Transcript Chapter 3 Stoichiometry: Calculations with Chemical

International University
for Science & Technology
College of Pharmacy
General Chemistry
(Students of Dentistry)
Prof. Dr. M. H. Al-Samman
CHEMISTRY
Chapter 4
Solutions and
Solution Stoichiometry
Solution :
Solution is a homogeneous distribution of the
components of a substance in the
components of another .
we call the substance with the higher amount :
a solvent (the substance doing the dissolving)
and, the substance with the lesser amount : a
solute ( the substance being dissolved)
As an exception :We consider water a solvent
in any system regardless of it’s amount.
Types of Electrolytes
We can divide the solutions according to the
conductivity into two types:
-Electrolytes
-Nonelectrolytes
Electrolytes : which are formed by ionic solute in
ionic or polar solvent such as the solution of
NaCl in water.
Electrolytes are divided into two types:
A)-Strong Electrolytes :which disassociate
completely in water to give ions (anions and
cations) and they are good conductors.
Types of Electrolytes
Strong electrolytes include:
-Strong acids (HCl, HBr, HI, HNO3, H2SO4,HClO3,
HClO4)
-Strong bases (IA and IIA hydroxides)
-Most water-soluble ionic compounds
B)-A weak electrolytes :which dissociate partially.
-Weak electrolyte solutions are poor conductors.
-Different weak electrolytes dissociate to different
extents.
A nonelectrolyte : which is formed by a
nonionic substance in a nonionic solvent ,
such as fat in diethyl ether .
Types of Electrolytes
Nonelectrolyte does not dissociate.
-A nonelectrolyte is present in solution almost exclusively as molecules. (example: fat in
chloroform).
-Nonelectrolyte solutions do not conduct electricity.
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Acids …
-Taste sour, if diluted with enough water to be
tasted safely.
-May produce a pricking or stinging sensation on
the skin.
-Turn the color of litmus or indicator paper from
blue to red.
-React with many metals to produce ionic
compounds and hydrogen gas.
-Also react with bases, thus losing their acidic
properties.
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Bases …
-Taste bitter, if diluted with enough water to be
tasted safely.
- Feel slippery or soapy on the skin.
-Turn the color of litmus or indicator paper from red
to blue.
- React with acids, thus losing their basic
properties.
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Acids and Bases:
The Arrhenius Concept
There are several definitions which may be used to
describe acids and bases.
An Arrhenius acid is a compound that ionizes in
water to give hydrogen ions H+ .strong acids
give high concentration of hydrogen ions.
An Arrhenius base is a compound that ionizes in
water to give hydroxyl ions OH– . Strong bases
give high concentration of hydroxyl.
Arrhenius’s theory in electrolytes:
Certain substances dissociate into cations and
anions when dissolved in water.
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Acids and Bases:
The Arrhenius Concept
- HClO₄ ,HClO₃,HBr, HI ,HCl, H₂SO₄ , HNO₃ , are strong
acids.(strong acids ionize completely in diluted
solutions).
- All bases of group one and group two are strong
bases.(strong bases ionize completely in diluted
solutions).
Neutralization it is the reaction between acid and
base to give salt and water.
A salt is the product of the neutralization reaction
between acid and base.
HCl + NaOH → NaCl + H₂O
Acid Base
Salt Water
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Acids and Bases:
The Arrhenius Concept
Strong acid: it is the acid that ionizes completely
in diluted solution.
Weak acid: it is the acid that ionizes partially in a
diluted solutions.
Strong base: it is the base that ionizes completely
in diluted solutions.
Weak base : it is the base that ionizes completely
in diluted solutions.
Solutions and
Solution Stoichiometry
Concentration of a solution: the quantity of a
solute in a given quantity of solution (or solvent).
A concentrated solution contains a relatively large
amount of solute vs. the solvent (or solution).
A dilute solution contains a relatively small
concentration of solute vs. the solvent (or solution).
“Concentrated” and “dilute” aren’t very quantitative …
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Molar Concentration
Molarity (M), or molar concentration: It is the
number of moles of the solute in one liter of the
Solution.
