chapter 8 acids and bases and oxidation-reduction denniston topping caret 5 th edition copyright ...
TRANSCRIPT
Chapter 8
Acids and Bases and Oxidation-Reduction
Denniston Topping Caret
5th Edition
Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
8.1 Acids and Bases
• Acids: Taste sour, dissolve some metals, cause plant dye to change color
• Bases: Taste bitter, are slippery, are corrosive
• Two theories that help us to understand the chemistry of acids and bases1. Arrhenius Theory
2. Brønsted-Lowry Theory
8.1
Aci
ds a
nd B
ases
• Acid - a substance, when dissolved in water, dissociates to produce hydrogen ions– Hydrogen ion: H+ also called “protons”
HCl is an acid:
HCl(aq) H+(aq) + Cl-(aq)
Arrhenius Theory of Acids and Bases
8.1
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ds a
nd B
ases
Arrhenius Theory of Acids and Bases
• Base - a substance, when dissolved in water, dissociates to produce hydroxide ions
NaOH is a base
NaOH(aq) Na+(aq) + OH-(aq)
8.1
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ds a
nd B
ases
Arrhenius Theory of Acids and Bases
• Where does NH3 fit?
• When it dissolves in water it has basic properties but it does not have OH- ions in it
• The next acid-base theory gives us a broader view of acids and bases
8.1
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nd B
ases
Brønsted-Lowry Theory of Acids and Bases
• Acid - proton donor
• Base - proton acceptor– Notice that acid and base are not defined
using water– When writing the reactions, both accepting
and donation are evident
HCl(aq) + H2O(l) Cl-(aq) + H3O+(aq)
What donated the proton? HClIs it an acid or base? Acid
What accepted the proton? H2OIs it an acid or base? Base
Brønsted-Lowry Theory of Acids and Bases
base
8.1
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ds a
nd B
ases
acid
.
base acidNH3(aq) + H2O(l) NH4
+(aq) + OH-(aq)
8.1
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ds a
nd B
ases
Brønsted-Lowry Theory of Acids and Bases
Now, let us look at NH3 and see why it is a
base.
Did NH3 donate or accept a proton? Accept
Is it an acid or base? Base
What is water in this reaction? Acid
Acid-Base Properties of Water
• Water possesses both acid and base properties– Amphiprotic - a substance possessing both acid
and base properties– Water is the most commonly used solvent for
both acids and bases– Solute-solvent interactions between water and
both acids and bases promote solubility and dissociation
8.1
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ds a
nd B
ases
8.1
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nd B
ases
Acid and Base Strength
• Acid and base strength – degree of dissociation– Not a measure of concentration– Strong acids and bases – reaction with water is
virtually 100% (Strong electrolytes)
8.1
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ds a
nd B
ases
Strong Acids and Bases
• Strong Acids:– HCl, HBr, HI Hydrochloric Acid,
etc.
– HNO3 Nitric Acid
– H2SO4 Sulfuric Acid
– HClO4 Perchloric Acid
• Strong Bases:– NaOH, KOH, Ba(OH)2
– All metal hydroxides
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)
H2CO3(aq) + H2O(l) HCO3-(aq) + H3O+(aq)
8.1
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ds a
nd B
ases
Weak Acids
• Weak acids and bases – only a small percent dissociates (Weak electrolytes)
• Weak acid examples:– Acetic acid:
– Carbonic Acid:
• Weak base examples:– Ammonia:
– Pyridine:
– Aniline:C6H5NH2(aq) + H2O(l) C6H5NH3
+(aq) + OH-(aq)
C5H5NH2(aq) + H2O(l) C5H5NH3+(aq) + OH-(aq)
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
8.1
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nd B
ases
Weak Bases
• The acid base reaction can be written in the general form:
• Notice the reversible arrows
• The products are also an acid and base called the conjugate acid and base
acid baseHA + B A– + HB+
8.1
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nd B
ases
Conjugate Acids and Bases
acid base
• Conjugate acid - what the base becomes after it accepts a proton
