ELECTROCHEMISTRY
Electrochemistry—the study of the interchange of chemical and electrical energy
GOAL 1: Balancing Redox Equations
Redox reactions: a reaction in which electrons are transferred, actually two separate processes (oxidation
and reduction) that always occur together
Oxidation – the loss of electrons, increase in charge
Reduction – the gain of electrons, reduction of charge
Oxidizing agent (OA) – the species that is reduced and thus causes oxidation
Reducing agent (RA) – the species that is oxidized and thus causes reduction
OIL RIG – oxidation is loss, reduction is gain (of electrons)
LEO goes GER – losing electrons is oxidation, gaining electrons is reduction
Example: write the reaction and identify what is being oxidized, reduced, the oxidizing agent, and the
reducing agent for the combustion of methane
Balancing Redox Equations (acidic):
1. Assign oxidation numbers to all elements
2. Break the reaction into two half reactions
3. Balance everything but H and O
4. Balance O using H2O
5. Balance H using H+
6. Balance charge by adding electrons to the more positive side
7. Make the # of electrons in both half reaction equal
8. Combine the reactions and cancel out duplicates, checking that the atoms and charges are balanced
Balancing Redox Equations (basic):
1. Do all the steps as above
2. Add a # of OH- equal to the H+ in the balance equation
3. Combine hydroxides and protons on the same side to make water; cancel out anything that will
Practice: On a separate piece of paper, identify what is being oxidized, reduced, the oxidizing agent, and
the reducing agent for the following reactions; balance the equation in an acidic environment (unless
otherwise indicated)
1.
2.
3.
4.
5.
6.
I- + SO42-  I2 + H2S
SO2 + MnO4-  SO42- + Mn2+
H2C2O4 + MoO42-  CO2 + Mo3+
Zn + NO3-  Zn2+ + NH4+
Al(s) + MnO4-(aq)  MnO2(s) +Al(OH)4-(aq)
Cl2(g)  Cl-(aq) + OCl-(aq)
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(BASIC SOLUTION)
(BASIC SOLUTION)
GOAL 2: Writing Balanced Redox Reactions
Common Oxidizing Agents
MnO4- in acidic solution
MnO2 in acidic solution
MnO4- in neutral or basic solution
Cr2O72- in acidic solution
HNO3, concentrated
HNO3, dilute
H2SO4, hot, concentrated
Metallic ions (higher oxidation #)
Free halogens
Na2O2
HClO4
H2O2
Products Formed
Mn2+
Mn2+
MnO2(s)
Cr3+
NO2
NO
SO2
Metallous ions (lower oxidation #)
Halide ions
NaOH
ClH2O
Common Reducing Agents
Halide ions
Free metals
C2O42Sulfite ions or SO2
Nitrite ions
Free halogens, dilute basic solution
Free halogens, concentrated basic solution
Metallous ions (lower oxidations #)
Products Formed
Free halogen
Metal ions
CO2
Sulfate ions
Nitrate ions
Hypohalite ions (ex: hypochlorite)
Halate ions (ex: chlorate)
Metallic ions (higher oxidation #)
Predict reactants and products for the following redox equations (you don’t need to balance then, but you
can later for practice). Be sure to notice whether reactions occur in acidic or basic conditions. Check the
reference sheet on the back for more possible reactions. Also, all spectator ions must be eliminated.
1. Potassium permanganate solution is added to concentrated hydrochloric acid.
2. Potassium dichromate solution is added to an acidified solution of sodium sulfite.
3. Manganese(IV) oxide is added to warm, concentrated hydrobromic acid.
4. Hydrogen peroxide solution is added to basic potassium iodide solution.
5. Hydrogen peroxide is added to an acidified solution of sodium nitrite.
6. Sulfur dioxide gas is bubbled through a basic solution of potassium permanganate.
7. A solution of tin(II) ion is added to an acidified solution of potassium dichromate.
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GOAL 3: Galvanic (aka Voltaic) Cells
A galvanic cell is an electrochemical cell that happens spontaneously and produces a voltage in the process. It is a
device that can convert chemical energy into electrical energy. A common name for a voltaic cell is a battery.
Another term we will use for voltage is Electromotive Force (emf). When voltages are measured in chemistry we
do them at standard conditions – solutions are 1 M, gases are 1 atm, and the temperature is 298 K.
Determining standard cell potential: Eocell
Eocell = Eocathode - Eoanode
Write the reaction that would occur in a voltaic cell prepared from copper metal, 1 M Cu(NO3)2, iron
metal, and Fe(NO3)3. What would be the voltage of cell?
 Step 1: Write the two half reactions exactly as they are written in the table of standard reduction
potentials (ions gaining electrons to form metals).

Step 2: Identify the reaction that will occur at the cathode and the anode. The reaction with the higher
reduction potential will occur at the cathode – think RED CAT. Plug in the potentials to calculate the
cell voltage.

