ELECTROCHEMISTRY
• The branch of chemistry dealing with interconversion between electrical
energy and chemical energy
• Electrochemistry is based on the principle of Redox reactions (Reductionoxidation reactions).
• Two major types of electrochemical cells are represented below:
• There are several applications of electrochemical process, examples:
1. Electrochemical cells e.g voltaic / galvanic cells and batteries
2. Electrolytic cells e,g electrochemical reduction of metal from their
ionic compounds
3. Electroplating: the use of electrochemical reduction to coat a shiny
onto surfaces of materials.
4. Electro-purification the use of electrochemical cells to purify metals.
QUICK REVIEW ON REDOX REACTIONS
•
Redox reactions are oxidation-reduction reactions where one or
more electrons are transferred between two reacting substances
like atoms or ions participating in it.
During redox reactions, oxidation numbers/ states of reacting substances
are changed.
Rules for Assigning Oxidation Numbers
1. The convention is that the cation is written first in a formula, followed by
the anion.
For example, in NaH, the H is H-; in HCl, the H is H+.
2. The oxidation number of a free element is always 0. The atoms in He and
N2, for example, have oxidation numbers of 0.
3. The oxidation number of a monatomic ion equals the charge of the
ion. For example, the oxidation number of Na+ is +1; the oxidation
number of N3- is -3.
4. The usual oxidation number of hydrogen is +1. The oxidation number of
hydrogen is -1 in compounds containing elements that are less
electronegative than hydrogen, as in CaH2.
5. The oxidation number of oxygen in compounds is usually -2. Exceptions
include OF2 because F is more electronegative than O, and BaO2, due to
the structure of the peroxide ion, which is [O-O]2-.
6. The oxidation number of a Group IA element in a compound is +1.
7. The oxidation number of a Group IIA element in a compound is +2.
8. The oxidation number of a Group VIIA element in a compound is -1,
except when that element is combined with one having a higher
electronegativity. The oxidation number of Cl is -1 in HCl, but the
oxidation number of Cl is +1 in HOCl.
9. The sum of the oxidation numbers of all of the atoms in a neutral
compound is 0.
Example: H2SO4
H = +1 S= +6 and O = -2
so the sum 2( +1) + (+6) + 4 (-2) = 0
10.The sum of the oxidation numbers in a polyatomic ion is equal to the
charge of the ion. For example, the sum of the oxidation numbers for
SO42- is -2.
See how rules for assigning oxidation numbers were applied in some
examples.
BALANCING REDOX REACTIONS
Take note that balancing a redox reaction requires that:
1. Atoms on both side of the equation are balanced.
2. Charges on both sides of the equation are balanced.
• Consider redox reactions equations below and check if they are
well balanced.
• Equation two cannot be balanced by our tradidional method of
balancing atoms only. It must be balanced by half reaction
method in acidic mediuam or basic medium.
• By half reaction, an unbalanced reaction redox reaction is first
split into its half reactions and then start we balancing each half
reaction separatley.
• After that we add up the two balanced half reactions to a get a
full balanced equation.
Balancing Redox Reactions in Acidic Solution
Consider the following unbalanced redox reaction:
Let’s try to balance this reaction in acidic solution.
STEP 1: DIVIDE THE REACTION INTO HALF REACTIONS
STEP 2: BALANCE THE ELEMENTS OTHER THAN H AND
O
Fortunately, all atoms other than H and O are already balanced, so we
can move on to the next step.
STEP 3: BALANCE THE O ATOMS BY ADDING H2O
STEP 4: BALANCE THE H ATOMS BY ADDING H+
STEP 5: BALANCE THE CHARGES BY ADDING E-
STEP 6: ADD THE HALF REACTIONS AND SIMPLIFY
Because the number of electrons in the first half reaction (3e-) does
not equal the number of electrons in the second half reaction (2e-), we
must multiply the reactions by coefficients so that the electrons will
cancel out when we add them. In this case, we multiply the first
reaction by 2 to yield 6e-, and the second reaction by 3 to yield 6e- as
well.
Carrying this out, we get:
Now we can add the two reactions, and we notice that the electrons on
both sides cancel out.