Example:
A solution that is 0.35 M sucrose contains 0.35
moles of sucrose in each liter of solution.
-Keep in mind that molarity signifies moles
of solute per liter of solution, not liters of
solvent.
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Molar Concentration
Molarity (M), or molar concentration: It is the
number of moles of the solute in one liter of
the Solution.
Example:
What is the molarity of ammonium sulphate
salt in a solution containing 26.4 g of the salt in
4 lit. of the solution?.
We can read formulas in terms
of moles of atoms or ions.
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Molarity
Q- Find the number of ammonium ions NH₄⁺ in
a solution of ammonium sulphate containing
10.56 g. in 4 Lit.? atomic masses: H=1, N=14,
O=16, S=32, thin calculate the molarity of the
compound.
A-The formula of ammonium sulphate= ( NH₄ )₂SO₄
-The molar mass = 132 g.
-The number of moles= 10.56/132 = 0.08 mol.
The number of moles of NH₄⁺ ions is twice the number of
moles of ammonium sulphate
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Molarity
The number of moles of NH₄⁺ ions :
2x0.08=0.16 mol.
- The number of ammonium ions NH₄⁺ =
0.16 x 6.023x 10²³ = 9.636 x 10²² ion
-The molarity of ammonium sulphate ( NH₄ )₂SO₄=
M= 0.08/4 = 0.02 mol/Lit
Example:
What is the molarity of a solution in which 333 g potassium
hydrogen carbonate ( KHCO₃ ) is dissolved in enough water
to make 10.0 L of solution?
Answer:
Number of moles = 333/100 = 3.33 mol.
The molarity = 3.33/10 = 0.333 mol/Lit
Example:
We want to prepare a 6.68 molar solution of NaOH (6.68 M
NaOH).
How many moles of NaOH are required to prepare 0.500 (a)
L of 6.68 M NaOH?
How many liters of 6.68 M NaOH can we prepare with (b)
2.35 kg NaOH?
Answer:
(a) 1 lit of NaOH contains 6.68 mol
0.500 lit
needs
x mol
Number of NaOH moles required :
0.500x6.68/1 = 3.34 mol
(b) Number of moles in 2.35 kg ( 2350 g):
2350/40 = 58.75 mol.
number of liters: 58.75/6.68 = 8.7949 lit
Example:
The label of a stock bottle of aqueous ammonia indicates
that the solution is 28.0% NH3 by mass and has a
density of 0.898 g/mL. Calculate the molarity of the
solution.
Example
A chemical reaction familiar to geologists is that
used to identify limestone. The reaction of
hydrochloric acid with limestone, which is largely
calcium carbonate, is seen through an
effervescence—a bubbling due to the liberation of
gaseous carbon dioxide:
CaCO3(s) + 2 HCl(aq)  CaCl2 (aq) + H2O(l) + CO2(g)
How many grams of CaCO3(s) are consumed in a
reaction with 225 mL of 3.25 M HCl?
Example cont.
Answer:
Number of moles of hydrochloric acid:
0.225 x 3.25 = 0.731 mol.
Mass of hydrochloric acid:
0.731 x 36.5 = 26.681 g
CaCO3(s) + 2 HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g)
1 mol
2mol
100 g
2x36.5
x
26.681
Mass of the consumed CaCO3(s) :
x = 0.365 g
Dilution of Solutions
Dilution: is the process of adding more solvent to
the solution to lower or decrease the
concentration.
- Addition of solvent does not change the amount
of solute in a solution but does change the
solution concentration.
-It is very common to prepare a concentrated
stock solution of a solute, then dilute it to
other concentrations as needed.
Conductivity Example
Example:
Which of the following is the best conducting
solution:
0.1M HCl, 0.2M CH₃COOH, 4M CH₃CH₂OH,
0.6M NH₄OH, 0.09M NaOH
Answer:
0.1 M HCl as strong acid with high molarity
Conductivity Example
Example:
Which of the following is the best conducting
solution:
0.1M HCl, 0.2M CH₃COOH, 4M CH₃CH₂OH, 0.6M
NH₄OH, 0.09M NaOH, 0.1M H₂SO₄
Answer:
0.1M H₂SO₄ as strong acid with high molarity(the
number of ions provided by H₂SO₄ molecule are 3
ions while the number of ions provided by HCl
molecule are 2).
