Download AP Ch. 20 Notes (2005)

Ch. 20: Electrochemistry
• Electrochemistry is the branch of chemistry that deals with
relationships between electricity and chemical reactions.
• We are going to study a type of reaction where electrons are
transferred between reactants.
• These reactions are called “oxidation-reduction reactions”, or redox
reactions.
•Before we begin studying redox reactions, we need to cover…
Oxidation Numbers
• Oxidation numbers allow us to keep track of the electrons gained and
lost during chemical reactions.
• The oxidation #, (or oxidation state), of an element in a compound is
a hypothetical charge based on a set of rules…
Rules for Assigning Oxidation Numbers
• Rule #1: An atom in its elemental form has an oxidation # of zero.
Examples: H2 , Ag, Br2, Pb…(Ox. #’s = zero for each element)
• Rule #2: Monatomic ions have an Ox. # equal to its charge.
Examples: K+ (Ox. # for K = +1); S2− (Ox. # for S = −2)
• Rule #3: Hydrogen is +1 when attached to nonmetals in a compound and −1 when
attached to metals in a “hydride”.
Examples: HCl (Ox. # for H= +1); NaH (Ox. # for H= − 1)
(* Since the compounds are electrically neutral, we can
then say that the Ox. # of Cl must be −1, and Na is +1.)
• Rule #4: Fluorine is −1 in all compounds, and Oxygen has an Ox. # of −2 in all
compounds except peroxides when it’s Ox. # is −1… O22−
Examples: Na2O… (Ox. # of oxygen = −2 while Na is +1)
H2O2 …(Ox. # of oxygen is −1 while H is +1)
• Rule #5: The sum of the Ox. #’s in a compound is zero, and for polyatomic ions,
the sum of the Ox. #’s equals the charge of the ion.
Example: HNO3…(H= +1, O= −2, so N = +5)
SO42−…(O= −2, so S= +6)
Redox Reactions
• Now we can explore a redox reaction in detail…
Consider the reaction of zinc with an acid:
Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)
• If we examine the oxidation state of zinc, we see that zinc started out
at zero and ended up at +2…it lost 2 electrons.
− The process in which a substance increases its oxidation state (by
losing electrons) is called “oxidation.”
• If we examine the oxidation state of hydrogen, we see that hydrogen
started out at +1 and ended up at zero…it gained an electron.
− The process in which a substance decreases its oxidation state (by
gaining electrons) is called “reduction.”
• Zinc reduced hydrogen, so zinc is called the reducing agent. (In other
words, the substance that is oxidized is the “reductant”.
• Hydrogen oxidized zinc, so hydrogen is called the oxidizing agent.
(In other words, the substance that is reduced is called the “oxidant”.
Redox Reactions
• The easiest way to remember the difference between reduction and
oxidation is…
“LEO goes GER”
Losing Electrons = Oxidation
Gaining Electrons = Reduction
Balancing Redox Reactions
• In order to balance redox reactions we need to remember the
following:
1) Conservation of Mass: the amount of each element present
at the beginning of the reaction must be present at the end.
2) Conservation of Charge: electrons are not lost in a chemical
reaction. They are transferred from one reactant to another.
• Half-reactions are a convenient way of separating oxidation and
reduction reactions.
Let’s look at an easy example…
Redox Reactions
Consider the reaction:
Sn2+(aq) + 2Fe3+(aq)  Sn4+ (aq) + 2Fe2+(aq)
• The oxidation half-reaction is:
Sn2+(aq)  Sn4+(aq) + 2e–
(LEO)
(*Note that electrons are shown as a product…they are lost.)
• The reduction half-reaction is:
2Fe3+(aq) + 2e–  2Fe2+(aq)
(GER)
(*Note that electrons are shown as a reactant…they are gained.)
Now we’ll look at one that is more complicated, but it is still simply a redox reaction.
Balancing Redox Reactions
• Consider the titration of an acidic solution of Na2C2O4 with
KMnO4...
+7 -2
+3 -2
+2
+4 -2
MnO4−(aq) + C2O42−(aq)  Mn2+(aq) + CO2(g)
• MnO4− is reduced to Mn2+…Manganese starts out at an oxidation
state of +7 and is reduced to +2.
• C2O42− is oxidized to CO2…Carbon starts out at an oxidation state
of +3 and ends up at +4.