• Conjugate base - what the acid becomes after it donates its proton
• Conjugate acid-base pair - the acid and base on the opposite sides of the equation
base acid
HA + B A- + HB+
8.1
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nd B
ases
HA + B A– + HB+
8.1
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ases
Acid-Base Dissociation
• The reversible arrow isn’t always written– Some acids or bases essentially dissociate 100%
– One way arrow is used
• HCl + H2O Cl- + H3O+ – All of the HCl is converted to Cl-
– HCl is called a strong acid – an acid that dissociates 100%
• Weak acid - one which does not dissociate 100%
8.1
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ases
Conjugate Acid-Base Pairs
• Which acid is stronger:
HF or HCN? HF
• Which base is stronger:
CN- or H2O? CN -
8.1
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ases
Acid-Base Practice• Write the chemical reaction for the following
acids or bases in water• Identify the conjugate acid-base pairs
1. HF (a weak acid)
2. H2S (a weak acid)
3. HNO3 (a strong acid)
4. CH3NH2 (a weak base)
Note: The degree of dissociation also defines weak and strong bases
• Pure water is virtually 100% molecular
• Very small number of molecules dissociate– Dissociation of acids and bases is often called
ionization
• Called autoionization
• Very weak electrolyte
H2O(l) + H2O(l) H3O+(aq) + OH-(aq)
8.1
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nd B
ases
The Dissociation of Water
• H3O+ is called the hydronium ion• In pure water at room temperature:
– [H3O+] = 1 x 10-7 M– [OH-] = 1 x 10-7 M
• What is the equilibrium expression for:
Remember, liquids are not included in equilibrium expressions
]OH][O[HK -3eq
+=
H2O(l) + H2O(l) H3O+(aq) + OH-(aq)
8.1
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nd B
ases
Hydronium Ion
• This constant is called the ion product for water and has the symbol Kw
• Since [H3O+] = [OH-] = 1.0 x 10-7 M, what is the value for Kw?
– 1.0 x 10-14
– It is unitless
]OH][O[HK -3w
+=
8.1
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ases
Ion Product of Water
8.2 pH: A Measurement Scale for Acids and Bases
• pH scale - a scale that indicates the acidity or basicity of a solution– Ranges from 0 (very acidic) to 14 (very basic)
• The pH scale is rather similar to the temperature scale assigning relative values of hot and cold
• The pH of a solution is defined as:
pH = -log[H3O+]
• Use these observations to develop a concept of pH– if know one concentration, can calculate the
other
– if add an acid, [H3O+] and [OH-]
– if add a base, [OH-] and [H3O+]
– [H3O+] = [OH-] when equal amounts of acid and base are present
• In each of these cases 1 x 10-14 = [H3O+][OH-]8.2
pH: A
Mea
sure
men
t S
cale
for
Aci
ds a
nd B
ases A Definition of pH
• pH of a solution can be:– Calculated if the concentration of either is
known• [H3O+] • [OH-]
– Approximated using indicator / pH paper that develops a color related to the solution pH
– Measured using a pH meter whose sensor measures an electrical property of the solution that is proportional to pH8.
2 pH
: A M
easu
rem
ent
Sca
le f
or A
cids
and
Bas
es Measuring pH
• How do we calculate the pH of a solution when either the hydronium or hydroxide ion concentration is known?
• How do we calculate the hydronium or hydroxide ion concentration when the pH is known?
• Use two facts:
8.2
pH: A
Mea
sure
men
t S
cale
for
Aci
ds a
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ases Calculating pH
pH = -log[H3O+]
1 x 10-14 = [H3O+][OH-]
8.2
pH: A
Mea
sure
men
t S
cale
for
Aci
ds a
nd B
ases
Calculating pH from Acid Molarity
What is the pH of a 1.0 x 10-4 M HCl solution?
– HCl is a strong acid and dissociates in water
– If 1 mol HCl is placed in 1 L of aqueous solution it produces 1 mol [H3O+]
– 1.0 x 10-4 M HCl solution has [H3O+]=1.0x10-4M
= -log [H3O+]
= -log [1.0 x 10-4]
= -[-4.00] = 4.00
pH = -log[H3O+]
8.2
pH: A
Mea
sure
men
t S
cale
for
Aci
ds a
nd B
ases Calculating [H3O+] From pH
What is the [H3O+] of a solution with pH = 6.00?
