Chemistry Lab 2010
Presenter:
John R Kiser, MS
Hickory Regional Director
State Supervisor, Chemistry Lab
Introductions
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Topics for 2010: Solutions and Redox
Regional vs State Topics
Safety Requirements – Long Sleeve Shirt!
Must bring calculator!
Need to know topics
• Formula Writing/Nomenclature
• Mole & Stoichiometry Calculations
Solution Terminology
• Solution: Homogeneous mixture
• Solvent: Component in greater/greatest amount
• Solute: Component(s) in lesser/least amount
Factors that influence solubility
• Polarity of Solute and Solvent –
“Like Dissolves Like”
Polar solutes dissolve in polar solvents
Nonpolar solutes dissolve in nonpolar solvents
Nonpolar solutes do not dissolve well in polar solvents
• Temperature
Solubility of most solids in water increases with temperature
Solubility of gases in water decreases with temperature
• Gas Pressure
As the pressure of a gas above a solution increases, the
solubility of the gas in the solution increases. (Henry’s Law)
Sweet Tea and Soft Drinks
Amounts of Solute in Solution
• Saturated: The maximum amount of solute is dissolved in the
solvent
• Unsaturated: Less than maximum amount of solute is
dissolved in the solvent
• Supersaturated: More than the maximum amount of solute is
dissolved in solvent
• To obtain a supersaturated solution, you heat solution until all
solute dissolves. Carefully and slowly cooling the solution
keeps all the solute dissolved in solvent.
• Solubility curves show maximum amount of solute that can be
dissolved in 100 mL (sometimes 100 g) of water at a particular
temperature. Above curve = supersaturated, Below curve =
unsaturated
Saturation
• Pay attention to units of solubility on Y axis!
• Suppose a saturated solution of sodium nitrate is prepared at
60 oC in 200 mL of water. The solution is quickly cooled to
20 oC. What mass of sodium nitrates crystallizes during
cooling?
Units of Concentration
Moles of solute
Molarity (M) =
Liters of solution
Mass of component
Mass Percent =
Total mass of solution
Moles of solute
Molality (m) =
Mass of solvent (kg)
Copyright © 2010 Pearson
Prentice Hall, Inc.
x 100%
Units of Concentration
Assuming that seawater is an aqueous solution of NaCl, what is its molarity?
The density of seawater is 1.025 g/mL at 20 °C, and the NaCl concentration is
3.50 mass %. 3.50 mass % = 3.50 grams of salt in 100.00 grams of solution
Assuming 100.00 g of solution, calculate the volume:
1 mL solution
100.00 g solution
x
1.025 g solution
1 L solution
x
= 0.09756 L solution
1000 mL solution
Convert the mass of NaCl to moles:
3.50 g NaCl
x
1 mole NaCl
= 0.0599 moles NaCl
58.4 g NaCl
Then, calculate the molarity:
0.0599 moles NaCl
= 0.614 M NaCl
0.09756 L solution
Units of Concentration
In the previous example, what was the MOLALITY (m) of sodium chloride in
seawater? Assume seawater contains only sodium chloride and water.
Calculate the mass of water (solvent) in kg:
100.00 g solution – 3.50 grams NaCl (solute) = 96.50 grams water (solvent) = 0.09650 kg
Convert the mass of NaCl (solute) to moles:
3.50 g NaCl
x
1 mole NaCl
= 0.0599 moles NaCl
58.4 g
Then, calculate the molality:
0.0599 moles NaCl
= 0.621 m NaCl
0.09650 kg Solvent
Concentrations - ppm
Mass of component
Parts per million (ppm)=
(Mass based) Total mass of solution
x 106
50 ppm means a solution contains 50 grams of solute in 106
grams of solution (50 mg in 1 kg solution)
In dilute aqueous solutions at 25 oC, ppm is also equivalent to
mg solute in 1 L solution
Solving For Unknown Concentration:
Titrations (Volumetric Analysis)
In a titration a solution of accurately known concentration is added
gradually added to another solution of unknown concentration until
the chemical reaction between the two solutions is complete.