• First, we have to write the 2 half-reactions:
MnO4−(aq)  Mn2+(aq)
(Reduction)
C2O42−(aq)  CO2(g)
(Oxidation)
• Then we balance each half-reaction by following these steps…
Balancing Redox Reactions
MnO4−(aq)  Mn2+(aq)
C2O42−(aq)  CO2(g)
a. First, balance the elements other than H and O…
(Nothing needs to be done yet for this half reaction….)
MnO4−(aq)  Mn2+(aq)
(Balance the carbon…)
C2O42−(aq)  2CO2(g)
b. Then balance O by adding water…
MnO4−(aq)  Mn2+(aq) + 4H2O(l)
(The oxygen is already balanced in this half-reaction…)
C2O42−(aq)  2CO2(g)
(Continued…)
MnO4−(aq)  Mn2+(aq) + 4H2O(l)
C2O42−(aq)  2CO2(g)
c. Then balance H by adding H+ (if in an acidic environment.)
8H+ + MnO4−(aq)  Mn2+(aq) + 4H2O(l)
(Nothing needs to be done yet to this half reaction…)
C2O42−(aq)  2CO2(g)
d. Finish by balancing charge by adding electrons….
8H+ + MnO4−(aq)  Mn2+(aq) + 4H2O(l)
(GER)
• The left side Mn is +7 while the right side Mn is +2. We add the
electrons to the left side of the equation…(How many are gained?)
5e− + 8H+ + MnO4−(aq)  Mn2+(aq) + 4H2O(l)
C2O42−(aq)  2CO2(g)
(LEO)
• The left side is −2 and the right side is 0. We put the electrons on
the product side which shows the electrons being “lost.”
C2O42−(aq)  2CO2(g) + 2e− (Each carbon lost one electron.)
(We are not done yet!!! )
• The half-reactions need to be multiplied so that the # of electrons
gained and lost is equal.
2 x [5e− + 8H+ + MnO4−(aq)  Mn2+(aq) + 4H2O(l)]
5 x [C2O42−(aq)  2CO2(g) + 2e−]
• Then you end up with…
10e− + 16H+ + 2MnO4−(aq)  2Mn2+(aq) + 8H2O(l)
5C2O42−(aq)  10CO2(g) + 10e−
• Now you add the reactions together and cancel out items wherever
possible…
16H+ + 2MnO4−(aq) + 5C2O42−(aq)  2Mn2+(aq) + 8H2O(l) + 10CO2(g)
• It is wise to double-check and see if everything is balanced,
including the charge!
What if the reaction were in a basic environment?
• Add [OH−] to both sides of the reaction to cancel out the [H+] ions…
16OH− + 16H+ + 2MnO4−(aq)+ 5C2O42−(aq)2Mn2+(aq) + 8H2O(l) +10CO2(g) + 16OH−
• This results in water being formed…
16H2O(l) + 2MnO4−(aq) + 5C2O42−(aq)  2Mn2+(aq) + 8H2O(l) + 10CO2(g) + 16OH−
• Now we can simplify the equation by canceling out 8H2O(l) that
appears on both sides of the equation…
8H2O(l) + 2MnO4−(aq) + 5C2O4 2−(aq)2Mn2+(aq)+10CO2(g) + 16OH−(aq)
Balancing redox reactions is one of the most important aspects in Ch. 20!
We will practice more examples later!
Voltaic Cells
• The energy released in a spontaneous redox reaction is
used to perform electrical work.
• Voltaic or galvanic cells are devices in which electron
transfer occurs via an external circuit.
• Voltaic cells are spontaneous.
Example: If a strip of Zn is placed in a solution of
CuSO4, Cu is deposited on the Zn and the Zn
dissolves by forming Zn2+.
•
•
•
Zn is spontaneously oxidized into Zn2+ by the Cu2+.
The Cu2+ is spontaneously reduced into Cu0 by the Zn.
The entire process is spontaneous!
Voltaic Cells
• In order for the redox reaction to do work, the halfreactions need to be separated, and the electrons need to
be able to flow from one half-reaction to the other.
• Voltaic cells consist of
–
–
–
Anode (where oxidation occurs): Zn(s)  Zn2+(aq) + 2e−
Cathode (where reduction occurs): Cu2+(aq) + 2e−  Cu(s)
Salt bridge (used to complete the electrical circuit): cations
move from anode to cathode, anions move from cathode to
anode.