Step 3: To complete the reaction, flip the reaction occurring at the anode so that it is an oxidation
reaction. Make the number of electrons transferred equal and add the two half reactions together,
cancelling out the electrons.
Parts of the voltaic or galvanic cell:
o
o
o
o
o
o
o
o
Anode--the electrode where oxidation occurs. After a period of time, the anode may appear to become
smaller as it converts from the metal into ions in the solution.
Cathode-- the anode where reduction occurs. After a period of time, it may appear larger due to ions from
solution plating onto it.
Inert electrodes—used when a gas is involved OR ion to ion involved such as Fe3+ being reduced to Fe2+
rather than Fe0. Made of Pt or graphite.
Salt bridge -- a device used to maintain electrical neutrality in a galvanic cell. This may be filled with agar
which contains a neutral salt or it may be replaced with a porous cup.
Electron flow -- always from anode to cathode. (through the wire) FAT Cat
Standard cell notation (line notation) - anode/solution// cathode solution/ cathode Ex. Zn/Zn2+ (1.0 M) //
Cu2+ (1.0M) / Cu
Cell potential(Ecell) or electromotive force(emf) – the ‘pull’ (or driving force) on electrons through a wire
from the reducing agent to the oxidizing agent
Voltmeter - measures the cell potential/emf. Usually is measured in volts.
Sketch the cell below. Be sure to show the anode, the cathode, the flow of the electrons through the
connecting wire, and the flow of ions from the salt bridge.
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Standard cell notation (line notation) – “Ion sandwich” in alphabetical order (Anode then Cathode)
Anode metal/anode ion//cathode ion//Cathode metal
For Reaction: M + N+  N + M+
Anode
||
Cathode
(alphabetical order!)
M(electrode)|M+ (solution)|| N+ (solution)|N(electrode)
| - indicates phase boundary
|| - indicates salt bridge
EX: Zn / Zn2+ (1.0M)// Cu2+ (1.0M) / Cu
What would be the short hand notation to describe the cell?
If you know that copper(II) ion has a blue color and iron(III) ion has an orange color, predict what will
happen to the colors of the cells as the reaction takes place.
Practice:
Consider a galvanic cell based on the reaction between Al3+ and Mg. Give the balanced cell reaction,
draw the cell (including ion and electron flow directions), calculate the cell potential, and give the short
hand notation for the cell
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GOAL 4: Nernst Equation – Galvanic Cells at Nonstandard Conditions
Voltaic cells at NONstandard conditions: LeChatlier’s principle can be applied. An increase in the
concentration of a reactant will favor the forward reaction and the cell potential will increase. The
converse is also true!
For a more quantitative approach…..
When a cell is not at standard conditions, use Nernst Equation:
RT
E = Eo - ------- ln Q
nF
E = Energy produced by reaction
T = Temperature in Kelvins
n = # of electrons exchanged in BALANCED
redox equation
R = Gas constant 8.315 J/K mol
F = Faraday constant
Q = reaction quotient
[productscoefficient]/[reactantscoefficient]
Rearranged, another useful form:
0.0592
NERNST EQUATION: E = Eo - ---------- log Q @ 25°C (298K)
n
As E declines with reactants converting to products, E eventually reaches zero.
Zero potential means reaction is at equilibrium [dead battery]. Also, Q =K AND G = 0 as well.
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Practice with Nernst & Amps
1.
2.
3.
8.
Calculate the Ecell for the following:
a.
Fe/Fe3+(0.75M)//Cu2+(1.15M)/Cu)
b.
Pt(s)/Cr2O72- (0.8M), Cr3+(1.3M), H+(aq)//MnO4-(0.2M),
Mn2+(2.0M), H+(aq)/Pt(s) when pH=0.90
c.
Pt(s)/Fe2+(1e-3M), Fe3+ (1e-2)//Sn4+(1e-1M), Sn2+(3e-3)/Pt(s)
Consider the cell described:
Al|Al3+(1.00M)||Pb2+(1.00M)|Pb
Calculate the cell potential after the reaction has operated long enough for
the [Al3+] to have changed by 0.60 mol/L (Assume T = 25̊ C.)
9.
An electrochemical cell consists of a standard hydrogen electrode
and a copper metal electrode.
a.
What is the potential of the cell at 25̊ C if the copper electrode
is placed in a solution in which [Cu2+] = 2.5 X 10-4 M?
b.
The copper electrode is placed in a solution of unknown [Cu2+].
The measured potential at 25̊ C is 0.195V. What is [Cu2+]?
(Assume Cu2+ is reduced)
10.
An electrochemical cell consists of a nickel metal electrode
immersed in a solution with [Ni2+] = 1.0 M separated by a porous
disk from an aluminum metal electrode.
a.