Dilution Calculations …
… couldn’t be easier.
Moles of solute does not change on
dilution.
Moles of solute = M × V
Therefore …
Mconc × Vconc = Mdil × Vdil
Indicators
Indicators : are organic compounds commonly
used in analytical analysis to tell:
that a neutralization is complete, or: if a solution is acidic or basic. Phenol red is … -
… yellow in
acidic solution
…
… orange in
neutral solution …
… and red in
basic solution.
Dilution Calculations …
Example:
How many ml of a concentrated solution of
sodium hydroxide 0.4 M , are needed to
prepare 1000 ml of 0.1 diluted solution.
Answer:
Mconc × Vconc = Mdil × Vdil
0.4 x Vconc = 0.1 x1000
Vconc= (0.1x1000)/0.4 = 250 ml
Dilution Calculations …
Example:
If we diluted 200 ml of 0.2M solution of
hydrochloric acid up to 1000 ml , what is
the molarity of the diluted solution.
Answer:
Mconc × Vconc = Mdil × Vdil
0.2 x 200 = Mdil x 1000
Mdil = 0.04 mol/lit
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PH of Solutions
The PH:
Is the minus logarithm of the hydrogen ions
concentration(in moles).
PH= -log [ H⁺ ]
Example:
The hydrogen ions concentration of hydrochloric
acid solution is 0.001 mole , calculate the PH
value of the solution.
Solution:
PH= - log [10⁻³]
PH = 3
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PH of Solutions
Example:
15- What the PH of 0.001 M solution of
hydrochloric acid?
Solution:
PH= - log [ H⁺ ]
PH= - log [ 10³⁻ ]
PH= 3
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PH of Solutions
Example:
16- Calculate the PH value of a solution of
sulphuric acid containing 4.9 g per liter of
the acid?
solution
-the molar concentration of the acid = 4.9/98=0.05 mol/lit
-the molar concentration of hydrogen ions:
[ H⁺ ]=2 [ M ] = 0.05 x 2 = 0.1 mol/lit
PH = - log [ 10⁻¹]
PH= 1
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PH of Solutions
Example:
17- Calculate the PH value of a solution of hydrochloric acid
containing 0.02 M of the acid?
solution
-since the hydrochloric acid is a strong acid, so the molar
concentration of acid is equal to the molar concentration
of hydrogen ions:
[ H⁺ ]=0.02
PH = - log [ 2x10⁻²]
PH=(-) [log2 + log10⁻²]
PH=(-) [ 0.30 – 2 ]
PH=(-) [ -1.7]
PH = 1.7
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PH of Solutions
- Since the ionic constant of water at 25C⁰ is :
[H⁺] . [OH⁻] = 10⁻¹⁴
- And by writing the expression of ( - log ) for the
above equation, we get :
PH + POH = 14
Where:
PH = - log[H⁺]
POH = - log[OH⁻]
14 = - log10⁻¹⁴
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PH of Solutions
Example:
18- Calculate the PH value of 0.001 M solution of
sodium hydroxide?
Solution:
[ OH⁻ ] = 10⁻³
POH = - log [ OH⁻ ]
POH = - log [ 10⁻³ ]
POH = 3
We know that : PH + POH = 14
So we write:
PH + 3 = 14
PH = 11
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Hydrates
A hydrate is an ionic compound in which the
formula unit includes a fixed number of water
molecules associated with cations and anions
Examples:
BaCl2 . 2 H2O
CuSO4 . 5 H2O
EOS
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Hydrates
Notice:
-The water molecules in the hydrate compounds are
part of the formula, so when we calculate the
molecular mass of the hydrate compound, we should
add the mass of the water molecules to the molar mass
of the compound.
-We name here some of the hydrates:
-BaCl₂. 2H₂O : barium chloride dihydrate.
-CuSO₄. 5H₂O : copper sulphate penta hydrate.
-CoCl₃. 6H₂O : cobalt (III)chloride hexa hydrate.
