Notes 8
Battery in your car. What does it do? How does it work?
Lead-acid storage battery
Take it apart, look at components.
One electrode - battery terminal is connected to spongy lead-filled grids
other terminal connected to lead(IV) oxide-filled lead grids
Both bathed in sulfuric acid
Reaction taking place in the cell
Pb(s) + 2H+ (aq) + 2HSO4- (aq) + PbO2(s) - > <- 2PbSO4(s) + 2H2O(l)
This can be written in terms of two half reactions:
1) Pb(s) + HSO4- -> PbSO4 + H+ +2e2) PbO2 + 3H+ + HSO4- + 2e- -> PbSO4 + 2H2O
Note that an electron is given up in rxn 1 and replaced used up in rxn 2
If we control or use the electron movement, we generate a current through a device like a
radio. Since there is resistance to driving the current, then there must be some driving
force to push the electron through the external circuit. We suspect that a spontaneous
rxn. essentially must be able to drive the electron, and a potential difference would drive
the current, and we expect that a spontaneous rxn must have some potential difference
associated with it.
Example:
Is the rxn above spontaneous under standard conditions?
The lead-acid storage battery is one type of a class of reaction cells known as a galvanic
or a voltaic cell.
Galvanic / voltaic cells are one type of an even more general classification called an
Electrochemical Cell
A typical electrochemical cell consists of two half cells connected together by a salt
bridge so that an oxidation reduction reaction may proceed!
In a Galvanic or Voltaic cell the rxn is harnessed to provide electrical work!A typical
electrochemical cell consists of two half cells connected together.
One of these half cells represents the _____________ half reaction and the other
half cell represents the ___________________ half reaction.
Remember how we determine which half reaction is the oxidation half reaction and
which is the reduction half reaction.
Assign oxidation numbers to the elements in each molecule.
Do this by using the rules in Table ____________
Rules:
1. The oxidation number of an atom in a pure elemental substance is zero.
2. The oxidation number of an atom in a monatomic ion equals the charge on the ion.
3. The oxidation number of oxygen is -2 i most of its compounds, except peroxides where
it its -1 (example the O in H2O2, hydrogen peroxide has an oxidation number of -1)
4. The oxidation number of hydrogen is +1 in most of its compounds. (The oxidation
number of hydrogen is -1 in binary compounds with a metal such as NaH or CaH2
5. The oxidation number of fluorine is -1 in all if its compounds. Each of the other
halogens (Cl, Br, I) has an oxidation number of -1 in binary compounds, except when the
other element is another halogen above it in the periodic table or the other element is
oxygen.
6. The sum of the oxidation numbers of the atoms in a compound is zero. The sum of the
oxidation numbers of the atoms in a polyatomic ion equals the charge on the ion.
In practice you will usually have a compound that contains elements whose oxidation
numbers you can assign based on rules 1-5. This in combination with rule 6 will allow
you to get the oxidation #s of other elements in the compound.
Again, oxidation numbers allow one to figure out which is the oxidation half reaction and
which is the reduction half reaction
Reduction half is the half reaction in which there is a reduction or decrease in the
oxidation number of some species or there is a gain of electrons by some species.
Oxidation half is the half reaction in which there is an increase in the oxidation number of
some species or loss of electrons by some species
Remember that since oxidation and reduction must both occur in a redox reaction, in a
sense the oxidation half reaction forces the reduction half rxn. to happen AND
the reduction half reaction forces the oxidation half rxn. to happen.
THUS the species containing the element being oxidized is called the reducing agent
and the species containing the element being reduced is called the oxidizing agent.
Examples:
What is the oxidation number of chlorine in each of the following:
a) HClO4
b) ClO3-
c) Cl2
Combustion reaction like:
2C4H10(g) + 13O2(g) -> 8CO2(g) + 10H2O(g)
or
Fe(s) + 3O2(g) -> 2Fe2O3(s)
Examples
Assign oxidation numbers to each atom in each of the following eqn.
What is the oxidizing agent and the reducing agent for the reaction?
2C4H10(g) + 13O2(g) -> 8CO2(g) + 10H2O(g)
or
Fe(s) + 3O2(g) -> 2Fe2O3(s)
Remember that oxidation-reduction reactions often involve charged species and for that
reason sometimes it is difficult to balance the oxidation-reduction reactions or redox
reactions.