Indicator – substance that changes color at (or near) the
equivalence point
Equivalence point – the point at which the reaction is complete
Sometimes called stoichiometric point
Endpoint – The point at which the indicator changes color
Slowly add
reactants
UNTIL
the indicator
changes color
4.7
Steps for Solving Titration Problems
Be sure you have the correct balanced equation before
beginning!
STEP 1 Determine moles of starting compound
STEP 2: Determine moles of desired compound
STEP 3: Solve the problem
Example Problem: Titration of Citric Acid in Fruit Juice
State Level Question, 2010, commonly missed at many
regionals!
3 NaOH (aq) + H3C6H5O7 (aq) → 3 H2O (l) + Na3C6H5O7 (aq)
Solving for Unknown Concentration:
Lambert-Beer
• Lambert-Beer Law: A = εbc
A = Absorbance (unitless)
b = path length (cm)
c = concentration (M)
ε = Molar absorbtivity (constant, units M-1 cm-1)
Lambert-Beer or Beer’s Law Plot:
Provided all absorbance measurements are made on the
same spectrophotometer with the same cell, a graph of
absorbance vs. concentration will be linear.
Lab activity at State in 2010!
Commonly missed: Incorrect graphing, use of SpectroVis,
and “connecting the dots” on the graph.
Solving for Unknown Concentration:
Density
• Density of solution increases as solute
concentration increases.
• The plot of density of solution versus
concentration of solution should be linear.
• Can be used to solve for an unknown
concentration.
Using Concentrations to Find
Molar Mass: FP Depression
• Adding a solute to a solvent decreases the freezing point
∆Tf = kf*m
∆Tf = decrease in freezing point
kf = Freezing point constant (1.86 oC m-1 for water)
m = molality (mole solute per kg solvent)
Assumes ideal behavior and that solute is NOT ionic.
How to use to find molar mass of solute:
• Use ∆Tf and kf to find molality of solution
• Use mass of solvent to find moles solute present
• Mass solute dissolved divided by moles solute gives the molar
mass of solute!
Freezing Point Depression Example
When 2.50 grams of a covalent compound is dissolved in 0.100
kg of water, the freezing point is determined to be
-0.750 ⁰C. What is the molar mass of the compound?
(Assume Ideal Behavior)
Molality =
∆Tf
=
0.750 ⁰C
= 0.403 m
Kf
1.86 ⁰C m-1
0.100 kg water * 0.403 moles solute = 0.0403 moles solute
1.000 kg water
Molar mass= 2.50 grams solute
= 62.0 grams per mole
0.0403 moles solute
Oxidation Reduction (Redox)
Fundamental Concepts
• Oxidation is
Loss of electrons, gain of O, loss of H
• Reduction is
Gain of electrons, loss of O, gain of H
Mnemonic Devices:
LEO the lion goes GER!
OIL RIG
The species being reduced is the oxidizing agent
The species being oxidized is the reducing agent
Activity Series of Elements
• Usually lists the best reducing agent (most easily
oxidized) at top.
• Metal above reacts with ion below.
• Sample lab activity: The metal that reacts with
everything else goes at the top; the ion that reacts
with everything goes (as element, not ion) at bottom
• Aluminum Demonstration
Determining if a species is oxidized or
reduced: Oxidation Numbers
Rules above take precedence over rules below!
• An atom in its elemental state has an oxidation number of 0.
Oxidation number of H in H2 = 0
• An atom in a monatomic ion has an oxidation number
identical to its charge.
Oxidation number of Fe3+ = +3
• The total sum of oxidation numbers in a polyatomic ion is
equal to the charge. Sum of oxidation numbers in a neutral
molecule is 0. Pay attention to subscripts!
•
In a compound or polyatomic ion:
1A metal, ox #= +1; 2A metal, ox # = +2; 3A Metal, Ox # = +3
H, ox # = -1 if bonded to metal or Boron; H, ox # = +1 if
bonded to another nonmetal
Oxygen, ox# = -2 (EXCEPT IN PEROXIDES, -1)
Fluorine, ox # = -1.