• 4.00 = -log [H3O+]
• Multiply both sides of equation by –1
• -4.00 = log [H3O+]
• Take the antilog of both sides
• Antilog -4.00 = [H3O+]
• Antilog is the exponent of 10
• 1.0 x 10-4 M = [H3O+]
pH = -log[H3O+]
8.2
pH: A
Mea
sure
men
t S
cale
for
Aci
ds a
nd B
ases Calculating the pH of a Base
What is the pH of a 1.0 x 10-3 M KOH solution?• KOH is a strong base (as are any metal hydroxides)• 1 mol KOH dissolved and dissociated in aqueous
solution produces 1 mol OH-
• 1.0 x 10-3 M KOH solution has [OH-] = 1.0 x 10-3 M
• Solve equation for [H3O+] = 1 x 10-14 / [OH-]• [H3O+] = 1 x 10-14 / 1.0 x 10-3 = 1 x 10-11
• pH = -log [1 x 10-11]
= 11.00
1 x 10-14 = [H3O+][OH-]
pH = -log[H3O+]
8.2
pH: A
Mea
sure
men
t S
cale
for
Aci
ds a
nd B
ases
Calculating pH from Acid Molarity
What is the pH of a 2.5 x 10-4 M HNO3 solution?
• We know that as a strong acid HNO3
dissociates to produce 2.5 x 10-4 M [H3O+]
• pH = -log [2.5 x 10-4]
• = 3.60
pH = -log[H3O+]
8.2
pH: A
Mea
sure
men
t S
cale
for
Aci
ds a
nd B
ases Calculating [OH-] From pH
What is the [OH-] of a solution with pH = 4.95?
• First find [H3O+] • 4.95 = -log [H3O+]
• [H3O+] = 10-4.95 • [H3O+] = 1.12 x 10-5
• Now solve for [OH-]• [OH-] = 1 x 10-14 / 1.12 x 10-5
= 1.0 x 10-9
pH = -log[H3O+]
1 x 10-14 = [H3O+][OH-]
The pH Scale8.
2 pH
: A M
easu
rem
ent
Sca
le f
or A
cids
and
Bas
es
1.0 x 100 0.001.0 x 10-1 1.001.0 x 10-2 2.001.0 x 10-3 3.001.0 x 10-4 4.001.0 x 10-5 5.001.0 x 10-6 6.001.0 x 10-7 7.00
For a strong acidHCl molarity pH
Mor
e A
cidi
c
1.0 x 100 14.001.0 x 10-1 13.001.0 x 10-2 12.001.0 x 10-3 11.001.0 x 10-4 10.001.0 x 10-5 9.001.0 x 10-6 8.001.0 x 10-7 7.00
For a strong baseNaOH molarity pH
Mor
e ba
sic
Each 10 fold change in concentration changes the pH by one unit
8.2
pH: A
Mea
sure
men
t S
cale
for
Aci
ds a
nd B
ases
8.3 Reactions Between Acids and Bases
• Neutralization reaction - the reaction of an acid with a base to produce a salt and water
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) Acid Base Salt Water• Break apart into ions:
H+ + Cl- + Na+ + OH- Na+ + Cl- + H2O• Net ionic equation
– Show only the changed components– Omit any ions appearing the same on both sides of
equation = Spectator ions
H+ + OH- H2O
8.3
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ases
• The net ionic neutralization reaction is more accurately written:
H3O+(aq) + OH-(aq) 2H2O(l)• This equation applies to any strong acid / strong
base neutralization reaction• An analytical technique to determine the
concentration of an acid or base is titration• Titration involves the addition of measured
amount of a standard solution to neutralize the second, unknown solution
• Standard solution - solution of known concentration
Net Ionic Neutralization Reaction
Buret – long glass tube calibrated in mL which contains the standard solution
Flask contains a solution of unknown concentration plus indicator
Indicator – a substance which changes color as pH changes
Standard solution is slowly added until the color changes
The equivalence point is when the moles of H3O+ and OH- are equal
8.3
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ases
Acid – Base Titration
8.3
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ases
Relationship Between pH and Color in Acid-Base Indicators
Determine the Concentration of a Solution of Hydrochloric Acid• Place a known volume of acid
whose concentration is not known into a flask
• Add an indicator, experience guides selection, here phenol red is good
• Known concentration of NaOH is placed in a buret
• Drip NaOH into the flask until the indicator changes color
8.3
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ases
Determine the Concentration of a Solution of Hydrochloric Acid
• Indicator changes color – Equivalence point is reached
– Mol OH- = Mol H3O+ present in the unknown acid
• Volume dispensed from buret is determined• Calculate acid concentration:
– Volume of Hydrochloric Acid: 25.00 mL
– Volume of NaOH added: 35.00 mL
– Concentration of NaOH: 0.1000 M
– Balanced reaction shows that HCl and NaOH react 1:1 8.3
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Determine the Concentration of a Solution of Hydrochloric Acid
• 35.00 mL NaOH x 1L NaOH x 0.1000 mol NaOH 103 mL NaOH 1L NaOH
= 3.500 x 10-3 mol NaOH• 3.500 x 10-3 mol NaOH x 1 mol HCl 1 mol NaOH
