Consider E°, ΔG° and Keq
NOTES:
Electrochemical Cell: If E° is positive, then the reaction favours _____________.
Thermodynamics: If ΔG°is negative, then the reaction favours ______________.
Obviously E° and ΔG° are related!
They are equated through the following relationship:
ΔG° = –n F E° (and ΔG = –n F E)
n = numbers of _________________ transferred in the reaction.
F = Faraday Constant
= the charge in Coulombs on 1 mole of ELECTRONS (C/mol or JV–1mol–1):
= 6.022×1023 e mol–1 × 1.6022 × 10–19 C e–1 = ____________________C/mol
E° = Standard Cell Potential (or Voltage) (V = JC–1)
You will recall that ΔG° = –RTlnKeq, and now we have
ΔG° = –nFE°, so we then get:
E° = (RT/nF)lnKeq
OR
Keq = e +(nFE°/RT)
At 298 K, RT/F = 0.025693 V and
therefore E° = (0.025693/n)lnKeq.
{At 298 K, 2.303RT/F = 0.0591 V and therefore E° = (0.0591/n)log10 Keq
–this constant is used when log base 10 is used}
What this also allows us to do is to calculate values of E° from ΔG° values or from
equilibrium constant values, i.e., given any ONE of E°, ΔG° or Keq, we can
calculate the others.
88
Example: Calculate the equilibrium constant and ΔG° at 298 K for the reaction:
NOTES:
CuSO4(aq) + Ni(s) → Cu(s) + NiSO4(aq) given access to an SRP table.
(Note: Clearly Cu2+ is being reduced and Ni(s) oxidised. Why?)
Ni2+(aq) + 2e– → Ni(s) E° =
V
Cu2+(aq) + 2e– → Cu(s) E° =
V
The overall net ionic equation is:
E°(reaction) =
Keq = e ( nFE°/RT )
= exp{
= exp (
ΔG°
}
) =
= –nFE°
=
=
J/mol =
kJ/mol
NERNST EQUATION:
You will recall from your thermodynamic studies that ΔG° = –RTlnKeq =
and ΔG = ΔG° + RTlnQ =
If we use the second equation and substitute in the first, we can arrive at
–nFE = –nFE° + RTlnQ
now divide by –nF to get:
E = E° –(RT/nF)lnQ
(This is the NERNST equation.)
Also,
E = E° – (0.0257/n) ln Q @ 298K
OR
E = E° – (0.0591/n) log10 Q @ 298K
89
Some things to note about this type of equation:
• E (not E°) INCREASES with _____________________ Q
• E DECREASES with INCREASING Q
NOTES:
Remember what Q looks like: ratio of products over reactants raised to the power of
the coefficients in the balanced net ionic equation.
Example:
At 25°C, what is the potential and what is the cell reaction for the following cell:
Ni|Ni2+(aq, 0.010M)||Cl–(aq, 0.20 M)|Cl2(g, 1atm) | Pt ?
First from SRP tables, deduce the E° value for the cell:
Ni2+(aq) + 2e–→ Ni(s)
E° =
V
Cl2(g) + 2e–→ 2Cl–(aq)
E° =
V
Note: The most positive half-reaction "goes" as written and the
other half-reaction is then the anode.
Overall Net Ionic Equation:
Q =
E°Rxn =
Now get E at NON–STANDARD conditions via the NERNST equation:
E = E° – (0.025693/n)lnQ
=
=
The reaction is (spontaneous/non-spontaneous)!
What is Keq at 298 K?
Keq = e ( nFE°/RT )
= exp{
}
=
90
Notice how even a relatively small positive E° gives rise to a large Keq!
NOTES:
Question #1: Which of the following are true for a voltaic cell?
A) ΔG < 0
B)
C)
D)
E)
K>1
Ecell > 0
All of the above are true.
None of the above are true.
Question #2: The numerical value of the Faraday constant is given as 96,500.
This value represents
A) the number of coulombs of charge carried by one mole of electrons.
B) the number of electrons corresponding to one coulomb of charge.
C) the number of electrons corresponding to one mole of electric charge.
D) the number of ions discharged by the passage of one mole of electrons.
E) 96,500 amp/s.