Thus, our balanced reaction is:
To summarize, the steps for balancing redox reactions in acidic
solution are as follows:
1. Divide the reaction into half reactions
2. Balance the elements other than H and O
3. Balance the O atoms by adding H2O
4. Balance the H atoms by adding H+
5. Balance the charges by adding e6. Add the half reactions and simplify.
Balancing Redox reactions in basic medium (Same procedure as in
acidic medium but you must neutralize H+ by adding equal amount
OH- on both sides). Same procedure as in acidic medium but you
must neutralize H+ by adding equal amount OH- on both sides.
Worked out example:
Br¯ + MnO4¯ ---> MnO2 + BrO3¯
Solution:
1) The two half-reactions, balanced as if in acidic solution:
3H2O + Br¯ ---> BrO3¯ + 6H+ + 6e¯
3e¯ + 4H+ + MnO4¯ ---> MnO2 + 2H2O
2) Make the number of electrons equal:
3H2O + Br¯ ---> BrO3¯ + 6H+ + 6e¯
6e¯ + 8H+ + 2MnO4¯ ---> 2MnO2 + 4H2O <--- multiplied by a factor
of 2
3) Convert to basic solution, by adding 6OH¯ to the first half-reaction
and 8OH¯ to the second:
6OH¯ + Br¯ ---> BrO3¯ + 3H2O + 6e¯
6e¯ + 4H2O + 2MnO4¯ ---> 2MnO2 + 8OH¯
4) The final answer:
H2O + 2MnO4¯ + Br¯ ---> 2MnO2 + BrO3¯ + 2OH¯
5) What happens if you add the two half-reactions without converting
them to basic?
You get this:
2H+ + 2MnO4¯ + Br¯ ---> 2MnO2 + BrO3¯ + H2O
Then, add 2OH¯ to each side:
2H2O + 2MnO4¯ + Br¯ ---> 2MnO2 + BrO3¯ + H2O + 2OH¯
Eliminate one water for the final answer:
H2O + 2MnO4¯ + Br¯ ---> 2MnO2 + BrO3¯ + 2OH
Activity: Balance the following redox reactions:
(a)
(b)
ClO3¯ + SO2 ---> SO42¯ + Cl¯ ( acidic medium)
Br¯ + MnO4¯ ---> MnO2 + BrO3¯ (basic medium)
VOLTAIC/ GALVANIC CELLS
• A Galvanic cell uses a spontaneous redox reaction between
reactants to generate electric current/energy.
• The apparatus shown above does not represent a voltaic/ Galvanic
cell. Do you know why ?
A simple diagram of a voltaic cell
• In voltaic cell, reactants are separated such that each half redox
reaction occurs in its own container.
• For example, in the diagram above electrons are spontaneously
transferred from zinc atoms of the zinc electrode to copper (II) ions at
the cathode electrode.
• In a voltaic cells lost electrons are transferred from the anode electrode
through the external wire to the cathode electrode.
• The voltmeter reading shows the maximum electrical potential
difference between the two electrodes.
• Such maximum potential difference is also known as Electromotive force
of a cell (Emfcell) or Cell potential ( Ecell).
• Ecell is measured in units of Volts (V) 1V = 1Joule/coulomb
• For spontaneous redox reaction Ecell is always positive
• Free energy change obviously must be negate (-ΔG ).
• Useful is work is done by the cell on the surrounding( external
circuit) ( - w)
MAJOR COMPONENTS OF A GALVANIC CELL
Anode: Electrode where always oxidation reaction takes place. Anode is
negative in a galvanic cell.
Cathode: Electrode where always reduction reaction takes place. Cathode is
positive in galvanic cell.
Salt bridge: A U-tube that connects the two electrode solutions. It contains ions
of a neutral salt e.g KCl
The main purpose of salt bridge:
• To neutralize the charges in both electrodes. Whereby positive ions move
towards the cathode to replace the consumed positive charges while
anions move towards the anode to neutralize the excess produced
positive charges.
• The salt bridge also complete circuit of the voltaic cell.
STANDARD REDUCTION POTENTIALS (E0red)
• Standard reduction potential is the maximum voltage of voltaic
cell obtained using a reference standard hydrogen electrode
(SHE).