Other halogens (if written to right in formula), ox # = -1
• Other atoms not on this list can be deduced by following the
rules above
Finding oxidation number of Cl in ClO4 –
Oxidation number of O = -2
Cl + 4*-2 = -1
Cl – 8 = -1
Cl = +7
If oxidation number increases, oxidation is taking place
If oxidation number decreases, reduction is taking place
Redox Half Reactions
In many complex redox reactions, H+ (or OH-) and H2O
are involved in the reaction and may not be obvious
at first.
Therefore, there is a systematic method to balancing
complex redox reactions.
First, separate into two half reactions, one for
oxidation, one for reduction.
Total Reaction:
Cl- + Cr2O72- →
Cl2 + Cr3+
Half Reactions:
Cl- → Cl2
and Cr2O72- → Cr3+
Balancing Half Reactions
•
•
•
•
•
Using Coefficients, balance all atoms BUT H and O: Cr2O72- →2 Cr3+
Balance O by adding H2O
Cr2O72- → Cr3+ + 7 H2O
Balance H by adding H+
14 H+ + Cr2O72- → 2 Cr3+ + 7 H2O
Balance charge by adding e- 14 H+ + Cr2O72- + 6e- →2Cr3+ + 4 H2O
Number of electrons should correspond to change in oxidation
number (keeping number of atoms in mind too!)
• No electrons in final answer, so multiply one (or both) half reaction(s)
by a number so that electrons are equal.
2 Cl- → Cl2 + 2 e2 Cl- → Cl2 + 2 ebecomes 6 Cl- → 3 Cl2 + 6 e-
Adding Half Reactions Together
6 Cl→
14 H+ + Cr2O72- + 6e- →
3 Cl2 + 6 e2Cr3+ + 7 H2O
Be sure to cancel out electrons, water, and H+ that appears on
both sides
6 Cl- + 14 H+ + Cr2O72- → 3 Cl2 + 2 Cr3+ + 7 H2O
Be sure to double check charges and numbers of atoms!
Electrons should not be left over!
If you are balancing a half-reaction in basic solution, add OH- to
both sides, convert H+ to H2O, and cancel out!
Example Problem
Balance the following half reaction in basic solution:
Oxidation or reduction?
OCl - → Cl-
Galvanic Cells
What happens when a redox reaction is spread out
over 2 beakers?
Oxidation at anode (-), reduction at cathode (+)
Electrons flow from anode to cathode
Standard Reduction Potential (Eored)
Eocell =Eo cathode – Eoanode
Positive Eocell means the reaction is spontaneous
More positive Eo means a more favored reaction
O Standard Conditions, 298 K, 1 atm, 1 M solutions
All these values are measured with respect to 2H+/H2
(These Eo values are technically Eoreduction. Eooxidation can
also be written for reverse reactions)
Shorthand Notation for Galvanic
Cells
Anode half-reaction:
Cathode half-reaction:
Overall cell reaction:
Zn2+(aq) + 2e-
Zn(s)
Cu2+(aq) + 2e-
Cu(s)
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
Salt bridge
Anode half-cell
Cathode half-cell
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
Electron flow
Phase boundary
Copyright © 2010 Pearson
Prentice Hall, Inc.
Phase boundary
Example Problem
• For the following Galvantic cell, determine the cathode,
anode, and Eocell
Ni (s) | Ni 2+ (aq) || Cu2+(aq) | Cu(s)
The Effect of Concentration on Cell E
What happens when conditions are not standard?
Nernst Equation:
E = E° -
RT
ln Q
nF
or
E = E° -
2.303RT
log Q
nF
or
E = E° -
0.0592 V
log Q
in volts, at 25°C
n
Q = Reaction Quotient (from Equilibrium topics)
Setup is just like an equilibrium constant, except the system is
not at equilibrium. Products on top, reactants on bottom.
Coefficients become exponents. Solutions in M, gases in atm,
pure liquids and solids omitted.
