Transcript Topic b Types of Chemical Reactions

TOPIC C
DRIVING FORCES, ENERGY
CHANGES, AND
ELECTROCHEMISTRY
Driving forces
• Evidence for chemical change can manifest itself in a
number of ways.
• The formation of precipitate
• a change of energy in the form of heat or light
• a color change
• the formation of a gas
• are all observations that can be made in the laboratory
• Sometimes these events are called driving forces, but are
they chemical or physical?
• an interruption in the inter-molecular forces ….. change is
physical
• a re-arrangement of the intra-molecular forces ….. change is
chemical.
• Practice:
• 1. Discuss the change in forces and bonds when water boils.
• 2. Discuss the change in forces and bonds when water
decomposes into its elements.
Gas producing reactions
• Several reactions produce gases as one of the products.
• These are worth learning, as well as the subsequent tests
for the gases produced.
• General gas producing reactions
• 1. Acid + Metal  Salt + Hydrogen
• For example: Zn(s) + H2SO4(aq)  ZnSO4(aq) + H2(g)
• Test for gas: “Squeaky pop” with lighted splint
• 2. Acid + Carbonate  Salt + Water + CO2
• For example:
• H2SO4(aq) + CaCO3(s)  CaSO4(aq) + H2O(l) + CO2(g)
• Test for gas: Extinguishes a glowing or lighted splint
•
Turns lime water (Ca(OH)2) milky
• Specific gas producing reaction
• 1. The production of O2 by the decomposition of H2O2
•
(with MnO2 catalyst)
• 2H2O2(aq)  2H2O(l) + O2(g)
• Test for gas: Relights glowing splint.
Energy Changes
• When reactants undergo a chemical change – products
are formed.
• The reaction is considered a part of the universe called
the system
• Everything else is called the surroundings
• Energy of Universe = Energy of System + Energy of Surroundings
• The system will often undergo an energy change where it
will either:
• release energy to the surroundings (exothermic reaction)
• absorb energy from the surroundings (endothermic reaction)
• For an exothermic reaction
•
- temperature of the surroundings increases
• For an endothermic reaction
•
- temperature of the surrounding decreases
• These changes can be shown graphically in energy
diagrams
Electrochemistry
• When a metal comes into contact with a solution
containing its own ions an equilibrium is set up.
•
Mx+(aq) + xe-   M(s)
• Some reactive metals (like Mg) will lose electrons readily
•
The equilibrium lies to the left.
• A large number of electrons are released which collect on
the surface of the metal giving a negative charge
•
Mg2+(aq) + 2e-   Mg(s)
• Less reactive metals (Ag) show less tendency to ionize
•
equilibrium lies to the right
• Fewer electrons will collect on the metal and the charge
will be much less negative
• In fact, if the aqueous ions remove electrons from the
metal it will develop a positive charge.
• Ag+(aq) + e-   Ag(s)
• Non-metals can also be considered, for example:
•
H+(aq) + e-   ½H2(g)
• When an element is placed in a solution containing its
own ions - an electric charge will develop on the metal
• (In the case of a non-metal – an inert conductor is used)
• The charge is called the electrode potential
• The system is called a half-cell.
• The sign and size of the charge will depend on the ability
of the element to lose or gain electrons.
Electrochemical series and electrode
potentials
• Species that appear at the top of the series:
• gain electrons most readily
• have the most positive Eo values
• are easily reduced and (best oxidizing agents)
• Species that appear at the bottom of the series:
• lose electrons most readily
• have the most negative Eo values
• are easily oxidized and (best reducing agents)
Electrochemical cells (batteries)
• An electrochemical cell
• Generates electrical energy from a spontaneous Redox reaction.
• Connecting two half-cells that have different electrode potentials
forms an electrochemical cell (battery).
• A voltmeter is used to measure the voltage
• A salt bridge connects the two half-cells.
• Connected this way the reaction starts
• Stops immediately because charge builds up.
• MnO4- + 1e-  MnO42• Fe2+ + 2e-  Fe
H+
MnO4-
Eo = 0.56
Eo = -0.44
Fe+2
Galvanic Cell - uses a spontaneous redox reaction to
produce a current that can be used to do work.