= 3.500 x 10-3 mol HCl– this amount of HCl is contained in 25.00 mL
• 3.500 x 10-3 mol HCl x 103 mL HCl 25.00 mL HCl 1 L HCl• = 1.400 x 10-1 mol HCl / L HCl = 0.1400 M HCl8.3
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Determine the Concentration of a Solution of Hydrochloric Acid
• Alternative strategy to solve the acid concentration
(Macid)(Vacid) = (Mbase)(Vbase)
• (Macid) = (Mbase)(Vbase) (Vacid)• (Macid) = (0.1000 M) (35.00 mL) (25.00 mL)• = 0.1400 M HCl
8.3
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8.3
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ases
Calculating the Concentration of Sodium Hydroxide
Calculate [NaOH] if 25.00 mL of this solution were required to neutralized 45.00 mL of 0.3000 M HCl•Alternative strategy to solve the acid concentration
(Macid)(Vacid) = (Mbase)(Vbase)
•(Mbase) = (Macid)(Vacid) (Vbase)
•(Mbase) = (0.3000 M) (45.00 mL) (25.00 mL)
= 0.5400 M NaOH
• The previous examples have the acid and base at a 1:1 combining ratio
– Not all acid-base pairs do this
• Polyprotic substance - donates or accepts more than one proton per formula unit
– Hydrochloric acid is monoprotic, producing one H+ ion for each unit of HCl
– Sulfuric acid is diprotic, each unit of H2SO4 produces 2 H+ ions
H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2 H2O(l)8.3
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Polyprotic Substances
Step 1.
H2SO4(aq) + H2O(l) HSO4-(aq) + H3O+(aq)
Step 2.
HSO4-(aq) + H2O(l) SO4
2-(aq) + H3O+(aq)
• In Step 1, H2SO4 behaves as a strong acid – dissociating completely
• In Step 2, HSO4-( behaves as a weak acid – reversibly dissociating,
note the double arrow
8.3
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ases
Dissociation of Polyprotic Substances
8.4 Acid-Base Buffers
• Buffer solution - solution which resists large changes in pH when either acids or bases are added
• These solutions are frequently prepared in laboratories to maintain optimum conditions for chemical reactions
• Buffers are also used routinely in commercial products to maintain optimum conditions for product behavior
8.4
Aci
d-B
ase
Buf
fers
• Buffers act to establish an equilibrium between a conjugate acid – base pair
• Buffers consist of either– a weak acid and its salt (conjugate base)– a weak base and its salt (conjugate acid)
– Acetic acid (CH3COOH) with sodium acetate (CH3COONa)
• An equilibrium is established in solution between the acid and the salt anion• A buffer is Le Chatelier’s principle in action
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)
The Buffer Process
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)8.4
Aci
d-B
ase
Buf
fers
Addition of Base (OH-) to a Buffer Solution
• Adding a basic substance to a buffer causes changes– The OH- will react with the H3O+ producing water– Acid in the buffer system dissociates to replace
the H3O+ consumed by the added base– Net result is to maintain the pH close to the initial
level
• The loss of H3O+ (the stress) is compensated by the dissociation of the acid to produce more H3O+
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)8.4
Aci
d-B
ase
Buf
fers
Addition of Acid (H3O+) to a Buffer Solution
• Adding an acidic substance to a buffer causes changes– The H3O+ from the acid will increase the overall
H3O+ – Conjugate base in the buffer system reacts with the
H3O+ to form more acid– Net result is to maintain the H3O+ concentration and
the pH close to the initial level
• The gain of H3O+ (the stress) is compensated by the reaction of the conjugate base to produce more acid
8.4
Aci
d-B
ase
Buf
fers
Buffer Capacity
• Buffer capacity - a measure of the ability of a solution to resist large changes in pH when a strong acid or strong base is added
• Also described as the amount of strong acid or strong base that a buffer can neutralize without significantly changing pH
• Buffering process is an equilibrium reaction described by an equilibrium-constant expression
– In acids, this constant is Ka
• If you want to know the pH of the buffer, solve for [H3O+], then calculate pH
COOH]CH[
]COOCH][OH[K
3
-33
a
+
=
8.4
Aci
d-B
ase
Buf
fers
Preparation of a Buffer Solution
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)
8.4
Aci
d-B
ase
Buf
fers
Calculating the pH of a Buffer Solution
Calculate the pH of a buffer solution in which – Both the acetic acid (acid) and sodium acetate (salt)