We need to be able to balance oxidation reduction reactions.
Use two methods:
1) The Half Rxn Method:
2) The Oxidation Number Method
Many different methods can be used for balancing redox reactions. In aqueous solutions,
the half-reaction method often works well since it focuses on the half-reactions which
individually could be paired up in other reactions. The different procedures used vary
slightly because of personal preference
Half Reaction Method
This depends on if the aqueous solution is acidic or basic. The half reaction method
depends on being able to separate the eqn. into the two half reactions as we mentioned
before using the oxidation numbers.
1) A).Illustrate the process by considering the oxidation of a solution of iron(II) sulfate to
iron(III) by employing potassium dichromate (which is reduced to chromium(III) in a
sulfuric acid solution.
1. Write the overall equation in an ionic form if it is not already in that form.
K+ + Cr2O72- + Fe2+ + H+ + HSO4- -> K+ + Cr3+ + Fe3+ + HSO4- + H2O
2. Reduce this equation to an unbalanced net ionic eqn.
Cr2O72- + Fe2+ + H+ -> Cr3+ + Fe3+ + H2O
3. Pick out the elements which are oxidized and reduced and write their chemical species
in separate oxidation and reduction half-reactions.
Reduction half - oxidation number is reduced
Reduction: Cr2O72- -> Cr3+
Oxidation half - oxidation number increases
Oxidation: Fe2+ -> Fe3+
4. Supply coefficients (as required) for each half reaction so that the same number of
atoms of the element oxidized or reduced appear on both sides of the half-reaction.
Red: Cr2O72- -> 2Cr3+
Ox: This half-reaction already has mass balance on oxidized elements
5. Add enough electrons to the appropriate side of the half-reaction to balance the change
in the oxidation state.
Red: Cr2O72- + 6e- -> 2Cr3+
Ox: Fe2+ -> Fe3+ + e-
If oxygen is not involved in the half-reaction, the equation for the half-reaction will be
balanced at the completion of step 5.
6. If oxygen is involved in the half-reaction, write the formula for hydrogen ions on the
side of the equation which has the electrons and place a coefficient in front of it which
will give charge balance.
Red: Cr2O72- + 6e- + 14H+ -> 2Cr3+
Ox: charge already balanced!
7. Now add a number of water molecules to the side of the equation opposite to the H+
ions to make the hydrogen balance.
Red: Cr2O72- + 6e- + 14H+ -> 2Cr3+ + 7H2O
Ox: no hydrogen present (already balanced half rxn.)
When this step is completed, the half reaction should be balanced with respect to both
mass and charge. Mass balance may be checked by verifying that the number of oxygen
atoms is the same on both sides of the equation.
8. Now the two half-reactions can be combined into the overall reaction. Each halfreaction must be multiplied by an appropriate coefficient before combination such that
the number of electrons lost in the oxidation half-reaction is equal to the number of
electrons gained in the reduction half-reaction
Red:
1 [ Cr2O72- + 6e- + 14H+ -> 2Cr3+ + 7H2O]
Ox:
6 [Fe2+ -> Fe3+ + e- ]
_________________________________________________________
Overall: 6Fe2+ + Cr2O72- + 14H+ -> 6Fe3+ + 2Cr3+ + 7H2O
1) B) In Basic solution, the same procedure may be used to balance redox reactions
taking place in basic solution except of steps 6 and 7. The following steps should be used
in their place.
6. If oxygen is involved in the half-reaction, write the formula for hydroxide ions on the
side of the equation opposite the electrons and place a coefficient in front of the formula
which will give charge balance.
7. Add sufficient number of water molecules to the side of the equation opposite to the
OH- ions to make the number of hydrogen atoms balance. Again, mass balance may be
checked by verifying that the number of oxygen atoms is the same on both sides of the
equation.
An alternate way to get the balanced eqn. in basic solution is simply to balance the
eqn. as if the soln. was acidic and then add enough hydroxide ions to both sides of the
eqn. to convert the H+ to water, leaving water and hydroxide.