2+
+
Question #3: For the cell Zn(s)⏐Zn ⏐⏐Ag ⏐Ag(s), the standard cell potential is
1.56 V. A cell using these reagents was made, and the observed potential was
1.35 V. A possible explanation for the observed voltage is that
A) there was 0.0100 mol of Zn2+ and 0.200 mol of Ag+.
B) the zinc-ion concentration was larger than the silver-ion concentration.
C) the Ag electrode was twice as large as the Zn electrode.
D) the volume of the Zn2+ sol’n was larger than the volume of the Ag+ sol’n.
E) the volume of the Ag+ sol’n was larger than the volume of the Zn2+ sol’n.
We have the ability to calculate the potential associated with a
non–standard half–cell on its own
Example: What is the potential of a half–cell which has Cu metal dipping into
1.4 × 10–4 M CuSO4(aq)?
SRP data:
Cu2+(aq) + 2e– → Cu(s), E° = + 0.34 V
From the Nernst equation:
E ½ cell = E°½ cell – (0.025693/n)lnQ
(Based on the chemical equation, what will Q equal?)
E ½ cell =
91
E ½ cell =
NOTES:
*******************************************************************
NOTE: The Nernst equation also shows you why the E° value is NOT doubled
if you double the equation.
Compare:
(1) Ag+(aq) + e–→ Ag(s)
(2) 2Ag+(aq) + 2e–→ 2Ag(s)
Use: E = E° –(0.025693/n)ln Q
For equation 1: E = 0.80 – (0.025693/1) lnQ
For equation 2: E = 0.80 – (0.025693/2) lnQ
What is the value of logQ in each case?
For equation 1: Q = 1/[Ag+]
For equation 2: Q = 1/[Ag+]2
We can then simplify equation 2 through the following process:
E = 0.80 –(0.025693/2)ln(1/[Ag+]2) = 0.80 –(0.025693)ln{(1/[Ag+])}
Note: An exponent inside a logarithm is the same as a constant in front,
i.e., E = 0.80 – (0.025693/2)(2) ln(1/[Ag+]), so we can then cancel out
the 2’s to arrive at the same equation as in 1,
i.e., E = 0.80 –(0.025693/1)ln(1/[Ag+])
*******************************************************************
PROBLEMS:
1. a. What is the voltage at 298K of the following cell: Co | Co2+(aq) || Ni2+(aq) | Ni(s)?
b. If [Ni2+] = 0.010 M and [Co2+] = 1.0 M, what is the cell voltage, the cell
reaction and the cell diagram?
2. At 25°C, a cell based on
Ni(s) + Cl2(g, 0.790 atm) → Ni2+(aq, 0.100 M) + 2Cl–(aq, ? M)
delivers an initial voltage of 1.791 V. Calculate the value of [Cl–(aq)] in the
cathode cell compartment given:
Ni2+(aq) + 2e– → Ni(s)
Cl2(g) + 2e
–
→ 2Cl
–
(aq)
E° = –0.231 V
E° = +1.361 V
ANSWERS:
1 a. Get SRP values:
Co2+(aq) + 2e– → Co(s); E° =
V
92
Ni2+(aq) + 2e– → Ni(s); E° =
NOTES:
V
Net Ionic equation:
based on the reaction: E°(cell) = E°(cathode) –E°(anode) =
1 b)
Non standard cell:
V
E = E° –(0.0591/n)log10Q where
Q =
E =
E =
i.e., at the specified (non–standard) concentrations the reaction
Ni2+(aq) + Co(s) → Ni(s) + Co2+(aq) has a NEGATIVE value of E, so the
spontaneous reaction (at these non–standard concentrations) is actually :
Ni(s) + Co2+(aq) → Ni2+(aq) + Co(s);
E = +0.03 V
and the cell diagram would be:
Ni(s) | Ni2+ (aq, 0.010 M) || Co2+( aq, 1.0 M) | Co(s)
(oxidation at the anode, reduction at the cathode).
2. Cell reaction: Ni(s) + Cl2(g) → Ni2+(aq) + 2Cl–(aq)
E ? E° ? n ? Q ?