• A standard hydrogen electrode is shown below.
• Standard reduction potential tells us about the tendency of a
substance to reduced.
• By definition reduction potential of a Standard hydrogen
electrode is exactly 0V.
• This serves as reference for determining standard reduction for
other substances.
• For voltaic cell given above, the voltmeter meter reading, is
the standard reduction potential of Cu2+ (aq) . Because for
SHE electrode by definition is 0V
• It is a positive value because Cu2+ has a greater tendency to be
reduced than hydrogen ion H+.
• We can use the same apparatus to determine standard reduction
potentials for other substances.
• Take note that the concentration of all reactant solutions under
standard state must be equal 1mol/L.
• Pressures for all gases 1bar or 100 kPa
• Standard reduction potentials for many reduction reactions are
tabulated in a table for reduction potentials.
• Substances with greater values of standard reduction potential
have greater tendency to be reduced than those with less values.
Example Standard Reduction Potentials for some common reduction
equations
CELL NOTATION
• A cell notation is symbolic representation of a Galvanic cell. All anode
electrode components and cathode electrode components separated by a
salt bridge are clearly represented in a cell notation.
• Example of cell notation is indicated in the diagram below:
• You are expected to know how to construct and interpret cell notations.
• Take note that different phases of electrode species are separated by a
single vertical line.
STANDARD CELL POTENTIAL E0CELL
• If standard reduction potentials of all electrodes are known we
can use them to calculate standard cell potential (E0cell).
E0cell = E0red(cathode) - E0red(anode)
OR
E0red + E0 oxi
E0cell = E0red(cathode) - E0red(anode)
……………………………….?
E0red + E0 oxi ………………………………………………………….?
Guidelines for determining standard cell potentials from standard
reduction potentials for a given redox reaction.
• Identify which substance is oxidised and which is reduced by
assigning oxidation numbers.
• Write down the half reactions (oxidation and reduction)
• Use the standard reduction potentials from the table to work the
standard cell potential.
Electrochemistry Review Questions
1. The net equation for a given voltaic cell is:
Sn (s) + 2 Ag+ → Sn2+ + 2 Ag (s)
a. Write the two half-reactions involved and identify each in terms of
(1) site of oxidation or reduction and (2) anode or cathode.
Sn → 2e + Sn2+
oxidation at the anode
+
2Ag + 2e → 2Ag reduction at the cathode
b. Calculate the net potential of the cell (the voltage), assuming
standard conditions.
E0cell = E0red (cathode) - E0red(anode) Try to obtain the values from the resource data table on internet
0.80 V - (-0.136V) = 0.936 V
Now proceed on your own!!
c. Draw a fully labeled diagram of the voltaic cell. Be sure to indicate
the flow of electrons in the external circuit (through the wire) and
the flow of ions in the solution.
(d) Write the cell notation for the galvanic cell in (c)
2. The following questions pertain to a voltaic cell in which the following
half-reactions occur:
S (s) + 2e- → S2- (aq)
Ca2+ (aq) + 2e- → Ca (s)
a) Which half-reaction will proceed as a reduction? Which will proceed as
an oxidation? How do you know?
b) Rewrite the half-reactions in the correct directions.
c) Write a balanced equation for the overall cell reaction.
d) Calculate the cell potential.
3. A voltaic cell is constructed using electrodes based on the following half
reactions:
Pb2+ (aq) + 2e- -> Pb(s)
Au3+(aq) +3e- -> Au(s)
a) Which is the anode, and which is the cathode in this cell?
b) Calculate the cell potential?
4 Using standard reduction potential calculate standard emf for each of the
following reaction equation. (You must identify the half reactions)
2Al3+ (aq) +3Ca (s) → 2Al (s) + 3Ca2+ (aq)
(a) Cl2 (g) + 2I- (aq) → 2Cl (aq) + I2 (s)
THERMODYNAMICS OF ELECTROCHEMICAL CELL
Free energy change and Cell Potential
• Free energy (G) is the total energy of a system that is available to perform
work.
• When a chemical reaction occur on its own without any external
intervention, we say it is spontaneous. Such reaction uses its free energy
to drive it.