Examples
Practice by balancing the following equations.
a. In acidic solution: BiO3- + Mn2+ -> BiO+ + MnO4-
b. In basic solution: Al + NO3- -> [Al(OH)4]- + NH3
2) The Oxidation Number Method
This method may be used for more complex reactions. The oxidation number methods is
useful since it is focused on the chemical changes involved.
The key to the oxidation number method is to realize that the net change in the total of all
oxidation numbers must be zero. That is, any increase in oxidation number for the
oxidized atoms must be matched by a corresponding decrease in the oxidation number for
the reduced atoms.
Consider the unbalanced net ionic eqn. in acidic solution
MnO4- + Br- -> Mn2+ + Br2 (acidic solution)
One can use these series of steps to balance a redox reaction in acidic solution by the
oxidation # method;
1. Write the unbalanced net ionic eqn.
2. Balance the equation for all atoms other than H and O.
3. Assign oxidation numbers to all atoms.
4. Decide which atoms have changed oxidation number and by how much.
5. Find the total net increase in oxidation number for the oxidized atoms and the total net
decrease in oxidation number for the reduced atoms. Multiply the net increase and the
net decrease by suitable numbers so that the two values become equal.
6. Balance the equation for oxygen by adding water to the side with less O, and then
balance for hydrogen by adding H+ to the side with less H. Check the answer by making
sure that the equation is balanced both for atoms and charge.
For basic solutions the steps are the same as for acidic solution, but OH- must be added at
the final step to neutralize any H+ ions that appear in the equation.
Example
Use the oxidation number method to give the balanced equation for the following in basic
soln:
Cr2O72- + Fe2+ -> Cr3+ + Fe3+
Now that we have a better idea of how to balance redox rxns let us return to
electrochemical cells.
Two types
1. Voltaic or Galvanic - Oxidation-Reduction Rxn occurs spontaneously
2. Electrolytic - redox rxn is not spontaneous but useful
Consider the following:
1) A solution of zinc sulfate with a piece of copper metal in it
A solution of copper sulfate with a piece of Zn metal in it.
A solution of zinc sulfate with a piece of Zn metal in it.
A solution of copper(II) sulfate with a piece of Cu metal in it.
From the observations in 1) write down the reaction that is taking place.
Cu(NO3)2(aq) + Zn(s) -> Cu(s) + Zn(NO3)2(aq)
Write the net ionic eqn. Is this a redox reaction?
Write down the half reactions. Label which is oxidation and which is reduction.
Is this rxn. spontaneous? Does it happen without outside influences, no human
intervention?
Remember G = wmax and G is negative for a spontaneous process
So wmax = negative value meaning work is out, the system can do work!
Can we harness the work? Basically this system is a redox reaction which is just a
movement of electrons.
We can harness the potential for this redox rxn to perform electrical work.
Electrical work = charge x potential difference
Electrical power = charge/time x potential difference
Now consider the second case 2):
If we separate the Cu2+ ions from the Zn, physically, and then connect the two, they have
a potential to perform work. This work manifests itself as a potential difference or a
voltage between the container containing the Zn(s) and the container containing the
Cu(aq) solution.
Think of a higher electrical potential as a higher electrical pressure (higher potential to
perform work) much like water at a higher height can gain more velocity in moving down
to a lower height and this will turn an turbine faster to generate more electricity.
So practically what we need to do is to control the flow of electrons that are moving from
the Zn to the Cu2+. We have put them in two different containers which stops the flow.
So we're in control now!
If we want to allow the electron flow to occur, the direction of electron flow must be
from Zn to Cu2+, BUT charge must also flow from the Cu to the Zn2+ as well. This is
done in the form of anions in the solution.
Must have a way for electrons to flow and then for anions also to flow. Can the electrons
flow through the solution?
To complete the circuit, we could allow the electrons to flow by connecting the metal
electrodes in the cell together with a ______________.
And to allow the anions to flow we use a ___________ _________________.
See figure showing Danielle Cell with salt bridge. Shows electron/charge movement.
One can measure the potential of the cell to do work by connecting each part of the cell to
a voltmeter with high internal resistance. Measure the potential difference between the
electrodes.
Potato Battery?
The MAX potential difference between the electrodes is referred to as the electromotive
force, the EMF of the cell or the cell potential, Ecell
We have a battery! Called a _____________ cell.
Potato battery?
A battery has a positive and negative terminal?