E = 1.791 V (given)
E° =
= 1.592 V (calculated)
Q = [Ni2+][Cl–(aq)]2 / P(Cl2) =
n = 2
E = E° – (0.0591/n)log10Q
[Cl–] = 1.2 × 10–3 M
93
NOTES:
Question #4: For any galvanic cell to operate,
A) ΔG° and E° are both positive.
B) ΔG° and E° are both negative.
C) ΔG° is positive and E° is negative.
D) ΔG° is negative and E° is positive.
Question #5: A cell consists of a magnesium electrode immersed in a solution of
magnesium chloride and a silver electrode immersed in a solution of silver
nitrate. The two half-cells are connected by means of a salt bridge. It is possible
to increase the voltage of the cell by
A) adding sodium chloride to both half-cells.
B) decreasing the Mg2+ and increasing the Ag+.
C) increasing the size of Mg electrode & decreasing the size of the Ag electrode.
D) decreasing the size of Mg electrode & increasing the size of the Ag electrode.
E) increasing the Mg2+ concentration.
EXAMPLES of PRACTICAL CELLS:
1) Measurement of [Ions] in solution
Because of the relationship between the cell’s potential (or emf) and Q, cell
potential can be used to measure ion concentration.
A common method of measuring Cl– is to use a silver/silver chloride
electrode (dipping into the solution containing Cl– ions) as one electrode and
a Saturated Calomel Electrode (SCE) as the other. It is observed that the
silver/silver chloride electrode is positive, i.e., it is the cathode (reduction)
and the cell voltage is +0.033 V.
The cell voltage is calculated by: Ecell = EAg/AgCl – ESCE
0.033 V = EAg/AgCl – 0.2458 (known value)
EAg/AgCl = +0.279 V
For the STANDARD silver/silver chloride electrode, the relevant SRP equation is:
AgCl(s) + e– → Ag(s) + Cl–(aq), E° = +0.222 V
Now use the Nernst eq’n to solve for [Cl–] within the silver/silver chloride half cell:
E = E° –(0.0591/n)log10Q
EAg/AgCl = E°Ag/AgCl –(0.0591/n)log10Q
log10[Cl–] = (0.279 – 0.222)/(-0.0591)
94
2) pH
[Cl–] = 10{(0.279 – 0.222)/(-0.0591)} = 0.11 M
meters (Read section in text)
NOTES:
A pH meter is simply an ion–selective electrode with a membrane that is a very thin
piece of special glass which is permeable to H+ ions. Inside the membrane is a sol’n
of HCl of a fixed concentration. A reference electrode completes the circuit.
Reference: Olmsted & Williams, Chemistry, 3rd ed., Figure 18-18 (Pg 889)
The voltage of such a system is given by the equation:
E = constant – 0.0591 × log {1/[H+]} OR since pH = –log[H+] = log {1/[H+]}
E = constant – 0.0591 × pH
What this means is that a pH meter will change its potential by 0.0591 V for each
change of 1 pH unit. The read–out voltmeter is set to read directly in pH units. The
meter is calibrated with a buffer of known pH before use. [Note the logarithmic scale.]
3) Determination of Ksp for an insoluble salt
1. AgBr(s) + e– → Ag(s) + Br–(aq);
2. Ag+(aq) + e– → Ag(s);
Given:
E° = +0.071
E° = +0.799 V
The reactions at the anode and cathode, respectively, are
and E° for AgBr(s) → Ag+(aq) + Br–(aq) is equal to
Therefore,
Ksp = 10(nE°/0.0591)
= 10 {
= 5 × 10–13
}
Note: The chemical rxn associated with a Ksp is defined as the solute dissolving into
its ions. (refer to section 17.1 and Appendix G for Ksp values)
95
Example: The standard potential of the cell: Pb(s) | PbSO4(s) | SO42–(aq) ||Pb2+(aq) |
Pb(s) is +0.23 V at 25ºC. Calculate the equilibrium constant for the reaction of
Pb2+(aq) with SO42–(aq). Determine the Ksp for PbSO4(s).