Consider a galvanic cell that uses the reaction:
Cu(s) + 2Fe3+(aq)
Cu2+(aq) + 2Fe2+(aq)
What is the potential of a cell at 25 °C that has the following ion
concentrations?
[Fe3+] = 1.0 x 10-4 M
[Cu2+] = 0.25 M
[Fe2+] = 0.20 M
Batteries
Lead Storage Battery
Anode:
Pb(s) + HSO4-(aq)
Cathode:
PbO2(s) + 3H+(aq) + HSO4-(aq) + 2e-
Overall:
Pb(s) + PbO2(s) + 2H+(aq) + 2HSO4-(aq)
Copyright © 2010
Pearson Prentice Hall,
Chapter 17/33
PbSO4(s) + H+(aq) + 2e-
PbSO4(s) + 2H2O(l)
2PbSO4(s) + 2H2O(l)
Batteries
Dry-Cell Batteries
Leclanché cell
Copyright © 2010
Pearson Prentice Hall,
Chapter 17/34
Fuel Cells
Hydrogen-Oxygen Fuel Cell
Electrolysis and Electrolytic Cells
Electrolysis: The process of using an electric current to bring about chemical
change.
Electrolysis of Molten Sodium Chloride
Anode:
2Cl-(l)
Cathode:
Overall:
Copyright © 2010
Pearson Prentice Hall,
Inc.
2Na+(l) + 2e2Na+(l) + 2Cl-(l)
Cl2(g) + 2e-
2Na(l)
2Na(l) + Cl2(g)
Electrolysis and Electrolytic Cells
Cathode has negative charge, connected to negative terminal on
battery; anode has positive charge, connected to positive terminal.
Reduction still at cathode, Oxidation still at anode
Electrolysis of Molten Sodium Chloride
Electrolysis and Electrolytic Cells
Electrolysis of Water
Anode:
2H2O(l)
O2(g) + 4H+(aq) + 4e-
Cathode: 4H2O(l) + 4e-
2H2(g) + 4OH-(aq)
Overall:
2H2(g) + O2(g) + 4H+ + 4OH-(aq)
6H2O(l)
Electrolysis of Aqueous Solutions
Electrolysis of NaI
• Consider reduction of water and reduction of cation
Na+ (aq) + e- → Na (s)
Eored = - 2.71 V
2 H2O (l) + 2e- → H2 (g) + 2 OH- (aq) Eored = - 0.83 V
Less negative (or more positive) reduction is preferred!
• Consider oxidation of water and oxidation of anion
2 H2O (l) → O2 (g) + 4 H+ (aq) + 4 e- Eo ox = -1.23 V
2 I- (aq) → I2 (s) + 2 eEoox = -0.54 V
Less negative (more positive) oxidation preferred!
(Note: reversing from Eored table. If reading directly from
table, without reversing oxidation, look for more negative
Eo red)
• Overvoltage makes predictions difficult if both
competing reactions have similar Eo values.
• In electrolysis of aqueous NaCl, even though
oxidation of water is less negative, Cl – is oxidized
to Cl2 gas.
• Poorly understood, mainly due to kinetic factors
and concentration effects.
• Prediction is difficult if values are close on table.
• Rules of Thumb for electrolysis of solutions
Look on left side of SRP table. If cation is higher
than water, cation will be reduced.
Look to right side of SRP table. If anion is below
water, anion will be reduced
Quantitative Aspects of
Electrolysis
Charge(C) = Current(A) x Time(s)
1 mol eMoles of e- = Charge(A) x
96,500 C
Faraday constant
Example Problem
How much Ca will be produced in an electrolytic cell of molten CaCl2
if a current of 0.452 A is passed through the cell for 1.5 hours?
19.8
Lab Time!
• Lab #1 – Beer’s Law
• Lab #2 – Redox Titration
• Lab #3 – Setup of a Galvanic cell and determining EO cell
and Eo for a half reaction
Thank you!
Please email me at jkiser@wpcc.edu
For follow-up questions, concerns, etc..