Now if we look at the way the electrons move spontaneously from the Zn to the Cu
electrode.
The _____________ electrode is the negative electrode AND
The _____________ electrode is the positive electrode.
since electrons flow from negative to positive.
Define the anode as the electrode where __________________ occurs.
Zn -> Zn2+ + 2e-
Define the cathode as the electrode where __________________ occurs.
Cu2+ + 2e- -> Cu
Most generally we call this an electrochemical cell, a system consisting of electrodes that
dip into an electrolyte and in which a chemical reaction either uses or generates an
electric current.
A charged up battery is an example of which type of electrochemical cell?
For a voltaic or galvanic cell, the cell generates may generate an electric current due to
the potential difference caused by the spontaneous redox reaction.
Do we use Potato Batteries?
Consider voltaic cells we use everyday!
Some nomenclature:
In a voltaic cell:
half cell - portion of electrochemical cell where half reaction takes place
anode - electrode at which ____________ occurs.
cathode - electrode at which ____________ occurs.
cell reaction - the net reaction that occurs n the voltaic cell.
Electrons flow from anode to cathode so anode is negative and cathode is positive
(voltaic or galvanic cell)
There is a short hand notation for designating a particular voltaic cell
1) Write the anode on the left side and the cathode on the right side.
2) Indicate salt bridge with double vertical line, ║.
3) Indicate phase boundary in half cells by single vertical line │.
4) cell terminals are at the extreme ends of this notation
So in the Zn - Cu2+, Danielle cell with a salt bridge reaction the cell notation would be:
Zn(s)│Zn2+(aq) ║ Cu2+(aq) │ Cu(s)
the terminals are the Zn(s) and the Cu(s)
When a half rxn. involves a gas - an inert metal or material serves as a terminal and the
electrode surface on which the reaction occurs (Pt or graphite is often used)
For instance:
2H+(aq) + 2e- -> H2(g)
So hydrogen electrode written as a cathode is:
H+(aq) │ H2(g) │ Pt
for the anode it would be written:
Pt │ H2(g) │ H+(aq)
The full notation should also give the concentration of ions or gas present in the solution.
For instance in the Danielle Cell, if the solutions were one molar, the cell notation would
be:
Zn(s)│Zn2+(1M) ║ Cu2+(1M) │ Cu(s)
Example: Write the notation for a cell in which the electrode reactions are:
A) 2H+(aq) + 2e- -> H2(g) (1 atm)
B) Zn(s) -> Zn2+(aq) + 2e- (1M soln)
Identify the anode and cathode half rxns.
Alkaline dry cell:
Overhead
Zn(s) + 2OH-(aq) -> Zn(OH)2 + 2e-
(this would occur at the _____________)
2MnO2(s) + H2O(l) + 2e- -> Mn2O3(s) + 2OH- (aq) (this occurs at the ______________)
The EMF, cell potential is 1.5 V. This works well in the cold and under current drain.
Note that when you use your IPOD or toy or radio or ?
work is performed by the current flow through the IPOD or toy? This means that the
potential that can be maintained by the battery is a bit ___________ than the EMF since
the current is flowing slowly.
Other batteries that we use include:
Lead Acid Storage Cell. We saw this previously.
The half rxn: Pb(s) + HSO4- -> PbSO4 + H+ + 2e- is the _____________
AND
half rxn: PbO2(s) + 3H+(aq) + HSO4-(aq) + 2e- -> PbSO4(s) + 2H2O is the ___________
Nickel-Cadmium Cell, Nicad batteries -
Cd(s) + 2OH-(aq) -> Cd(OH)2(s) + 2eNiOOH(s) + H2O(l) + e- -> Ni(OH)2(s) + OH-(aq)
Nicads are used in drills, toys, razors etc.
Note: The Lead acid and Nicad batteries are rechargeable batteries
Another type of "battery" that differs in that it has a constant supply of energetic fuel is a
FUEL CELL
The hydrogen-oxygen or methane oxygen fuel cell:
overall is combustion hydrogen H2(g) + O2(g) -> 2H2O (l)
CH4(g) + 2O2(g) -> CO2(g) + 2H2O
O2(g) + 2H2O(l) + 4e- -> 4OH-(aq)
2H2(g) + 4OH-(aq) -> 4H2O(l) + 4e-
(_____________ electrode)
(_____________ electrode)
used in the space shuttle for poser and for drinking water?