NOTES:
ASIDE: Dentistry & Chemistry:
The Pain of a Dental Voltaic Cell
If one has an amalgam dental filling and accidentally bites down on a scrap of foil, they will feel a jolt of pain. Here’s why: The aluminum foil acts as an active anode (E°(Al) = ‐1.66 V), saliva as the electrolyte, and the filling (usually a Ag/Sn/Hg alloy) as an inactive cathode. O2 is reduced to water, and the short circuit between the foil in contact with the filling creates a current that is sensed by the nerve of the tooth. Reference: Silberberg, Chemistry – The Molecular Nature of Matter and Change,
3rd ed., (McGraw-Hill, New York, 2003), p. 913.
4) Commercial Voltaic Cells, i.e., BATTERIES:
Voltaic cells are used commercially as portable energy sources (batteries).
Two main kinds –rechargeable and disposable
RECHARGEABLE:
1. Lead–Acid battery (or Lead Storage Cell)
Reference: Olmsted & Williams, Chemistry, 3rd ed., Figure 18-20 (Pg 892)
Rxns:
PbO2(s) + HSO4–(aq) + 3H3O+(aq) + 2e– ↔ PbSO4(s) + 5H2O(l); E° = +1.6913 V
PbSO4(s) + H3O+(aq) + 2e– ↔ Pb(s) + HSO4–(aq) + H2O(l); E° = –0.3588 V
96
2. Nickel-Cadmium Batteries
NOTES:
Cathode Reaction: NiO(OH)(s) + H2O(l) + e– → Ni(OH)2(s) + OH–(aq)
Anode Reaction: Cd(s) + 2OH–(aq) → Cd(OH)2(s) + 2e–
NET: 2NiO(OH)(s) + Cd(s) + 2H2O(l) → 2Ni(OH)2(s) + Cd(OH)2(s)
DISPOSABLE BATTERIES: (Refer to text for further details)
1. Leclanche batteries
2. Alkaline batteries
3. Mercury batteries
DETAILS:
1. Leclanche batteries
These are the batteries used to power flashlights, radios, etc. Each has a voltage
(E not E°) of 1.5 volts when new and the voltage gradually falls off during use
as the battery reactants are used up.
Anode: A ZINC cup to hold the reactants. This is the negative connection –
usually at the bottom of the battery.
Cathode: An INERT GRAPHITE rod – the Positive terminal (usually top centre)
The Zinc cup is filled with a paste of MnO2(s), NH4Cl(s) and water.
Rxns are quite complex.
Anode Rxn:
Zn(s) → Zn2+ + 2e–
Cathode Rxn: 2MnO2(s) + 2NH4+(aq) + 2e– → Mn2O3(s) + 2NH3(aq) + H2O(l)
Cell Notation:
Zn(s) | Zn(NH3)4Cl2(s), NH4Cl(aq), MnO2(s) | C(s)
Note: Material in centre is separated by commas since it exists
in the form of a paste (same phase).
Disadvantages: Not very good shelf life and when the battery is 'dead', real
chance of corrosion.
97
NOTES:
2. Alkaline batteries
Reference: Olmsted & Williams, Chemistry,
3rd ed., Figure 18-21 (Pg 893)
Very similar to the Leclanche cell, but
replacing NH4Cl with NaOH.
More expensive, has a steel casing.
Anode usually a ZINC rod.
OH– is produced at the cathode and
consumed at the anode, so its conc’n is
effectively constant → steady voltage.
Anode Reaction:
Zn(s) + 2OH–(aq) → ZnO(s) + H2O(l) + 2e–
Cathode reaction: 2MnO2(s) + H2O(l) + 2e– → Mn2O3(s) + 2OH–(aq)
Cell Notation: Zn | Zn(OH)2(s), NaOH(aq), MnO2(s) | C(s)
3. Mercury batteries:
Reference: Olmsted & Williams, Chemistry, 3rd ed., Figure 18-22 (Pg 894)
Widely used for cameras, watches, hearing aids, pacemakers, etc.
Small size and constant voltage.
Anode Reaction:
Zn(s) + 2OH–(aq) → ZnO(s) + H2O(l) + 2e–
Cathode Reaction: HgO(s) + H2O(l) + 2e– → Hg(l) + 2OH–(aq)
Overall reaction:
Zn(s) + HgO(s) → ZnO(s) + Hg(l)
Note: In the overall equation, we have all solids or liquids, so Q in the
Nernst equation is 1 and E = E° = 1.35 volts for the life of the battery. The
migration of OH– from the Hg electrode to the Zn electrode carries current
and maintains a uniform concentration.