• The two solid metals are the electrodes… [cathode, (+),
and anode, (–).]
• Next, we will look at a picture of this voltaic cell and
then we will analyze what is going on…
Voltaic Cells
• As oxidation occurs, Zn is converted to Zn2+ and 2e−.
These electrons flow towards the cathode where they are
used in the reduction reaction which turns Cu2+ into Cu0.
• We expect the Zn electrode, (the anode), to lose mass
and the Cu electrode, (the cathode), to gain mass.
• Electrons flow from the anode (−) to the cathode (+).
• Electrons cannot flow through the solution; they have to
be transported through an external wire.
•
Anions and cations move through a porous barrier or
salt bridge.
Voltaic Cells
Voltaic Cells
• Anions in the salt bridge move into the anode
compartment to neutralize the excess Zn2+ ions formed
by oxidation.
• In the same way, cations from the salt bridge move into
the cathode compartment to neutralize the excess
negative charge formed during reduction.
• Summary: Anode, (-), LEO, anions move towards anode
Cathode, (+), GER, cations move towards cathode
Electrons move from the anode to the cathode.
A way to remember them…
ANO
anode negative oxidation
CPR
cathode positive reduction
Voltaic Cells
The zinc anode gets smaller and the copper cathode gets larger.
ANO LEO
CPR GER
Voltage
• The flow of electrons from anode to cathode is
spontaneous. What is the “driving force”?
• Electrons flow from anode to cathode because the cathode
has a lower electrical potential energy than the anode.
• Potential difference: difference in electrical potential.
• The potential difference is measured in volts…
One volt (V) is the potential difference required to impart one
joule (J) of energy to a charge of one coulomb (C).
Volts=Joules/Coulomb
• A Coulomb of charge is ≈ 6.25 x 1018 electrons.
• One mole of electrons is defined as having 96,500 C of
charge. This is called a faraday, F…
1 F = 96,500 C/mole of e–
EMF
• Electromotive force (emf) is the force required to push
electrons through the external circuit.
• Cell potential: Ecell is the emf of a cell.
• This is known as the cell voltage.
• Ecell is > 0 for a spontaneous reaction.
• For 1 Molar solutions & 1 atm pressure for gases, at
25º C (standard conditions), the standard emf (standard cell
potential) is called Eºcell.
Example: For the zinc/copper voltaic cell: Eºcell = +1.10 V
• The emf of a cell depends on the particular cathode and
anode half-cells that are used.
Standard Reduction (Half-Cell) Potentials
• In order to determine the emf of a particular cell, you need to know
the Eºred of each half cell involved.
• Standard reduction potentials, Eºred , are measured relative to a
standard.
• We use the following half-reaction as our standard:
2H+(aq), (1 M) + 2e–  H2(g), (1 atm)
Eºcell = 0 V.
• This electrode is called a standard hydrogen electrode, (SHE).
• The SHE is assigned a standard reduction potential of zero.
• The Eºred for other half-reactions can be measured relative to the
SHE.
• The emf of any cell can then be calculated from all of the standard
reduction potentials that have been tabulated…
Eºcell = Eºred (cathode) – Eºred (anode)
SHE and Zinc Anode
Zinc is oxidized. Since the cell’s emf is + 0.76 V, that means the Eºred
for [Zn2+(aq) + 2e− Zn(s)] would be –0.76 V.
Eºcell
• Practice Problem: The standard emf for the following cell is 1.46 V.
In+(aq) + Br2(l)  In3+ (aq) + Br−(aq)
Calculate the Eºred for the reduction of In3+ to In+.
• First, let’s break this up into the two half-reactions…
(LEO ANO)
In+(aq)  In3+(aq) + 2e−
Eºred = ?
(GER CPR)
2e− + Br2(l)  2Br−(aq)
Eºred = +1.06 V
• Plug and Chug… E ºcell = Eºred (cathode) − Eºred (anode)
1.46 = 1.06 – [Eºred (anode) ]
Therefore… Eºred (anode) = – 0.40 V
• Note: This reaction is spontaneous. Also, In+ was oxidized in this
reaction, so you would expect the value for Eºred to be (−).
Eºcell
• Changing the coefficients of a half-reaction does not change Eºred.