Salt
Bridge
allows
current
to flow
H+
MnO4-
Fe+2
• Electricity travels in a complete circuit
• Oxidation occurs at the anode
• Reduction occurs at the cathode
Anode
Cathode
H+
MnO4-
Fe+2
Porous
Disk
H+
MnO4-
Fe+2
Overview in General
ee-
e-
e-
Anode
e-
Reducing
Agent
Cathode
e-
Oxidizing
Agent
Cell Potential
• Oxidizing agent pushes the electron.
• Reducing agent pulls the electron.
• The push or pull (“driving force”) is called the cell
potential Ecell
• Also called the electromotive force (emf)
• Unit is the volt(V) = 1 J/C (joule of work/coulomb of
charge) measured with a voltmeter
• A coulomb is the SI unit of quantity of electricity, (the
charge transferred in one second with a constant current
of one ampere)
0.76
e
-
Pt metal
Zn metal
Anode
Zn+2
SO4-2
1 M ZnSO4
Cathode
H+
Cl1 M HCl
H2 in
Standard Hydrogen Electrode
• This is the reference all
other oxidations are
compared to
• Eº = 0
• º indicates standard states of
25ºC,
1 atm, 1 M solutions.
H2 in
H+
Cl-
1 M HCl
Standard Reduction Potentials
• It is universally accepted that the half-reaction
potential for 2H+ + 2e- → H2 assigned a value of
zero volts.
• The overall cell potential (Ecell) is then assigned to
the other half reaction.
• This reaction is always written as a reduction
potential
• Table 17.1 has a list of the most common Standard
Reduction Potentials we will use.
Finding Cell Potentials using Standard
Reduction Potentials
• Fe3+(aq) + Cu(s)  Cu +2(aq) + Fe
2+(aq)
• Two half reactions:
 Fe3+ + e- → Fe2+
E o = 0.77 V
 Cu2+ + 2e- → Cu
Eo = 0.34 V
• Two rules apply:
 1. The half reaction with the largest potential is written as a
reduction, the other must be reversed (change its sign).
 2. The number of electrons lost must equal the number of electrons
gained (multiply half reactions).
 Do not multiply the standard potentials! They do not change!
• The total cell potential is the sum of the potential
at each electrode.
• Eocell
= Eo(cathode) + Eo(anode)
• Eocell = 0.77 V + (– 0.34 V) = 0.43 V
• We can look up reduction potentials in table
17.1.
Line Notation
•
solidAqueousAqueoussolid
• Anode on the leftCathode on the right
• Single line different phases.
• Double line porous disk or salt bridge.
• If all the substances on one side are aqueous, a
platinum electrode is indicated.
• For the last reaction
• Cu(s)Cu+2(aq)Fe+2(aq),Fe+3(aq)Pt(s)
Galvanic Cell


1)
2)
3)
4)
The reaction always runs spontaneously in the
direction that produced a positive cell potential.
Four things for a complete description.
Cell Potential
Direction of flow
Designation of anode and cathode
Nature of all the components- electrodes and ions
• Practice:
• Using the SERP table, write cell diagrams for the following
combinations of electrodes. Remember to include state
symbols and inert electrodes where appropriate
•
•
•
•
(i) Zn/Zn2+ and fluorine
(ii) Sn/Sn2+ and hydrogen
(iii) Cu/Cu2+ and Fe/Fe2+
(iv) Fe2+/Fe3+ and hydrogen
• Practice:
• A cell formed from a silver standard electrode and a
hydrogen standard electrode generates a voltage of +
0.80 V. Hydrogen is the more negative electrode and has
a value of 0.00 V.
• (a) Write the cell diagram and calculate the standard potential of
the silver electrode.
• (b) When the silver electrode is combined with a standard
aluminum electrode the voltage of the cell is +2.46 V and aluminum
is the more negative electrode.
• Write the cell diagram and calculate the standard
electrode potential of the aluminum electrode.
Potential, Work and DG
 ΔG = Gibbs Free Energy

– the energy available for the cells to do work
• As long as the cell has a potential (+Eo) it can do work
• As the cell runs is approaches equilibrium
• At Equilibrium, Eo = 0, ΔG = 0, and K can be calculated
The Relationship between
Gibbs Free Energy and Ecell
• Summarized by the expression:
•
ΔGo = -nFEo
• Where F = Faraday = 96,485 C / mol (coulombs / mols e-)
• and n = # of moles of electrons transferred
• Gibbs free energy can also be expressed as:
•
ΔGo = -RT lnK
(R = 8.31 J/K · mol)
• Setting the two equal to each other and solving for Eo
•
Eo = (RT / nF) lnK
Potential, Work and DG
• DGº = -nFE º
• If E º < 0, then DGº > 0 nonspontaneous
• if E º > 0, then DGº < 0 spontaneous
• In fact, reverse is spontaneous.