98
FUEL CELLS: are essentially a battery, too, but it operates with a continuous
NOTES:
supply of energetic reactants, or fuel.
Refer to the following web site to learn more about the Ballard© fuel cell:
http://www.ballard.com/About_Ballard/Resources/How_Fuel_Cells_Work.htm
(Ballard has manufacturing facilities in Burnaby, British Columbia,
Michigan, Massachusetts and Nabern, Germany.)
Corrosion
Corrosion is the natural process that returns refined metals to more stable metal
oxides. Corrosion is particularly a problem with iron and steel, where this natural
process cuts its lifetime. This process is not a problem for Au, Al, Pb or Cu. Why?
In the case of iron:
Fe2+(aq) + 2e– → Fe(s), E° –0.44 V
O2(g) + 4H+(aq) + 4e– → 2H2O(l), E° +1.23 V
Once in solution the Fe2+ is oxidised to Fe3+ which is converted to hydrated Fe2O3
(i.e., rust). This does NOT adhere to the metal surface and corrosion and pitting
occurs. See below.
Reference: Olmsted & Williams, Chemistry, 3rd ed., Figure 18-23 (Pg 895)
A water droplet on an iron surface is a miniature electrochemical cell. The oxidation of
iron occurs in an interior region of the droplet, whereas the reduction of oxygen
preferentially occurs near the air-droplet interface. Ionic charges carriers are required to
complete the circuit and allow the redox reactions to proceed.
Question: A piece of iron half-immersed in a sodium chloride sol’n will corrode
more rapidly than a piece of iron half-immersed in pure water because
A) the sodium ions oxidize the iron atoms.
B) the chloride ions oxidize the iron atoms.
C) the chloride ions form a precipitate with iron.
D) the chloride ions increase the pH of the solution.
E) the sodium ions and chloride ions carry a current through the solution.
99
To avoid corrosion:
NOTES:
1) Protect the surface of the metal from exposure to air and water by
• Paint or coating (e.g., rust-proof paint or “undercoating” on vehicles)
• Coat with another metal, such as Zn (this is called galvanizing)
Zn2+(aq) + 2e– → Zn(s); E° = –0.76 V
O2(g) + 4H+(aq) + 4e– → 2H2O(l); E° = +1.23 V
Zinc lies below iron in the SRP table and, if a scratch exposes the iron below,
the more strongly reducing zinc is oxidized not the iron.
2) For larger objects, such as ships and pipe lines, a sacrificial anode is used.
A block of zinc or magnesium is attached to the object to supply electrons.
The sacrificial anode is cheap to replace instead of the object it is protecting.
In the same way, vehicles sometimes have a negative ground system as part of
their electrical circuitry. The body of the vehicle is connected to the anode of the
battery. The anode of the battery forms a sacrificial anode.
Another Example of redox reactions: Blue Sapphires
•
If trace amounts of Ti4+ is present in a crystal of corundum (Al2O3), no colour change.
•
If Fe2+ or Fe3+ impurities are present in corundum, a pale yellow colour results.
•
When both Ti4+ and Fe2+ present, a transparent blue colour seen due to redox:
Ti4+ + Fe2+ → Ti3+ + Fe3+
•
Reaction requires energy → white light supplies red-orange-yellow light.
•
Reverse reaction occurs quickly without emitting light.
Problems:
1. Predict the effect on the cell voltage (bigger, smaller, no change) for a cell with the reaction:
2H+(aq) + SO42–(aq) + Pb(s) → H2(g) + PbSO4(s)
a) Increase in pressure of H2(g)
b) Increase in size of the Pb electrode
c) Decrease in pH
d) Dilution of the electrolyte with water
e) Addition of Na2SO4 to the cell electrolyte
f) Decrease in amount of PbSO4(s)
g) Addition of a small amount of NaOH
100
2. You are told the following:
Np3+ + Sm → Np + Sm3+
V2+ + Th → V + Th4+
V2+ + Np → V + Np3+
Th4+ + Np → no reaction
Sm3+ + Th → no reaction
Put the relevant half–reactions in the correct order, with the one with the most negative SRP
last and the most positive first, and the others in the correct sequence.