• During a spontaneous reaction, free energy of a reaction decreases, So
free energy change ΔG must be negative.
• Take note that both free energy change ΔG and Ecell
determine
whether a reaction is spontaneous or not.
• Voltaic cells uses spontaneous redox reactions therefore ΔG must be
negative and
Ecell must be positive.
Enthalpy change ΔH and Entropy change ΔS.
• Enthalpy change ΔH is a measure of heat change of a system at a
constant pressure.
• ΔH0 symbolizes standard enthalpy change.
• Entropy change is measure of degree of randomness or disorder of a
system.
•
ΔS0 symbolizes standard entropy change.
•
Both ΔHrxn and ΔSrxn are used to calculate free energy change ΔGrxn
•
The subscript “rxn” denotes reaction.
ΔGrxn = ΔHrxn - T ΔSrxn
ΔG0rxn = ΔH0rxn - T ΔS0rxn ( when standard state values are used)
• To calculate ΔH0rxn and ΔS0rxn we use the tabulated standard values
in the following general equations that are based on how a chemical
equation is balanced.
• ΔG > 0 means nonspontaneous reaction
• ΔG < 0 means spontaneous reaction (the key requirement for voltaic
cell)
Effect of change in concentration on ΔG
• Let us first derive some equations that relate ΔG and ΔG0. ΔG (
without a superscript) is the free energy change under non-standard
condition state. (I hope you recall what is meant by standard state).
• ΔG of any reaction changes with change concentrations of both reactants
and products.
• ΔG and ΔG0 are related as follow:
ΔG = ΔG° + RT InQ
• Where Q is the reaction quotient. The ratio of concentrations at any time
in the reaction mixture. Used to predict how far is the reaction from
equlibrium.
Under standard state Q = 1 and ΔG = ΔG° beacouse all
concentrations and pressures at this point of recation must be equal to 1.
• We use the above equation to calculate free energy change (∆G) using
the concentrations or pressures of products and reactants in the reaction
mixture , provided that ΔG° is known.
Activity :
Given the reaction equation below .
(a) Calculate the G0 for the following reaction at 298 K?
2H2(g)
+
N2 (g) → 3NH3 (g)
H0 = 92.4 kJ
S0 = 198 J/
(b) Using the following sets of concentrations to calculate reaction Q
and G for each set at 298 K . Interpret your results.
[N2]
[H2]
[NH3]
a) 1.0 M
1.0 M
1.0M
Q
G
b) 0 M
0M
1.0 M
c) 0.01M
0.01 M
0.1M
d) 1.0 M
1.0 M
845M
Relationship between Free energy change G and Equilibrium constant
(Kequ)
• All chemical reactions always proceed in such a way that eventually they
reach a state of equilibrium , whereby three will be no more significant
changes in concentrations of both products and reactants, meaning Q
become a constant.
• This constant value is known as equilibrium constant Kequ to signify that
the reaction is at equilibrium state
ΔG = ΔG° + RT InQ
• At equilibrium, ∆G=0 and Q become equal to the equilibrium constant K.
Hence the equation above becomes,
ΔG° = –RT In K(eq)
The factor 2.303 is used to convert from ln to log
ΔG° = –2.303 RT log Keq
R = 8.314 J mol-1 K-1 or 0.008314 kJ mol-1 K-1.
T is the temperature on the Kelvin scale
At 298K (250C) and we substitute in the value of R the equation is simplified
to
ΔG° = –0.05916 log Keq
• We use the above equation to work out the equilibrium constant K
or ΔG° at a given temperature.
Activity: For the reaction
CO (g) + 2 H2O (g) → CO2 (g) + H2 (g) at
770 K,
Keq = 5.10
Calculate the G0 for the reaction at 770 K.
RELATIONSHIP BETWEEN GIBBS FREE ENERGY AND EMF OF A
CELL
• In the case of galvanic cells, Gibbs energy change ΔG is related to
the electrical work done by the cell.