What is the potential of this cell at 25̊ C if the aluminum
electrode is placed in a solution in which [Al3+] = 7.2 X 10-3 M?
b.
When the aluminum electrode is placed in a certain solution in
which [Al3+] is unknown, the measured cell potential at 25̊ C is
1.62 V. Calculate [Al3+] in the unknown solution. (Assume Al is
oxidized)
Find the equilibrium constant for the reaction in 1a.
Consider the two half reactions:
M2+ + 2e-  M
E0= -2.10V
+
0
NO3 + 4H + 3e  NO + 2H2OE = 0.96V
a.
Write the balanced reaction and find the E0cell
b.
Determine the identity of metal M if the mass of the “M”
electrode changed by 0.3286g when 40mL of NO gas at 1.2 atm
and 27C were used.
4.
Consider the cell:
Ni(s)/Ni2+ (1.0M)// Al3+ (1.0M)/ Al(s)
Determine the Ecell when the aluminum concentration has changed by 0.4M.
Can permanganate ion oxidize Fe2+ to Fe3+ at 25̊ C under the
following conditions?
[Mn2+] = 1 X 10-6 M
[MnO4-] = .001 M
[Fe2+]= 1X 10-3 M
[Fe3+]= 1 X 10-6
pH = 4.0
5.
6.
The overall reaction in the lead storage battery is
Pb(s) + PbO2(s) + 2H+(aq) + 2HSO4-(aq)  2PbSO4(s) + 2H2O(l)
Calculate E at 25̊ C for this battery when [H2SO4] = 4.5 M that is, [H+] = [HSO4] = 4.5 M.
At 25̊ C, E̊ = 2.04 V for the lead storage battery.
7.
Consider the cell described:
Zn|ZN2+(1.00M)||Cu2+(1.00M)|Cu
Calculate the cell potential after the reaction has operated long enough for
the [Zn2+] to have changed by 0.20 mol/L (Assume T = 25̊ C.)
11.
Consider the galvanic cell based on the following half-reactions:
Au3+ + 3e-  Au
E̊ = 1.50 V
+
Tl + e  Tl
E̊ = -0.34 V
a.
Determine the overall cell reaction and calculate E̊cell.
b.
Calculate delta G° and K for the cell reaction at 25̊ C.
c.
Calculate Ecell at 25̊ C when [Au3+] = 1.0 X 10-2 M and [Tl+] =
1.0 X 10-4 M.
Consider the following galvanic cell at 25° C: Pt|Cr2+(0.30M),
Cr3+(2.0M)||Co2+(0.20M)|Co
The overall reaction and equilibrium constant value are
2Cr2+(aq) + Co2+(aq)  2Cr3+(aq) + Co(s) K = 2.79 X 107
Calculate the cell potential, E, for this galvanic cell and delta G for the cell
reaction at these conditions
12.
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GOAL 5: Electrolytic Cells
A galvanic cell produces current when an oxidationreduction reaction proceeds spontaneously. A similar
apparatus, an electrolytic cell, uses electrical energy to
produce
chemical change. The process of electrolysis involves
forcing a
current through a cell to produce a chemical change for
which the
cell potential is negative (ΔG>0); that is, electrical work
causes an
otherwise nonspontaneous chemical reaction to occur.
Electrolysis has great practical importance; for example, charging a battery, producing metals like aluminum
from ores (rocks with aluminum ions as part of their structure), and chrome plating an object are all done
electrolytically.
Electrolytic cells [NON spontaneous cells]:
Used to separate ores or plate out metals.
Important differences between a voltaic/galvanic cell and an electrolytic cell:
1) Voltaic cells are spontaneous and electrolytic cells are forced to occur by using an electron pump or
battery or any DC source.
2) A voltaic cell is separated into two half cells to generate electricity; an electrolytic cell occurs in a single
container.
3) A voltaic [or galvanic] cell IS a battery, an electrolytic cell NEEDS a battery
4) AN OX and RED CAT still apply BUT the polarity of the electrodes is reversed. The cathode is
Negative and the anode is Positive (remember E.P.A – electrolytic positive anode). Electrons still flow
FATCAT.
(a) Usually use inert electrodes
****Calculating the Electrical Energy of Electrolysis
How much metal could be plated out?
How long would it take to plate out?
Helpful links for electrolysis calculations:
1 Volt = 1 Joule/Coulomb (not on equation sheet)
96500 Coulombs = 1 mole of electrons
1 Amp = 1 Coulombs/second
I = q/t
(A)
(C) / (s)
A. How man moles of electrons are required to produce 5.00 A for 2.00 hours?
B. What mass of copper metal is produced if 10000 C are passed through a solution of copper(II) sulfate?
C. If it is necessary to deposit 1.50 g of Al on an object, how many minutes must 5.00 A flow through a solution
of aluminum nitrate?
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