Why might the fuel cell be more efficient than just regular combustion of hydrogen or
methane?
G = wmax
What happens when a battery runs down?
For the Danielle cell, the potato battery, when the iron is used up then the cell no longer
functions!
Rechargeable Batteries are recharged by causing the non-spontaneous reverse reaction to
occur by putting a large voltage across the battery terminals and causing the electrons
(current) to flow in the opposite direction.
The non-spontaneous reaction proceeds because we put in some ___________ WORK!
This is a form of a more general type of electrochemical cell
Called an Electrolytic Cell.
An electrolytic cell is an electrochemical cell in which an electric current drives an
otherwise non-spontaneous rxn.
The process of producing a chemical change in an electrolytic cell is called electrolysis.
Electrolysis is an important process used in industry for the production of many
important materials and for metal plating.
Downs Cell - commercial electrochemical cell used to obtain sodium metal and chlorine
gas by the electrolysis of molten sodium chloride.
Na+(l) + e- -> Na(l)
Cl-(l) -> ½ Cl2(g) + e-
( _______________ half reaction)
( _______________ half reaction)
In electrolytic cells, the anode is still associated with oxidation
and the cathode is still associated with reduction.
Figure:
The anode is connected to the positive terminal of the external current source AND
the cathode is connected to the negative terminal of the external current source.
Calcium chloride is sometimes added to the sodium chloride so that this process can run
at lower temperatures. (How does that work?)
Example:
Write half reactions for electrolysis of the following molten compounds:
KCl
KOH
Electrolysis in aqueous solutions is similar except that the water can be reduced or
oxidized in half reactions, other species involved include H2O, O2, H+, and OH-, Only H2
and O2 involve a change of oxidation state.
Example: NaCl Solutions:
Na+(aq) + e- -> Na(s)
OR
2H2O + 2e- -> H2 + 2OHCATHODE
___________________________________________
Cl-(aq) -> ½ Cl2(g) + eOR
2H2O(l) -> O2(g) + 4H+(g) + 4eANODE
IT TURNS OUT that the electrolysis involves:
2H2O + 2e- -> H2 + 2OH(Cathode) negative terminal
Cl (aq) -> ½ Cl2(g) + e
(Anode) positive terminal
COULD THIS HAVE BEEN PREDICTED
YES! How? See in a bit!
Electrolytic cells also allow plating of one metal with another. Zinc coatings are often
used to protect steel. We'll see why shortly.
To get the zinc plating:
The steel is placed in a bath of zinc salts and mad the ____________ in an electrolytic
cell. The half reaction here then is:
Zn2+ + 2e- -> Zn(s)
Electrolysis may also be used to purify some metals. Slabs of impure copper serve as
anodes, with pure copper sheets serving as cathodes.
During electrolysis, copper(II) ions leave the anode slabs and plate on the pure copper
cathode sheets.
Quantifying the amount of material deposited
1831 Michael Faraday showed that the amount of substances released at the electrodes
during electrolysis are related to the charge that flowed in the electric circuit.
THERE IS A STOICHIOMETRIC RELATIONSHIP
Essentially if you know how many _______________ are moving from one electrode to
the other, and you know how many electrons are involved in the redox reaction, then you
can figure out how much product is formed from the # of electrons that moved.
Example:
For the electrolysis of molten NaCl(l)
Na+ + e- -> Na(l)
Cl- -> ½ Cl + eWhen one mole of electrons reacts with sodium ions, _________ mole of sodium metal is
deposited at one electrode, and _____ ________ mole of chlorine gas evolves at the other
SO one can measure the # of electrons to find the amount of product.
1) Use the fact that one Faraday 9.65 x 104 Coulombs = charge on one mole of electrons
2) Electric charge = electric current (Amperes (Amps = C/s)) x time elapsed.
So measure the current and the elapsed time during the electrolysis.
Use the charge in Coulombs with 1mol e- = 96,500 Coulombs to find the moles of
electrons transferred. and thus the moles of product formed
Example:
A constant current deposits 365 mg of silver in 216 min from an aqueous silver nitrate
solution. What is the current?