concentrations are 2.0 x 10-2 M– Sodium acetate, the salt, is also the conjugate base
– Ka = 1.75 x 10-5
– [H3O+] = [acid]Ka [conjugate base] = {(2.0 x 10-2 M) x (1.75 x 10-5)} / 2.0 x 10-2 M = 1.75 x 10-5
– pH = -log 1.75 x 10-5 = 4.76
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)
COOH]CH[
]COOCH][OH[K
3
-33
a
+
=
8.4
Aci
d-B
ase
Buf
fers
Henderson-Hasselbalch Equation
• Solution of equilibrium-constant expression and pH can be combined into one operation
• Henderson-Hasselbalch equation is this combined expression
• Using these two equations:
– pKa = -log Ka just as pH = -log[H3O+]
– pKa = pH – log ( [CH3COO-] / [CH3COOH] )
– Henderson-Hasselbalch –
– pH = pKa + log( [CH3COO-] / [CH3COOH] )
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)
COOH]CH[
]COOCH][OH[K
3
-33
a
+
=
8.4
Aci
d-B
ase
Buf
fers
Henderson-Hasselbalch Equation
• pH = pKa + log ([CH3COO-] / [CH3COOH]) can be rewritten
– pH = pKa + log ([conjugate base] / [weak acid])
8.5 Oxidation-Reduction Processes
• Oxidation-reduction processes are responsible for many types of chemical change
• Oxidation - defined by one of the following – loss of electrons
– loss of hydrogen atoms
– gain of oxygen atoms
• Example: NaNa+ + e-
– Oxidation half reaction
8.5
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sses
• Reduction - defined by one of the following:– gain of electrons– gain of hydrogen– loss of oxygen
• Example: Cl + e- Cl-
– Reduction half reaction
• Cannot have oxidation without reduction
Oxidation-Reduction Processes
Na + Cl Na+ + Cl-
Oxidizing Agent• Is reduced• Gains electrons• Causes oxidation
Reducing Agent• Is oxidized• Loses electrons• Causes reduction
8.5
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Oxidation and Reduction as Complementary Processes
Na Na+ + e-
Cl + e- Cl-
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Applications of Oxidation and Reduction
• Corrosion - the deterioration of metals caused by an oxidation-reduction process
– Example: rust (oxidation of iron)
4Fe(s) + 3O2(g) 2Fe2O3(s)
• Combustion of Fossil Fuels
– Example: natural gas furnaces
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
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• Bleaching
• Most bleaching agents are oxidizing agents
• The oxidation of the stains produces compounds that do not have color
– Example: Chlorine bleach - sodium hypochlorite (NaOCl)
Applications of Oxidation and Reduction
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Biological Processes
• Respiration– Electron-transport chain of aerobic
respiration uses reversible oxidation and reduction of iron atoms in cytochrome c
• Metabolism– Break down of molecules into smaller pieces
by enzymes
• Is Zn oxidized or reduced?• Oxidized
• Copper is reduced8.5
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Voltaic Cells
• Voltaic cell - electrochemical cell that converts stored chemical energy into electrical energy
• Let’s consider the following reaction:
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
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Voltaic Cells
• If the two reactants are placed in the same flask they cannot produce electrical current
• A voltaic cell separates the two half reactions
• This makes the electrons flow through a wire to allow the oxidation and reduction to occur
Zn Zn2+ + 2e-
Oxidationanode – electrode
where oxidation occurs
Cu2+ + 2e- CuReduction
cathode – electrode where reduction occurs
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Voltaic Cell Generating Electrical Current
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Voltaic Cell Generating Electrical Current
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Silver Battery• Batteries use the concept of the voltaic cell
• Modern batteries are:– Smaller
– Safer
– More dependable
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Electrolysis
• Electrolysis reactions - use electrical energy to cause nonspontaneous oxidation-reduction reactions to occur
• These reactions are the reverse of a voltaic cell
– Rechargeable battery• When powering a device behaves as a voltaic cell• With time, the chemical reaction nears completion• Battery appears to “run down”• Cell reaction is reversible when battery attached to
charger
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Comparison of Voltaic and Electrolytic Cells