• Calculate DGº for the following reaction:
• Cu+2(aq)+ Fe(s) Cu(s)+ Fe+2(aq)
• Fe+2(aq) + e-Fe(s)
• Cu+2(aq)+2e- Cu(s)
Eº = 0.44 V
Eº = 0.34 V
• Possible combinations of K, E, and ΔGo leads to the
following conclusions:
K
Eo
ΔGo
Conclusion
 1
Positive
Negative
Spontaneous
cell reaction
=1
0
0
At equilibrium
<1
Negative
Positive
Reaction is
spontaneous in
the reverse
direction
Cell Potential and Concentration
• Qualitatively - Can predict direction of change in E from
LeChâtelier.
• 2Al(s) + 3Mn
+2(aq)  2Al+3(aq) + 3Mn(s)
• Eocell = 0.48v
• Predict if Ecell will be greater or less than Eºcell
•
If [Al+3] = 1.5 M and [Mn+2] = 1.0 M
•
if [Al+3] = 1.0 M and [Mn+2] = 1.5M
•
if [Al+3] = 1.5 M and [Mn+2] = 1.5 M
The Nernst Equation
• DG = DGº +RTln(Q)
• -nFEE = -nFEº + RTln(Q)
• E = Eº - RT ln(Q)
nF
 This is not commonly used !
• 2Al(s) + 3Mn+2(aq)  2Al+3(aq) + 3Mn(s)
•
Eº = 0.48 V
• Consider a cell at 25oC where:
 [Mn2+] = 0.50 M and [Al3+] = 1.5 M
• Use the Nernst equation on the next slide to solve.
The Nernst Equation
• E = Eº -
0.0591 log(Q)
n
used at 25oC
• As reactions proceed concentrations of products
increase and reactants decrease.
• The cell will discharge until it reaches
equilibrium.
• At this point: Q = K (the equilibrium constant) and
Ecell = 0
• At equilibrium, the components in the two cells
have the same free energy and ∆G=0.
• The Cell no longer has the ability to do work.
• Practice:
• If the reaction Zn(s) + Cu2+(aq)  Cu(s) + Zn2+(aq) is
carried out using solutions that are 5.0M Zn2+ and 0.3M
Cu2+ at 298 K, predict the effect on the voltage of the cell,
when compared to the voltage generated under standard
conditions.
Electrolysis
• Running a galvanic cell backwards.
• Put a voltage bigger than the potential and reverse the
direction of the redox reaction.
• Used for electroplating.
1.10
e-
e-
Zn
Cu
1.0 M
Zn+2
Anode
1.0 M
Cu+2
Cathode
e-
e-
A battery
>1.10V
Zn
Cu
1.0 M
Zn+2
Cathode
1.0 M
Cu+2
Anode
Steps:
• 1. Current and time → charge
• amps (C/s) x time (s) to get coulombs
• 2. Quantity of charge → mol e• coulombs (C) x 1/F (mol e- / C) to get mol e• 3. Moles of e- → mol of element
• mol e- x 1 mol of element / mols of e- (needed to form neutral
element from ion)
• 4. Moles of element → mass of element
• mole of element x molar mass / 1 mol
Calculating plating
• Have to count charge.
• Measure current I (in amperes)
• 1 amp = 1 coulomb of charge per second
•q=Ixt
• q/nF = moles of metal
• Mass of plated metal
 What mass of copper is plated out when a current of
10.0 amps is passed for 30.0 minutes through a solution
containing Cu2+.
 How long must 5.00 amp current be applied to produce
15.5 g of Ag from Ag+
Other uses
• Electroysis of water.
• Separating mixtures of ions.
• More positive reduction potential means the reaction proceeds
forward.
• For metals this is typically gaining electrons and forming the solid
metal – this removes the ion from solution.
• Ions with the more positive the reduction potential will “plate out”
first.