Zn2+(aq), (1 Molar)  Zn(s)
Eºred = – 0.76 V
2Zn2+(aq), (1 Molar)  2Zn(s)
Eºred = – 0.76 V
• As you may suspect, the concentration will change Eºcell as we will
see later on in the notes.
• The cathode reaction (CPR) of a voltaic cell will always have the
more positive Eºred than the reaction at the anode (ANO).
• In essence, the greater driving force of the cathode half-reaction is
used to force the anode reaction to occur “in reverse” as an oxidation.
• The entire process ends up being spontaneous.
Eºcell
Practice Problem: A voltaic cell is based on the following two
standard half-reactions…
Cd2+(aq) + 2e−  Cd(s)
Sn2+ (aq) + 2e−  Sn(s)
• Which reaction takes place at the cathode and which is at the anode?
Look up the Eºcell for each in Appendix E…
Cd2+… – 0.403 V
Sn2+… – 0.136 V
• Since Sn2+ is “more positive”, then it will be reduced at the cathode.
• Therefore, Cd2+ will be oxidized at the anode.
• What is the value for Eºcell?
Eºcell = Eºred (cathode) – Eºred (anode)
Eºcell= –0.136 V – (– 0.403) = +0.267 Volts
• Note: The reaction had to be positive (and spontaneous) for a voltaic cell!!
Oxidizing and Reducing Agents
• We can use this table to
determine the relative
strengths of reducing (and
oxidizing) agents.
• The more positive Ered the
stronger the oxidizing agent
on the left.
• The more negative Ered the
stronger the reducing agent
on the right.
• We can use this to predict if
one reactant can
spontaneously oxidize
another.
Oxidizing and Reducing Agents
EMF and Free Energy
• How does emf relate to ∆G?
∆G = – nFE
where...
n = # of electrons transferred during the process
F = 96,500 C/mole
E = cell potential
• Since n and F are positive, if E > 0 then G < 0 and the reaction
would be spontaneous.
• If the reactants and products are in their standard states, then…
∆Gº = – nFEº
Effect of Concentration on Cell EMF
The Nernst Equation
• A voltaic cell will function until E = 0, at which point
equilibrium has been reached.
• The point at which E = 0 is determined by the
concentrations of the species involved in the redox
reaction.
• The Nernst equation relates emf to concentration…
E = Eº – RT lnQ
nF
or expressed as log base 10…
E = Eº – 2.303 RT log Q
nF
(See p.799 in the text for the derivation.)
The Nernst Equation
• When T = 25º C, the equation is simplified to…
E = Eº – 0.0592 V log Q
n
• Let’s see how this equation can be used…
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
At standard conditions, the Zinc/Copper cell has an emf of
+1.10 V. What is the voltage of the cell at 25º C when
[Cu2+]= 5.0 M and [Zn2+] = .050 M?
• Recall that Q = [products]/[reactants]. (Remember, pure solids and
liquids do not appear in the expression for Q!)
• Therefore… E= 1.10 V – 0.0592 V log [0.050]/[5.0]
2 e–
• Since Q < 1, then log Q < 0, and this will increase the voltage…
E = 1.16 Volts
Concentration Cells
• The Nernst equation shows us that using the same species
in the anode and cathode compartments of a cell but at
different concentrations then a voltage will be generated.
• For example…
• If the concentrations of [Ni2+] are unequal, a voltage will
be generated.
Concentration Cells
• At the anode, Ni(s)  Ni2+ increasing the concentration of
nickel ions in this compartment.
• At the cathode, Ni2+  Ni(s) and the concentration of
nickel ions in this compartment is reduced.
• Therefore…
Q = [Ni2+]dilute/[Ni2+]concentrated
• At 25º C…
E = 0 – 0.0592 log [0.001]/[1] = 0.0888 V
2 e−
• When the compartments reach equilibrium, emf = zero.
• The pH meter functions on this principle, and so does your
heartbeat!
Cell EMF and Chemical Equilibrium
• A voltaic cell is functional until E = 0 at which point equilibrium
has been reached. The cell is then “dead.”
• A system is at equilibrium when G = 0.
• From the Nernst equation, at equilibrium and 298 K (E = 0 Volts
and Q = Keq):
0 = Eº – RT ln Keq
nF
• This equation can be simplified at 25º C to…
log Keq = nEº/0.0592
• So you can calculate Keq of a redox reaction by knowing Eº or vice
versa!