3. Given the following data at 298 K, Ag(NH3)2+(aq) ↔ Ag+(aq) + 2NH3(aq);
and Ag+(aq) + e– → Ag(s); E° = +0.799 V, calculate E° for
K = 6.21 x 10–8
the half–reaction: Ag(NH3)+(aq) + e– → Ag(s) + 2NH3(aq).
4. Consider the following cell: Pt |H2(g, 1 atm) | H+(aq, ? M)|| Ag+(aq, 1.0 M) |Ag(s)
If the voltage of this cell is +1.04 V at 25oC and the standard potential of the Ag+/Ag couple is
+0.80 V, calculate the pH in the anode compartment.
5. Which of the following are true for a electrolytic cell?
A) ΔG < 0
B) K > 1
C) Ecell > 0
D) All of the above are true.
E) None of the above are true.
6. a) The reaction occurring in the voltaic cell shown above a has free energy,
A) ΔG = –212 kJ that becomes more negative with time.
e–
Voltmeter
e
–
K+
B) ΔG = 212 kJ that becomes more negative with time.
Cl–
C) ΔG = –212 kJ that becomes more positive with time.
NO3 –
Zn2+
NO3 –
Zn
NO3
I
–
NO3 –
Zn2+
SO4 2–
Cu2+
Cu
Cu2+
SO4 2–
D) ΔG = 212 kJ that becomes more positive with time.
E) ΔG = –212 kJ that does not change with time.
II
b) The reaction occurring in the voltaic cell shown below has an equilibrium constant,
A) Kc = 2 × 1037 that decreases with time.
D) Kc = 2 × 10–37 that increases with time.
B) Kc = 2 × 10–37 that decreases with time.
E) Kc = 2 × 1037 that does not change with time.
C) Kc = 2 × 1037 that increases with time.
101
7. All the following statements concerning zinc are true EXCEPT
A) Zn is a good reducing agent.
B) Zn2+ is a poor oxidizing agent.
C) Zn2+ is a colorless ion.
D) Zn2+ is a diamagnetic ion.
E) Zn is used as a cathode in the protection of Fe from corrosion.
8. Why is aluminum protected from corrosion?
(Note: The standard reduction potential for Al3+ is −1.66 V)
A) Al is not protected from corrosion.
B) Oxygen and Al have no affinity for one another.
C) The oxidation of Al is not a favored process as seen by the std. reduction potential for Al3+.
D) Aluminum forms a protective oxide coating.
Answers to some of the problems:
1. Nernst Equation for reaction: E = 0.360 –(0.0591/2) × log{PH2(g)/[H+]2[SO42–]}
Anything that makes the reaction go L → R gives a bigger cell voltage.
2. The four half–reactions are (in random sequence) are:
Np3+ + 3e–→ Np, (1)
V2+ + 2e–→ V, (2)
Th4+ + 4e–→ Th, (3)
Sm3+ + 3e–→ Sm, (4)
From the observations, we get (1) more positive than (4), (2) more positive than (3),
(2) more positive than (1), (1) more positive than (3) and (3) more positive than (4).
This gives (2) > (1) > (3) > (4).
3. Ag(NH3)2+(aq) + e– → Ag(s) + 2NH3(aq); E°1
Ag+(aq) + e– → Ag(s);
Ag(NH3)2+(aq)
Ag+(aq) + 2NH3(aq)
E° = +0.799 V
E°cell = E°1 – 0.799
but E°cell = (0.0591/n)log10Keq = (0.0591/1)log10(6.21×10–8) = –0.426
∴E°1 = 0.373 V
4. pH = –log [H+] = 4.06; Note that E = E° – (0.0591/2)log10[H+]2 (pull through the exponent)
= E° – (0.0591/2) × 2 log10[H+] (the 2’s cancel)
= E° + (0.0591) × (–log10[H+]) (pull the negative thru’)
= E° + (0.0591) × (pH)
(simplify for pH)
102