Download Electrochemistry I: Electrochemical cells Reading: Moore chapter 19

Electrochemistry I: Electrochemical cells
Reading: Moore chapter 19, sections 15.1-15.5
Questions for Review and Thought: 6, 14, 16, 20, 24, 28, 30, 32
Key Concepts and Skills:
•
definition of anode, cathode, cell voltage, electrochemical cell, half-reaction, half
cell, standard hydrogen electrode, salt bridge, standard reduction potential,
electrode, voltaic cell
•
Balance redox equations in acidic media and basic media, identify oxidizing and
reducing agents in a redox reaction, determine which reaction occurs at the anode,
which reaction occurs at the cathode, construct an electrochemical cell
diagrammatically, using standard reduction potentials for half reactions, calculate
cell potential.
Lecture Topics:
I.
Oxidation-Reduction (Redox) Reactions
•A redox reaction involves a transfer of electrons from one element in a reactant toa
second element; one element undergoes an increase in oxidation number (i.e., an
oxidation process); another element undergoes a decrease in oxidation number (i.e. a
reduction process)
a reducing agent (reductant) causes the reduction of another substance while it itself is
oxidixed
an oxidizing agent (oxidant) causes the oxidation of another substance while it itself is
reduced.
Reduction: (a) decrease in oxidation state of an element; (b) gain of electrons by a
chemical substance (c) a gain of hydrogen by a chemical substance.
Oxidation: (a) increase in the oxidation state of an element (b) loss of electrons by a
chemical substance (c) a gain of oxygen by a chemical substance.
See additional handout for assigning oxidation numbers.
Electrochemistry is the study of the relationship between electron flow and redox
reaction.
Redox reactions are often signaled by the presence of an uncombined element in the
reactants or products, as well as by the presence of oxidizing agents and reducing agents,
and by changes in the oxidation number of elements.
+4
0
+3
+1
Example:
2MnO2(s) + H2(g) ==>Mn2O3 (s) + H2O
1e- + MnO2(s) ==>1/2 Mn2O3 (s)
reduction occurs
1/2 H2 (g) ==> 1/2 H2O + 1eoxidation occurs
• H2 is the reducing agent; it gets oxidized to H2O (0 => +1)
• MnO2 is an oxidizing agent; it gets reduced to Mn2O3 (+4 => +3)
An overall redox reaction can thus be thought of as the sum of two simultaneous halfreactions.
Example:
3 Br2 (aq) + 2 Al(s) ==> 2 Al3+ (aq) + 6Br-
2e- + Br2 (aq) ==> 2Brreduction half reaction
3+
Al(s) ==> Al + 3 e
oxidation half reaction
Br2is the oxidizing agents which itself is reduced to Br- (each Br2 molecule gains 2
electrons) ; Al is the reducing agent which itself is oxidized (each Al atom loses 3
electrons).
Note that in the balanced net reaction, no electrons appear. To combine the half
reactions, each must be multiplied by a factor so that the number of electrons cancels out
in the net reaction
3(2e- + Br2 (aq) ==> 2Br-)
2(Al(s) ==> Al3+ + 3 e-)
net: 3 Br2 (aq) + 2 Al(s) ==> 2 Al3+ (aq) + 6BrII.
Balancing Redox reactions in Acidic and Basic media
Strategy: for each half reaction: balance atoms first, then balance charge
For overall reaction: multiply half-reactions by integers so that upon adding
half reactions, electrons cancel
For reactions in acidic or neutral solution
1. Identify the species being oxidized and the species being reduced (use oxidation
#’s)
2. Separate the overall process into half-reactions, omitting H+ and H2O as reactants
and/or products: provide an oxidation half-reaction and a reduction half reaction.
3. For each half-reaction:
a. Balance all atoms other than H and O first
b. Balance oxygen by adding H2O to the side deficient in O
c. Balance hydrogen by adding H+ to the side deficient in H
d. Balance charge by adding e-‘s
4. If necessary, multiply half-reactions by integers to make electrons
gained=electrons lost.
5. Add half reactions, canceling like species.
6. For each H+, add H2O to both sides, and make each H+ + H2O into a H3O+.
7. Check charge and atom balance.
For reactions in basic solution
1. Follow the same procedure as for acidic and neutral solution to obtain an
overall reaction (steps 1-5).
2. Then, for every H+ that appears in the overall equation, add one –OH to BOTH
sides.
3. Convert H+ to H2O by combination with –OH.
4. Check charge and atom balance.
Example: Balance MnO4- + Zn(s) ==> Mn2+ + Zn2+ in acidic media.
Answer: 16H+(aq) + 5 Zn (s) +2 MnO4-(aq) ==> 5 Zn2+(aq) + 2 Mn2+ (aq) + 8 H2O(l)
Also: 16H3O+(aq) + 5 Zn (s) +2 MnO4-(aq) ==> 5 Zn2+(aq) + 2 Mn2+ (aq) + 24 H2O(l)
Net charge: +14
Net charge: +14
2Balance: Cr(OH)3(s) + Br2(aq) ==> CrO4 (aQ) + Br-(aq) in basic media.