Acid–Base Reactions:
Net Ionic Equations
HCl + NaOH  H2O + NaCl + Q
In the reaction above, the HCl, NaOH, and
Na+ and
NaCl all are strong electrolytes and
Cl– are
dissociate completely.
spectator
ions.
The actual reaction occurs between ions.
H+ + Cl– + Na+ + OH–  H2O + Na+ + Cl–
H+ + OH–  H2O + Q
Q= 57.3 J/lit , it is the same for all neutralization reactions.
Calculating Ion Concentrations in Solution
In 0.010 M Na2SO4:
-two moles of Na+ ions are formed for
each mole of Na2SO4 in solution, so
[Na+] = 0.020 M.
-one mole of SO42– ion is formed for each
mole of Na2SO4 in solution, so [SO42–] =
0.010 M.
An ion can have only one
concentration in a solution, even if
the ion has two or more sources.
Calculating Ion Concentrations in Solution
How many moles of Na+ ions are in a solution
containing 0.24 mole of Na2SO4 and 0.46 mole
of NaCl.
Na2SO4  2 Na+ + SO42–
1 mol
0.24 mol
X= 0.48 mol
NaCl
1 mol
0.46 mol
Y= 0.46 mol
2mol
x mol
 Na+ + Cl –
1 mol
y mol
The number of moles of Na+
: 0.48 + 0.46 = 0.94 mol
Reactions that Form
Precipitates
There are limits to the amount of a solute that
will dissolve in a given amount of water.
-If the maximum concentration of solute is
less than about 0.01 M, we refer to the
solute as insoluble in water.
When a chemical reaction forms such a
solute, the insoluble solute comes out of
solution and is called a precipitate.
Guide Line in Precipitation
-all compounds of nitrate ion(NO₃⁻) , acetate (ch₃coo⁻ )
and perchlorate (ClO₄⁻) are soluble.
-all compounds of group (l) in periodic table and
ammonium ion (NH₄⁺) are soluble.
- most compounds of chloride (Cl⁻), bromide (Br⁻) and
iodide (I⁻) are soluble except of Ag⁺, Pb²⁺, and Hg₂²⁺
- most compounds of sulfate are soluble, except of
group ll and Ag⁺, Pb²⁺, and Hg₂²⁺ are insoluble.
- most compounds of hydroxide ( OH⁻), sulfide (S²⁻) and
phosphate ( PO₄³⁻ ) are insoluble.
Guide Line in Precipitation
-With these guidelines we can predict precipitation reactions:
Example:
-When solutions of sodium carbonate and iron(III) nitrate are
mixed, a precipitate of Fe(CO₃)₃ will be formed .
Guide Line in Precipitation
-With these guidelines we can predict precipitation reactions:
Example:
-When solutions of lead acetate and calcium chloride are mixed, a
precipitate of PbCl₂ will be formed.
Guide Line in Precipitation
Example:
Assign the insoluble compounds out of the
following compounds:
(a)Ca(OH)₂ , (b)Na₂SO₄, (c)K₂S, (d)PbCl₂,
(e)HgCl₂
Answer:
(a) and (d) only
Guide Line in Precipitation
Example:
One cup (about 240 g) of a certain clear chicken broth
yields 4.302 g AgCl when excess AgNO3(aq) is added to
it. Assuming that all the Cl– is derived from NaCl, what is
the mass of NaCl in the sample of broth?
Answer:
AgNO₃ + NaCl → AgCl + NaNO₃
1 mol
1 mol 1 mol
X
4.302/143.32
Moles of NaCl : x= 0.03 mol
Mass of NaCl = 0.03 x 58.5 = 1.755 g
Reactions Involving
Oxidation and Reduction
Oxidation: it is loosing one electron or more
(Loss of electrons) , and the positive increase
of the oxidation number.
Reduction: it is gaining one electron or more
(Gain of electrons), and the decrease of the
oxidation number.
Both oxidation and reduction must occur
simultaneously.
.
Historical: “oxidation” used to mean “combines with oxygen”;
the modern definition is much more general.
Oxidation Numbers
An oxidation number is the charge on an ion, or a
hypothetical charge assigned to an atom in a
molecule or polyatomic ion.