Example:
How many grams of oxygen are liberated by the electrolysis of water after 185s with a
current of 0.0565 Amps
So one can calculate how much substance is formed in an electrolysis process if the
current and time elapsed are known.
Talked now about the two types of Electrochemical Cells, but look closely at how one
knows if a particular redox rxn. is SPONTANEOUS!
Nomenclature reminder:
In a voltaic cell:
half cell - portion of electrochemical cell where half reaction takes place
anode - electrode at which ____________ occurs.
cathode - electrode at which ____________ occurs.
cell reaction - the net reaction that occurs n the voltaic cell.
Electrons flow from anode to cathode so anode is negative and cathode is positive
(voltaic or galvanic cell)
There is a short hand notation for designating a particular voltaic cell
1) Write the anode on the left side and the cathode on the right side.
2) Indicate salt bridge with double vertical line, ║.
3) Indicate phase boundary in half cells by single vertical line │.
4) cell terminals are at the extreme ends of this notation
So in the Zn - Cu2+, Danielle cell with a salt bridge reaction the cell notation would be:
Zn(s)│Zn2+(aq) ║ Cu2+(aq) │ Cu(s)
the terminals are the Zn(s) and the Cu(s)
When a half rxn. involves a gas - an inert metal or material serves as a terminal and the
electrode surface on which the reaction occurs (Pt or graphite is often used)
For instance:
2H+(aq) + 2e- -> H2(g)
So hydrogen electrode written as a cathode is:
H+(aq) │ H2(g) │ Pt
for the anode it would be written:
Pt │ H2(g) │ H+(aq)
The full notation should also give the concentration of ions or gas present in the solution.
For instance in the Danielle Cell, if the solutions were one molar, the cell notation would
be:
Zn(s)│Zn2+(1M) ║ Cu2+(1M) │ Cu(s)
Example: Write the notation for a cell in which the electrode reactions are:
A) 2H+(aq) + 2e- -> H2(g) (1 atm)
B) Zn(s) -> Zn2+(aq) + 2e- (1M soln)
Identify the anode and cathode half rxns.
As noted previously, the cell EMF is a measure of the driving force of the cell
reaction. This reaction occurs in the cell as separated half-rxns.
Thus the cell EMF is composed of a contribution from the anode (whose value depends
on the ability of the oxidation half reaction to lose electrons), an oxidation potential, Eox
and a contribution from the cathode (the ability to gain electrons in reduction), Ered
Thus
Ecell = Eox + Ered
Consider the Danielle Cell:
Zn(s)│Zn2+(1M) ║ Cu2+(1M) │ Cu(s)
Need Eox for Zn(s) and Ered for Cu2+
Values for the half reactions are tabulated for standard conditions as std. reduction
potentials.
Can find Ered for Cu2+ + 2e- -> Cu directly in table but not Eox for Zn -> Zn2+ + 2eHowever can find Zn 2+ + 2e- -> Zn (Ered) We can use this since Eox = - Ered
So can state a set of rules for Determining the Eocell from tabulated Eored.
Rules to get Eocell
1) The oxidation potential for the oxidation is = - reduction potential for the reverse rxn.
2) Since the electrode potential is intensive, its value is not dependent on the amount of
the species present in the cell.
Thus Ered for the half cell rxn: Cu2+ + 2e- -> Cu(s) is the same numerical value as for the
half cell rxn: 2Cu2+ + 4e- -> 2Cu(s)
(This is different than what we are used to doing when employing Hess's Law for H and
G.
3) The standard EMF of a voltaic cell operating under standard state conditions (solute
conc. each 1M gas pressures each 1atm. Temp. specified (usually 25 is specified) is
E0cell = Eored (reduction half rxn) + Eoox (oxidation half rxn)
= Eored (reduction half rxn) - Eored (oxidation half rxn)
Example:
What is Eocell for the Danielle Cell?
The standard reduction potential table works very similar to the Activity Series Table
Metals at the top of the table could reduce metals below them. They are the best reducing
agents. This is because they are related to their reducing strength.
REMEMBER THAT THE BEST REDUCING AGENT is the one that is most easily
oxidized. So the metal at the top of the table is the most easily oxidized metal and the
one hardest to reduce.