Batteries
• A battery is a portable, self-contained
electrochemical power source consisting
of one or more voltaic cells.
• The (+) end of the battery is called the
cathode, and the (–) end is called the
anode.
• If you wanted to generate 9 V, you will
have to link together multiple cells in
series since no single redox reaction can
produce that much voltage.
• Primary cells cannot be recharged.
• Secondary cells are rechargeable.
Types of Batteries
• Lead-Acid Battery (12 V car battery):
Cathode: lead dioxide plate
PbO2(s) + HSO4−(aq) 3H+(aq) + 2e−  PbSO4(s) + 2H2O
Anode: (lead plate):
Pb(s) + HSO4−(aq)  PbSO4(s) + H+(aq) + 2e−
• The overall electrochemical reaction is
PbO2(s) + Pb(s) + 2SO42−(aq) + 4H+(aq)  2PbSO4(s) + 2H2O(l)
• Ecell = Ered (cathode) – Ered (anode)
Ecell = (+1.685 V) – (– 0.356 V)
Ecell = +2.041 V.
• Six of these cells are linked in series to make 12 V, & wood or glassfiber spacers are used to prevent the electrodes from touching.
Lead Acid Batteries
Lead-Acid Batteries
• Lead-Acid batteries can be recharged.
• [H2SO4] decreases as the battery discharges and increases as the cell
is charged back up while the reactions run in reverse.
• H2SO4 is much more dense than H2O, so the condition of the battery
can be monitored by testing the specific gravity, (density), of the
solution…
* specific gravity = 1.25 to 1.30 charged up and ready to go!
* specific gravity < 1.20 needs charging
• These batteries are called “wet cells” because the electrolyte is a
liquid.
Alkaline Batteries (Dry Cells)
• The most common nonrechargeable battery is the alkaline battery.
• Powdered zinc metal is immobilized in a gel in contact with a
concentrated solution of KOH.
• The electrolyte is a base, thus these batteries are alkaline.
•The electrolyte is a paste, hence the name “dry cell”.
• The reaction at the anode is:
Zn(s) + 2OH−(aq) Zn(OH)2(s) + 2e−
• The reaction at the cathode is the reduction of MnO2:
2MnO2(s) + 2H2O(l) + 2e−  2MnO(OH)(s) + 2OH−(aq)
• The cell potential of these batteries is 1.55 V at room temperature.
Alkaline Batteries (Dry Cells)
Other Batteries
• A common rechargeable battery is the nickel–cadmium (NiCad)
battery.
• The reaction at the cathode is:
2NiO(OH)(s) + 2H2O(l) + 2e−  2Ni(OH)2(s) +2OH−(aq)
• The reaction at the anode is:
Cd(s) + 2OH−(aq)  Cd(OH)2(s) + 2e−
• The cell potential of this battery is about 1.30 V at room temp.
• Cadmium is a toxic heavy metal.
• There are environmental concerns to be addressed with respect to
disposal of such batteries.
• Other rechargeable batteries have been developed.
- NiMH batteries (nickel–metal–hydride).
- Li–ion batteries (lithium–ion batteries)…(greater energy density)
Corrosion
• An example of an undesirable redox reaction is the
corrosion of metals.
• A metal is attacked by a substance in the environment and
converted to an unwanted compound.
• Consider the corrosion (rusting) of iron:
• Since Eºred(O2) > Eºred(Fe2+), iron can be oxidized by O2.
• Cathode:
O2(g) + 4H+(aq) + 4e−  2H2O(l)
Eºred = +1.23 V.
• Anode:
Fe(s)  Fe2+(aq) + 2e−
Eºred = −0.44 V.
Corrosion
• Dissolved oxygen in water usually causes the oxidation of
iron.
• The Fe2+ initially formed can be further oxidized to Fe3+,
which forms rust, Fe2O3·xH2O(s).
• Oxidation occurs at the site with the greatest concentration
of O2.
• Other factors to consider are the pH, presence of salts,
stress on the iron, and contact with other metals.
Corrosion of Iron
Preventing the Corrosion of Iron
• Corrosion can be prevented by coating the iron with paint or another
metal to keep it away from O2 and water.
• Another way to protect the iron is by a process called cathodic
protection.
• Coating iron with a layer of zinc, “galvanized iron”, can protect the
iron from corrosion even after the surface coat is broken.