Examples: in NaCl, the oxidation number of Na is +1,
that of Cl is –1 (the actual charge).
In CO2 (a molecular compound, no ions) the oxidation
number of oxygen is –2, because oxygen as an ion
would be expected to have a 2– charge.
The carbon in CO2 has an oxidation number of +4
(Why?)
Rules for Assigning Oxidation Numbers
- The oxidation number of any atom in the neutral state is 0
- The oxidation number of any compound is 0 and it is equal
to the sum of the oxidation numbers of the atoms in the
formula.
- The sum of the oxidation numbers of the atoms in an ion is
equal to the charge on the ion. Such as NH₄+
- In compounds, the group 1A metals all have an oxidation
number of +1 and the group 2A metals all have an
oxidation number of +2.
- In binary ionic compounds, the oxidation number of
fluorine is –1.( F, Cl, Br, I )
- In compounds, hydrogen has an oxidation number of
+1.(except in the Hydrides like CaH₂ , Hydrogen has an
oxidation number -1 ).
Rules for Assigning Oxidation Numbers
-In most compounds, oxygen has an oxidation number of
–2. ( except in hydrogen peroxide and its compounds
where oxygen has oxidation number -1, such as H₂O₂ ,
Na₂O₂ )
-In binary compounds with metals, group 7A elements have
an oxidation number of –1, group 6A elements have an
oxidation number of –2, and group 5A elements have an
oxidation number of –3.
- The maximum oxidation number of a nonmetal is equal to the group
number.
For nitrogen, +5.
For sulfur, +6.
For chlorine, +7.
The minimum oxidation number is equal to the (group number – 8).
Examples in assigning the oxidation number
Example:
What are the oxidation numbers assigned to the under lined
atoms of the following:
(a)KClO4
(b) Cr2O72–
(c) CaH2
Answer:
(a)K+1,O-2,
so Cl = +7
(d) Na+-O-O-Na+ just like H2O2
(b) 2Cr + ( 7x-2 ) = -2
2Cr-14 = -2
2Cr = +12
Cr = 6
(d) Na2O2
(e) Fe3O4
(e) Fe3O4 = FeO+ Fe2O3
Fe=2 ,Fe=3
(c) Ca = +2 , so H = -1
Oxidizing and Reducing Agents
An oxidizing agent causes another substance
to be oxidized.
The oxidizing agent is reduced.
A reducing agent causes another substance to
be reduced.
The reducing agent is oxidized.
Mg + Cu2+  Mg2+ + Cu
What is the oxidizing agent? What is the
reducing agent? Mg  Mg²⁺ + 2eoxidation
Cu²⁺ +2e = Cu
reduction
Oxidation Numbers of Nonmetals
The maximum
oxidation number of
a nonmetal is equal
to the group
number.
For nitrogen, +5.
For sulfur, +6.
For chlorine, +7.
The minimum
oxidation number is
equal to the (group
number – 8).
Activity Series
of Some Metals
In the activity series, any metal above
another can displace that other metal.
Mg metal can
react with …
… Cu2+ ions to
form Cu metal.
Will lead metal react
with Fe3+ ions?
Will iron metal
dissolve in an acid
to produce H2 gas?
Oxidation–Reduction
OxidationReduction
Example:
Balance the following equation using the redox method:
FeSO4 + KMnO4 + H2SO4 → K2SO4 + MnSO4 + H2O+Fe2 (SO4) 3
Answer :
To balance the equation , let’s display first the oxidation and the reduction
in the equation :
- then we have to multiply each oxidized or reduced element by the
number of oxidation change of the other, then we balance the
equation:
Oxidation–Reduction
OxidationReduction
- then we have to multiply each oxidized or reduced element by the
number of oxidation change of the other, then we balance the
equation:
-Then :
Typical Example
Example:
Balance the following equation using the redox method:
HNO 3 + Zn → Zn(NO3) 2 + NO + H2O
Answer:
1- We have to assign the elements where the oxidation and reduction
are occurred, we notice that nitrogen element changed from oxidation
number 5 to oxidation number 2 , and zinc changed from oxidation
number 0 to oxidation number 2.
