those in a galvanic cell.
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Electrolysis of Molten Sodium Chloride:
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Electrolysis of Water
• Anode: Water is oxidized to oxygen gas.
2 H2O(l) → O2(g) + 4 H+(aq) + 4e–
• Cathode: Water is reduced to hydrogen gas.
4 H2O(l) + 4e– → 2 H2(g) + 4 OH–(aq)
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Quantitative Electrolysis:
The amount of substance produced at an electrode by electrolysis depends on the
quantity of charge passed through the cell.
Reduction of 1 mol of sodium ionsrequires 1 mol of electrons to pass
through the system.
The charge on 1 mol of electrons is 96,500 Coulombs.
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To determine the moles of electrons passed measure current and time
Charge (C) = Current (A) x Time (s)
• Because the charge on 1 mol of e– is 96,500C, the number of moles of e– passed through the cell is:
C 95,000e 1mole (C) Chargee of Moles
-×=−
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Electrolysis
A constant current of 30.0A is passed through an aqueous solution of NaCl for a time of 1.00 hour. How many grams of NaOH and how many liters of Cl2 gas at STP are produced?
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Electrolysis
A constant current is passed through a cell containing molten MgCl2 for 1 hour. If 71.0 g of Cl2 is obtained, How many grams of Mg is obtained and what is the current in amperes?
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Electrolysis Applications: Electroplating
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External source to drive reaction
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Ni electrode and another metallic electrode in a solution of Ni2+ (aq)
• Anode: Ni(s) → Ni2+(aq) + 2e-
• Cathode: Ni2+(aq) + 2e- → Ni(s)
• Ni plates on the inert electrode
Electrolysis Electrolysis Electroplating
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Balancing OxidationBalancing Oxidation--Reduction Reduction ReactionsReactions
Balancing Equations by the Method of Half Reactions
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1. Write down the two half reactions.2. Balance each half reaction:
a. First with elements other than H and Ob. Then balance O by adding water.c. Then balance H by adding H+.d. Finish by balancing charge by adding
electrons
Balancing OxidationBalancing Oxidation--Reduction ReactionsReduction Reactions
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3. Multiply each half reaction to make the number of electrons equal.
4. Add the reactions and simplify.5. Check!
Balancing OxidationBalancing Oxidation--Reduction ReactionsReduction Reactions
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• Reaction of an acidic solution of Na2C2O4(sodium oxalate, colorless) with KMnO4 (deep purple).
• MnO4- is reduced to Mn2+ (pale pink) while the
C2O42- is oxidized to CO2.
Balancing OxidationBalancing Oxidation--Reduction ReactionsReduction Reactions
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For KMnO4 + Na2C2O4
1. The two incomplete half reactions areMnO4
-(aq) → Mn2+(aq)C2O4
2-(aq) → 2CO2(g)
Balancing OxidationBalancing Oxidation--Reduction Reduction ReactionsReactions
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1. For the MnO4- half reaction
MnO4-(aq) → Mn2+(aq)
2. Adding water and H+ yields
8H+ + MnO4-(aq) → Mn2+(aq) + 4H2O
There is a 7+ charge on the left and 2+ on the right. So 5 electrons need to be added to the left:
5e- + 8H+ + MnO4-(aq) → Mn2+(aq) + 4H2O
For KMnO4 + Na2C2O4 �
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• In the oxalate reaction, there is a 2- charge on the left and a 0 charge on the right, so we need to add two electrons:
C2O42-(aq) → 2CO2(g) + 2e-
3. To balance the 5 electrons for permanganate and 2 electrons for oxalate, we need 10 electrons for both. Multiplying gives:
10e- + 16H+ + 2MnO4-(aq) → 2Mn2+(aq) + 8H2O
5C2O42-(aq) → 10CO2(g) + 10e-
Balancing ReactionsBalancing Reactions
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4. Adding gives:
16H+(aq) + 2MnO4-(aq) + 5C2O4
2-(aq) →2Mn2+(aq) + 8H2O(l) + 10CO2(g)
5. Which is balanced!
Balancing ReactionsBalancing Reactions
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Balancing Equations for Reactions Occurring in Basic Solution
• We use OH- and H2O rather than H+ and H2O.• The same method as above is used, but OH- is
added to “neutralize” the H+ used.
Balancing OxidationBalancing Oxidation--Reduction ReactionsReduction Reactions