Zn2+(aq) + 2e−  Zn(s)
Eºred = − 0.76 V
Fe2+(aq) + 2e−  Fe(s)
Eºred = − 0.44 V
• The standard reduction potentials indicate that Zinc is easier to
oxidize than iron, so the Zinc will be oxidized!
• Zinc is the “sacrificial anode” and is slowly destroyed. Iron acts as
the cathode where the oxygen is reduced.
• We can use something similar to protect underground pipelines.
• Often, Mg is used as a sacrificial anode…(Eºred = − 2.37 V)
Preventing the Corrosion of Iron
Preventing the Corrosion of Underground Pipes
Electrolysis
• It is possible to cause a nonspontaneous redox reaction to occur.
Such processes are driven by an outside source of electrical energy.
• These reactions are called “electrolysis reactions” and take place in
“electrolytic cells”.
• In voltaic and electrolytic cells, reduction occurs at the cathode, and
oxidation occurs at the anode.
• However, in electrolytic cells, electrons are forced to flow from the
anode to the cathode.
• In electrolytic cells the anode is positive and the cathode is negative.
• (In voltaic cells the anode is negative and the cathode is positive.)
• Here’s what an electrolytic cell looks like…
Electrolytic Cell
Electrolysis of Molten NaCl
• Example of Electrolysis:
Cathode (GER): 2Na+(l)+ 2e−  2Na(l)
Anode (LEO):
2Cl−(l)  Cl2(g) + 2e−
• A battery or some other source of direct current acts as an
electrical pump pushing electrons to the cathode and
pulling them from the anode.
• As Na(s) is produced at the cathode (–) , additional Na+
migrates in.
• As Cl2(g) is produced at the anode (+), additional Cl−
migrates in.
• Industrially, electrolysis is used to produce metals like
aluminum and gases like chlorine.
Electrolysis of Aqueous Solutions
• Do we get the same products if we electrolyze an aqueous solution
of the salt?
• No! Water complicates the issue.
• Example: Consider the electrolysis of NaF(aq):
Cathode (GER): Na+(aq) + e−  Na(s)
Eºred = − 2.71 V
Cathode (GER): 2H2O(l) + 2e−  H2(g) + 2OH−(aq) Eºred = − 0.83 V
• Thus water is more easily reduced than the sodium ion!
Anode (LEO): F−(aq)  F2(g) + 2e−
Eºred = +2.87 V
Anode (LEO): 2H2O(l)  O2(g) + 4H+(aq) + 4e−
Eºred = +1.23 V
• Thus it is easier to oxidize water than the fluoride ion.
• What you end up doing is electrolyzing water into H2(g) and O2(g)!
Electrolysis with Active Electrodes
• Active electrodes: electrodes that take part in electrolysis.
Example of active electrodes: electroplating
• Consider an active Ni electrode and another metallic electrode
(steel) placed in an aqueous solution of NiSO4…
Anode:
Ni(s)  Ni2+(aq) + 2e−
Eºred = – 0.28 V
Cathode:
Ni2+(aq) + 2e−  Ni(s)
Eºred = – 0.28 V
• Ni gets transferred from the anode to the cathode and the Ni plates
on the inert (steel) cathode.
• (Of course a small emf will be needed to “push” the reaction along.
• Electroplating is important in protecting objects from corrosion…
Example: Stainless steel
Electrolysis with Active Electrodes
Quantitative Aspects of Electrolysis
• We want to know how much material we obtain with
electrolysis.
• Consider the reduction of Cu2+ to Cu…
Cu2+(aq) + 2e−  Cu(s)
• 2 moles of electrons will plate 1 mole of Cu.
• Using this information as well as the following
conversion factors, we can calculate how much copper
can be obtained from electrolysis given a certain amount
of current…
1 mole of electrons = 96,500 Coulombs = 1 faraday
1 Ampere of current = 1 Coulomb/second …(1A = 1C/s)
Quantitative Aspects of Electrolysis
Practice Problem: How much copper (in grams) can be obtained
from electrolysis by using 3.5 amps for 5 hours?
• First, determine the amount of coulombs…
3.5C x 3600 s x 5 hr = 63,000 C
1 sec
1 hr
• Then convert coulombs to grams of copper…
1mole e− 1 mole Cu
63.5 grams 20.7 g of Cu
x
=
x
63,000 C x 96,500 C
−
2 moles e
1 mole Cu