2- we have to multiply the number of change of the zinc by the
nitrogen monoxide NO , and the number of change of the nitrogen by
the zinc nitrate Zn(NO3) 2
Typical Example
Oxidation–Reduction
3- we have to multiply the zinc in the left side of the equation by 3 to
get it equal to the zinc in the right side of the equation, and since the
total number of the nitrogen atoms in the right side of the equation is
8 we should multiply nitric acid by 8.
4- we have to multiply the water in the right side of the equation by
four to get the number of hydrogen atoms equal to the number of
hydrogen atoms in the left side.
Class Practice Problem
Balance the following equations: •
(a) Na(s) + H2O(l)
2Na(s) + 2H2O(l)
(b) Al(s) + HCl(aq)
2Al(s) + 6HCl(aq)
(c) C2H4(g) + O2(g)
C2H4(g) + 3O2(g)
NaOH(aq) + H2(g)
2NaOH(aq) + H2(g)
AlCl3(aq) + H2(g)
2AlCl3(aq) + 3H2(g)
CO2(g) + H2O(l)
2CO2(g) + 2H2O(l)
Half-Reactions
In any oxidation–reduction reaction, there are two
half-reactions
Oxidation takes place when a species loses
electrons to another species
Cu(s) 
Cu2+(aq) + e–
Reduction takes place when a species gains
electrons from another species
Zn2+(aq) + e–  Zn(s)
Redox Rxn
EOS
Half-Reaction Method of
Balancing Redox Equations
Separate an oxidation–reduction equation into two halfequations, one for oxidation and one for reduction
Cu(s) 
Cu2+(aq) + e–
Ag+(aq) + e–  Ag(s)
oxidation
reduction
As with all chemical equations, one must have
mass AND charge balance
EOS
Half-Reaction Method of
Balancing Redox Equations
Balance the atoms and the electric charge in each
half-reaction
Cu(s) 
Cu2+(aq) + 2 e–
2 Ag+(aq) + 2 e–  2 Ag(s)
oxidation
reduction
Adjust the coefficients in the half-equations so that
the same number of electrons appears in each halfequation
EOS
Half-Reaction Method of
Balancing Redox Equations
Add together the two adjusted half-equations to
obtain an overall oxidation–reduction equation
Cu(s) 
Cu2+(aq) + 2 e–
2 Ag+(aq) + 2 e– 
oxidation
2 Ag(s) reduction
Cu(s) + 2 Ag+(aq) + 2 e–  Cu2+(aq) + 2 e– + 2 Ag(s)
Write the net balanced chemical equation …
Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s)
EOS
A Qualitative Description of
Voltaic Cells
- A voltaic cell uses a spontaneous redox reaction to
produce electricity.
- A half-cell consists of an electrode (strip of metal or
other conductor) immersed in a solution of ions.
This Zn2+ becomes
a Zn atom.
Both oxidation and
reduction occur at the
electrode surface, and
equilibrium is reached.
This Zn atom leaves the
surface to become a Zn2+ ion.
Important Electrochemical Terms
- An electrochemical cell consists of two half-cells with the
appropriate connections between electrodes and
solutions.
-Two half-cells may be joined by a salt bridge that permits
migration of ions, without completely mixing the
solutions.
-The anode is the electrode at which oxidation occurs.
- The cathode is the electrode at which reduction occurs.
- In a voltaic cell, a spontaneous redox reaction occurs and
current is generated.
- Cell potential (Ecell) is the potential difference in volts
between anode and cathode.
Ecell is the driving force that moves electrons and ions.
A Zinc–Copper
Voltaic Cell
… the electrons
produced move
through the wire …
Zn(s) is oxidized to
Zn2+ ions, and …
Positive and negative
ions move through the
salt bridge to equalize
the charge.
… to the Cu(s) electrode,
where they are accepted
by Cu2+ ions to form
more Cu(s).
Reaction: Zn(s) + Cu2+(aq)  Cu(s) + Zn2+(aq)
Activity Series of
Some Metals
In the activity series, any metal
above another can displace that
other metal.