Zn reduces Cu2+ This is spontaneous!!!
Zn is above Cu on the activity series or on the Std. Reduction Potential Table.
NOTE that the std reduction table are arranged from the most negative value to the most
positive value of the std reduction potential. Zn has a more negative std. reduction
potential than Cu.
Apparently a more negative reduction potential makes a better reducing agent.
AND note that if the reduction potential is more negative then the oxidation of Zn should
be more positive. That is Zn -> Zn2+ + 2e- is more positive that Cu -> Cu2+ + 2e-. So the
fact that Zn has a more positive oxidation makes it a good reducing agent as well.
SUMMARIZING:
Lithium is at the top of both the metal activity series table and the standard electrode
reduction potential table meaning it has the most negative standard reduction potential.
This means that Li is the best reducing agent since it is most readily oxidized. Its
oxidation potential is Eoox = -Eored so Eoox(Li) = +3.04V for the half rxn:
Li(s) -> Li+(aq) + eTHIS LEADS ONE TO BELIEVE that the more positive the Eocell, the more spontaneous
the rxn!
General Rule:
The more positive the value of Eocell the more spontaneous the reaction!
Since this is the case then Eocell must be related to Go
remember G = wmax
For a voltaic cell, this is electrical work!
Electrical Work = - n F E
n = # of electrons transferred int he balanced rxn.
F = Faradays Constant 96,500 Coulombs/mole eThus
Go = -nFEocell
A negative Go means the reaction is spontaneous, and thus a positive Eocell indicates that
the reaction is spontaneous.
Example
Use Eocell to determine Go at 25o for the rxn:
Sn2+(aq) + 2Hg2+(aq) -> Sn4+(aq) + Hg22+(aq)
Example
Use Gorxn to get Eocell for the rxn: Mg(s) + Cu2+(aq) -> Mg2+(aq) + Cu(s)
Since Go = -RT lnKeq then also
-nFEocell = -RT ln Keq
Eocell = RT/(nF) ln Keq
So Keq =
Example:
For the half reactions: Fe(s) -> Fe2+(aq) + 2e- and 2e- + Sn4+ -> Sn2+(aq)
a) Determine the overall balanced reaction:
b) Determine if the reaction is spontaneous
c) Determine the value of Keq
So far we've talked about Eocell at std. concentration or std. pressure conditions. and const
temp.
However know that as battery runs down its potential changes.
Since running down of a battery involves concentration changes in the reactants, then the
EMF or Ecell must also depend on conc. Also know batteries depend on temp.
Scientist named NERNST studied the conc. dependence of Ecell.
Since Grxn = Gorxn + RTlnQ
= -nFEcell
AND
Gocell = -nFEocell
Then:
-nFEcell = -nFEocell + RTlnQ
Ecell = Eocell - RT lnQ (This is called the NERNST EQN!)
Q is the reaction quotient like in equilibrium chapter
Q=
At 25oC and plugging in F and R on finds
Ecell = Eocell - 0.0592/ n x log10Q in Volts at 25oC
n = # electrons in balanced reaction
can be used for half reactions as well as overall reactions
Examples:
What is the EMF of the following cell at 25oC?
Zn(s) │ Zn2+(0.200M) ║ Ag+(0.00200M) │Ag(s)
Example
What is the potential of the copper electrode at 25oC?
Cu2+(0.0350M) │Cu(s)
Determination of pH
pH can also be obtained very accurately from EMF measurements using the Nernst Eqn..
Use soln. you know the pH of as the electrolyte for a hydrogen electrode. Connect this to
a standard zinc electrode giving
Zn│Zn2+(1M) ║ H+(test soln M) │H2(1atm) │Pt
Zn(s) + 2H+(test soln.) -> Zn2+(1M) + H2(1 atm)
EMF depends on the hydrogen ion concentration in test soln. by Nernst Eqn.
Q = Zn2+ PH2 / [H+]2
Std EMF = 0.176 so
Ecell = 0.76 - 0.0592/2 log(1/[H+]2 = 0.76 + 0.0592 log[H+]
Where [H+] is hydrogen ion conc. of the test soln.
pH = -log[H+]
so Ecell = 0.76 - 0.592 x pH
pH = (0.76 - Ecell ) / 0.0592
in Volts