• ΔG = -wmax (maximum work done by the reaction)
• In a galvanic work Wmax = electrical work = nFEcel
ΔG = -nFEcell
Where:
n = number of moles of electrons involved in a redox reaction
F = the Faraday constant = 9.65 x 10 4 C/mol = 96500 C/mol
E = emf of the cell
F=1 Faraday =96500 coulombs
If reactants and products are in their standard states,
ΔG°= –nFE°cell
• We use the above equation to calculate ΔG or Ecell provided that
one of them is known
Activity: Consider the a redox reaction equation represented below:
2Fe3+ (aq) + H2 (g) → 2F2+ (aq) + 2H+ (aq)
(a) Calculate Emf of this cell under standard state.
(b) Calculate standard free energy change for this reaction.
(c) Based on your calculations is this reaction spontaneous or not under
standard state? Explain your reasoning.
THE NERNST EQUATION
•
The Nernst equation relates the standard cell potential E0cell and cell
potential Ecell. (Cell potential under non-standard state)
• The Nernst equation is derived from the equation for ΔG as
shown below.
If you substitute:
R, with R = 8.314 J mol-1 K-1
T = 298.15 K a
F= 96500
C/mol
you will obtain the last derived Nernst equations.
Relationship between E0cell and Equilibrium constant K
• When an electrochemical cell runs down the cell voltage is
zero. The system is now in a state of equilibrium. The
Nernst equation can be used to estimate the value of the
equilibrium constant.
Ecell = Eocell - (RT/nF) ln Q
• R is the thermodynamic constant, 8.31 Joules/mole K. F is
Faraday's constant, 96,485 C/mole e-. At 25°C, or 298K
and 1 atm pressure this equation simplifies to
Ecell = Eocell - (0.0257/n) ln Q
• At equilibrium the cell potential is zero and Q =
K. Substituting these into the equation we get
0 = Eocell - (0.0257/n) ln K
• In terms of log10 the equation becomes
0 = Eocell - (0.0592/n) log K
• Rearranging this equation to solve for Eocell we get
Eocell = (0.0592/n) log K
By substituting into the equation the values for Eocell and n, we can
solve for log K. The anitlog can then be calculated with the aid of a
calculator.
log K = x
K = 10x
The above equation srelate equilibrium constant K to standard cell
potential E0cell
The triangle below shows how K, ΔG0 and E0 for an electrochemical
reaction are related
Practice Activities
1.) Consider this overall redox process:
Cu2+(aq)+Ba(s)→Cu(s)+Ba2+(aq)
a.) Split the reaction into half reactions and determine their standard
reduction potentials. Indicate which would be the anode and cathode.
b.) Construct a cell diagram.
c.) Give the line notation for this cell.
d.) Calculate E°cell, the standard cell potential, which is given by E°red
- E°ox.
2.) Consider this overall redox process:
Al(s)+Sn2+(aq)→Al3+(aq)+Sn(s)
a.) Split the reaction into half reactions and determine their standard
reduction potentials. Indicate which would be the anode and cathode.
b.) Construct a cell diagram.
c.) Give the line notation for this cell.
d.) Calculate E°cell.
3.) Consider a galvanic cell with Zn(s) and 0.25 M Zn(NO3)2(aq) in
one compartment and Cu(s) and 0.25 M Cu(NO3)2(aq) in the other
compartment.
a.) Give the half-reactions and their standard reduction potentials.
Reduction (cathode):
Eocathode
Oxidation (anode):
Eoanode =
=
b.) Give the net, overall cell reaction and the standard cell potential:
Eocell =
c.) Now calculate the overall cell potential at the specific
concentrations given. You’ll need to use the Nernst equation: Ecell =
Eocell – (0.0592/n)log Q
d.) What would be the cell potential if the concentration of Zn(NO3)2
was increased to
2.5 M?
4.) Consider a voltaic cell with Cr(s) and Cr3+(aq) in one compartment
and Zn(s) and Zn2+(aq) in the other compartment.
a.) Draw and label a diagram of this cell.
b.) Calculate the standard cell potential for the overall reaction.
c.) If zinc is the anode and [Cr3+] = 0.010 M, what must the [Zn2+] be
to achieve 0.050V?
5.) A Zn/Zn2+ half-cell was coupled to a hydrogen electrode in which
PH2 = 1 bar. The [Zn2+] in the anode compartment was 0.10 M and
the cell potential was 0.542 V. Calculate the pH in the cathode
compartment.