Mg metal can
react with …
… Cu2+ ions to
form Cu metal.
Will lead metal react
with Fe3+ ions?
Will iron metal
dissolve in an acid to
produce H2 gas?
Cell Diagrams
- A cell diagram is “shorthand” for an electrochemical cell.
- The anode is placed on the left side of the diagram. (oxidized)
element
- The cathode is placed on the right side.
(reduced one)
- A single vertical line ( | ) represents a boundary between phases,
such as between an electrode and a solution.
- A double vertical line ( || ) represents a salt bridge or porous barrier
separating two half-cells.
Standard Electrode Potentials
Since an electrode represents
only a half-reaction, it is not
possible to measure the
absolute potential of an
electrode.
The standard hydrogen
electrode (SHE) provides a
reference for measurement
of other electrode potentials.
The SHE is arbitrarily assigned
a potential of 0.000 V.
Standard Electrode Potentials
(cont’d)
-The standard electrode potential, E°, is based on the
tendency for reduction to occur at an electrode.
- E° for the standard hydrogen electrode is arbitrarily
assigned a value of 0.000 V.
- All other values of E° are determined relative to the
standard hydrogen electrode.
- The standard cell potential (E°cell) is the difference
between E° of the cathode and E° of the anode.
E°cell = E°(cathode) – E°(anode)
Criteria for Spontaneous
Change in Redox Reactions
- If Ecell is positive, the forward reaction is spontaneous.
- If Ecell is negative, the forward reaction is
nonspontaneous (the reverse reaction is _____).
- If Ecell = 0, the system is at equilibrium.
- When a cell reaction is reversed, Ecell and DG change
signs.
Measuring the Standard Potential
of the Cu2+/Cu Electrode
The voltmeter reading
and the direction of
electron flow tell us
that …
… Cu2+ is more easily
reduced than H+, by
0.340 volts.
Standard
hydrogen
electrode
Cu2+ + 2e  Cu
E° = +0.340 V
Measuring the Standard Potential
of the Zn2+/Zn Electrode
The voltmeter reading
and the direction of
electron flow tell us
that …
… Zn2+ is harder to
reduce than H+, by
0.763 volts.
Standard
hydrogen
electrode
Zn2+ + 2e  Zn
E° = – 0.763 V
Voltic Cells Examples
Example:
Will copper metal displace silver ion from aqueous
solution? That is, does the reaction
Cu(s) + 2 Ag+(1 M)  Cu2+(1 M) + 2 Ag(s)
occur spontaneously from left to right?
{Ag+ + e --------Ag}……….reduction
{ Cu ---------- Cu+2 + 2e}……oxidation
E = Ecathod - E anode
= E Ag+/Ag - E Cu+2 /Cu
= 0.800v – 0.340v = 0.460 v because the E positive
So displacement is spontaneous
Typical Voltic Cell Example
Draw the short hand diagram of the cell which does the
reaction:
Cu(s) + 2 Ag+(1 M)  Cu2+(1 M) + 2 Ag(s)
If the electrode potential of Ag⁺/Ag=0.800v and the electrode
potential of Cu/Cu⁺⁺=0.340v :
1- Calculate the cell potential of the cell
2-Does the reaction occur spontaneously from left to right?
Answer:
12E = E cathode - E anode
E= E Ag+/Ag - E Cu+2 /Cu
E= 0.800v – 0.340v = 0.460 v
3- because the E positive value, So displacement is
spontaneous
Other Secondary Cells
The nickel–cadmium (NiCd) cell uses a cadmium
anode and a cathode containing Ni(OH)2.
- A NiCd cell can be recharged hundreds of times. It
produces 1.2 V (a Leclanché cell produces 1.5 V).
Nickel–metal hydride cells (NiMH) use a metal alloy
anode that contains hydrogen.
- In use, the anode releases the hydrogen, forming
water. Like the NiCd cell, a NiMH cell produces 1.2 V.
Lithium-ion cells use a lithium–cobalt oxide or lithium–
manganese oxide material as the anode. The
electrolyte is an organic solvent containing a dissolved
lithium salt.
- Many modern laptop computers and cellular
phones use lithium-ion cells.