Answer: 10 HO-(aq) + 2 Cr(OH)3 + 3Br2(aq) ==> 6 Br-(aq) + 2CrO42-(aq) + 8H2O(l)
Net charge: -10
Net charge: -10
III.
Electrochemical Cells
An electrochemical reaction is redox reaction designed to either produce or utilize
electrical energy. An electrochemical cell is an arrangement of an oxidizing agent and a
reducing agent pair in such a way that they can react only if electrons flow through an
outside conductor (a wire). Such cells are also called Voltaic or Galvanic cells.
Characteristics of Electrochemical cells:
Consider: Cu2+(aq) + Zn(s) ==> Cu(s) + Zn2+(aq)
• The overall redox process must be spontaneous in the forward (or reverse) direction.
• The two half-cells solutions are not in physical contact except via the conducting wire
and a salt-bridge;
• an electrode is the metal surface (Cu(s) and Zn(s) in this case) in a half-cell where
electrons enter or leave; the conducting wire is attached to the electrode.
• The electrode where oxidation takes place is called the anode (where negative charge is
produced): Zn(s) ==> Zn2+(aq) + 2e• The electrode where reduction takes place is called the cathode (where positive charge
is consumed): Cu2+ + 2e- ==> Cu (s)
• a salt bridge prevents charge build up in either compartment of the cell during
operation: it completes the circuit by allowing ions to flow into the half cells. It is a Ushaped tube filled with a salt solution (an electrolyte, like KNO3) and an aqueous gel with
porous plugs at the ends to prevent appreciable mixing with the half cell solutions except
to permit ion flow.
Phase boundary
Salt bridge Phase boundary
Shorthand:
Anode | anode electrolyte || cathode | cathode electrolyte
Thus:
Zn(s) | Zn2+ || Cu2+ | Cu(s)
IV.
Cell Voltage
Cell Voltage (Ε°cell) is a measure of how much work a cell can produce for each coulomb
of charge the reaction produces; it is a measure of the electrical potential energy
difference between two half-cell electrodes. Also called cell potential, cell electromotive
force, cell emf . The Standard cell voltage
•Cell voltage depends on concentration; standard conditions are defined for voltage
measurements: pure solids, liquids and gases, solutes at 1M concentration, @25°C: E°.
The Standard cell voltage is the cell voltage of a galvanic cell with reactants and
products in their standard states.
•Cell voltages for product-favored reactions are positive; condition of spontaneity for an
electrochemical reaction written in a given direction: E°>0
•The cell voltage is the sum of the two half-cell potentials:
E°cell = E°ox + E°red
E°cell>0 reaction is product favored
E°cell<0 reaction is reactant favored
How to measure half-cell potentials? Calibrate/compare with a standard half reaction.
The standard hydrogen electrode half cell is assigned a voltage of 0:
2 H3O+ (aq) + 2 e- ==> H2 (g, 1bar) + 2 H2O(l) E°red = 0.0 V
H2 (g, 1bar) + 2 H2O(l)==> 2 H3O+ (aq) + 2 e- E°ox = 0.0 V
Compare all half-cells against this standard to obtain a table of standard reduction
potentials.
Anode:
Zn(s) ==>Zn2+(aq, 1M) + 2 eE°ox= ?V
+
Cathode:
2 H3O (aq, 1M) + 2e ==> H2(g, 1bar) + 2H2O(l)
E°red=0V
Measure for the galvanic cell:
Zn(s) + 2 H3O+ (aq, 1M) ==> H2(g, 1bar) + 2H2O(l)+ Zn2+(aq, 1M)
E°cell=0.76V
Thus,
Zn(s) ==>Zn2+(aq, 1M)+ 2 eE°ox= 0.76V
Note that Zinc is a stronger reducing agent than H2, so the reaction is spontaneous in the
direction written.
Connect a copper electrode to the standard hydrogen electrode: copper is a weaker
reducing agent than H2, thus
Cathode:
Cu2+(aq, 1M) + 2e- ==>Cu(s)
E°red= ?V
Anode:
H2(g, 1bar) + 2H2O(l)==> 2 H3O+ (aq, 1M) + 2eE°ox=0V
Measure for the galvanic cell:
Cu2+(aq, 1M) + H2(g, 1bar) + 2H2O(l)==> Cu(s) + 2 H3O+ (aq, 1M) + 2e- E°cell=0.34V
Thus,
Cu2+(aq, 1M) + 2 e- ==>Cu(s)
E°red= 0.34V
Combining the copper and zinc half cells gives:
Zn(s) + Cu2+(aq, 1M) ==> Zn2+(aq, 1M) + Cu(s)
We expect the cell voltage to be E°cell = E°ox + E°red = 0.76V + 0.34V= 1.10V
One can use cell voltages to obtain half-cell potentials:
Example: if the cell Cu2+ (s) + Fe(s) ==> Fe2+(aq) + Cu(s)
Calculate the half-cell potential for Fe(s) ==> Fe2+(aq) + 2e-
E°cell= 0.78
E°ox = ?V
Many such measurements lead to tables of standard reduction potentials. All values
are relative to the Standard Hydrogen Electrode
•Each half reaction is written as a reduction, e.g. Fe2+(aq) + 2e- ==> Fe(s)
•Depending on conditions, each half reaction can occur in either direction
•The more positive E°red, the more easily an ion is reduced. Positive values of E°red means
that H2 is a stronger reducing agent than the reduced form of the ion/metal.
•Substances easily reduced are strong oxidizing agents: F2, Cl2, H2O2 (possess highly
electronegative elements)
•Negative reduction potentials indicate that the favorable direction of the reaction is in
reverse. Thus Li, Na, K metals tend to be oxidized while reducing other species.
•In the table, any species on the left of a half reaction will oxidize any species on the
right, further down in the table. Thus, the metals below O2 in the table will tend to rust in
the presence of air and water.
•Electrode potentials depend on concentration, but not on the quantity of reactants or
products, since a half-cell voltage is energy per unit charge: 1 V=1 J/C.
Example: Will Sn4+ oxidize aluminum?
Sn4+(aq) + 2e- ==> Sn2+(aq) E°red= 0.15V
Al3+ + 3e- ==> Al(s)
E°red=-1.66V; thus Sn4+ is a stronger oxidant than Al3+
So,
2(Al(s)==> Al3+ + 3e-)
E°ox= 1.66V
4+
2+
3(Sn (aq) + 2e ==> Sn (aq))
E°red= 0.15V
net: 3Sn4+(aq) + 2Al(s) ==> 2Al3+(aq) + 3Sn2+(aq) E°cell= 1.81 V, Yes.
Problem Set #13
1. Complete and balance the following equations for reactions taking place in acidic
solution:
a. VO2+(aq) + SO2(g) ==> VO2+(aq) + SO42b. Zn(s) + NO3-(aq) ==> Zn2+(aq) + NH4+(aq)
c. HCOOH(aq) + MnO4-(aq) ==> CO2(g) + Mn2+(aq)
2. Complete and balance the following equations for reactions taking place in basic
solution
a. N2H4 (aq) + CO32-(aq) ==> N2(g) + CO(g)
b. Sn(OH)62-(aq) + Si(s) ==> HSnO2-(aq) + SiO32-(aq)
c. OCl-(aq) + I-(aq) ==> IO3-(aq) + Cl-(aq)
3. A galvanic cell is constructed to carry out each of the following reactions. What is
the standard cell voltage (E°cell) at 25°C in each case? Use data from table 19.1 in
your text.
a. 2Ag (s) + Cl2(g) ==> 2 Ag+(aq) + 2 Cl-(aq)
b. Pb2+ (aq) + 2 Cr2+(aq) ==> Pb(s) + 2 Cr3+ (aq)
Note : Pb2+ (aq) + 2e- ==> Pb(s)
E°red= -0.1263V
Cr3+(aq) + 2e- ==> Cr2+(aq)
E°red= -0.424V
2+
3+
c. Cr2O7 (aq) + 14H3O + 6 I ==> 2 Cr (aq) + 3 I2(s) + 21 H2O
4. Diagram the following cells, indicating the direction of flow of electrons in the
external circuit and the motions of ions in the salt bridge and in each half cell:
a. Ni(s)|Ni2+(aq) || HCl (aq) | H2(g) | Pt(s)
b. Ni(s)|Ni2+(aq) || Ag+ | Ag(s)
c. Pt(s) | Cr2+(aq), Cr3+(aq) || Cu2+(aq) | Cu(s). What is the function of Pt(s) in
this cell? Note: Cr3+ + e- ==> Cr2+ E° = -0.424 V
5. Answer the following questions for the following galvanic cell:
Ni
Fe
1M Fe2+
a.
b.
c.
d.
e.
f.
g.
h.
i.
In which direction do electrons flow when the switch is closed?
In which half-cell does oxidation occur?
In which half-cell do electrons enter the cell?
At which electrode are electrons consumed?
Which electrode will decrease in mass during operation of the cell?
Suggest a solution for the electrolyte with the Ni electrode
In which net direction do cations flow within each cell?
Write balanced half-reactions and an overall cell reaction.
Show the direction of flow and charge of ions entering the cell from the
salt bridge.
j. What will be the reading on the voltmeter if a 1M solution of your chosen
electrolyte is used in the Ni half-cell? Assume the temperature is 25°C.