chem1612 - pharmacy week 9: nernst equation dr. siegbert schmid school of chemistry, rm 223 phone:...
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CHEM1612 - PharmacyWeek 9: Nernst EquationDr. Siegbert Schmid
School of Chemistry, Rm 223
Phone: 9351 4196
E-mail: siegbert.schmid@sydney.edu.au
Unless otherwise stated, all images in this file have been reproduced from:
Blackman, Bottle, Schmid, Mocerino and Wille, Chemistry, John Wiley & Sons Australia, Ltd. 2008
ISBN: 9 78047081 0866
Lecture 25-3
Electrochemistry Blackman, Bottle, Schmid, Mocerino & Wille:
Chapter 12, Sections 4.8 and 4.9
Key chemical concepts: Redox and half reactions Cell potential Voltaic and electrolytic cells Concentration cells
Key Calculations: Calculating cell potential Calculating amount of product for given current Using the Nernst equation for concentration cells
Lecture 26 - 4
The measured voltage across the cell under standard conditions is the standard cell potential E0
cell (also called emf).
Recap: Standard cell potential
Zn(s) + Cu2+(aq) Zn2+
(aq) + Cu(s)
E0cell = E0
cathode – E0anode = 0.34 - ( - 0.76)= 0.34 +0.76 = 1.10 V
Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s)
Lecture 26 - 5
Tricks to memorise anode/cathode 1. Anode and Oxidation begin with a vowels,
Cathode and Reduction with consonants.
2. Alphabetically, the A in anode comes before the C in cathode, as the O in oxidation comes before the R in reduction.
3. Think of this picture:
AN OX and a RED CAT
(anode oxidation reduction cathode)
Lecture 26 - 6
Standard cell potential and free energy For a spontaneous reaction, E0
cell > 0 and also ΔG0 < 0
For a non-spontaneous reaction, E0cell <0 and also ΔG0 > 0
So there is a proportionality between E0 and -ΔG0.
You also know that the maximum work done on the surroundings is
-wmax = ΔG
Electrical work done by the cell is w = Ecell × charge
Lecture 26 - 7
Standard cell potential and free energy The emf E0
cell is related to the change in free energy of a reaction:
∆G0 = –nFE0cell
∆G0 = Standard change in free energy
n = number of electrons exchanged
F = 96485 C/mol e- (Faraday constant)
∆G = –nFEcell
Also, away from standard conditions:
But what is Ecell?
Lecture 26 - 8
ExampleCalculate ∆G0 for a cell reaction:
Cu2+(aq) + Fe (s) Cu(s) + Fe2+ (aq)
Is this a spontaneous reaction?
This process is spontaneous as indicated by the negative sign of G0 and the positive sign for E0
cell.
Cu2+ + 2e– Cu E0 = 0.34 V
Fe2+ + 2e– Fe E0 = –0.44 V
E0cell = 0.34 - (-0.44) = 0.78 V
∆G0 = –nFE0cell
∆G0 = – 2 · 96485 · 0.78 = – 1.5 · 105 J
Lecture 26 - 9
Al is very easily oxidised,
Al3+ + 3e− Al Eo = -1.66V.
The filling is an inactive
cathode for the
reduction of oxygen,
O2 + 4H+ + 4e− 2H2O
and saliva is an electrolyte.
Put the three together (biting on a piece of foil) results in generation of a current and possible pain.
Al(s)|Al3+(aq) || O2(aq), H+(aq), H2O(aq)|Ag,Sn,Hg
Example: a Dental Galvanic Cell
Lecture 26 - 10
Example: a Dental Galvanic CellAl(s)|Al3+(aq) || O2(aq), H+(aq), H2O(aq)|Ag,Sn,Hg
O2 + 4H+ + 4e– 2H2O E0 = 1.23 V
Al3+ + 3e– AlE0 = –1.66 V
12H+ + 3O2 + 4Al 6H2O + 4Al3+
E0cell = 2.89 V
∆G0 = –nFE0cell=
–12 · 96485 · 2.89 = –3346 kJmol–1
Lecture 26 - 11
Recall ΔG = ΔG0 + RT ln(Q) (1)
Since ΔG0 = -nFE0cell and ΔG = -nFE
Equation (1) becomes
-nFE = -nFE0cell + RTln(Q) Q = [products] / [reactants]
dividing both sides by –nF gives:
E = E0 – RT ln(Q) nF
Nernst Equation
Walther Nernst
Nobel Prize 1920
But what is Ecell?
Ima
ge
from
no
be
lprize
.org
Lecture 26 - 12
Ecell = E0 – RT ln(Q)nF
E = cell potentialE0 = Standard cell potentialR = Real Gas Constant= 8.314 JK-1mol-1
T =Temperature (K)n = no. of e- transferredF = Faraday constant = 96485 C mol-1
Q = Reaction quotient(Q = K at equilibrium)
Nernst Equation
Since ln (x) = 2.303 log (x)
E = E0 – 2.303 · RT log(Q) nF
At 25°C, (2.303·R·298)/96485 = 0.0592
)log(0592.00 Qn
EEcell Nernst equation more commonly
written like this (note: only at 25°C)
Lecture 26 - 13
Calculate the expected potential for the following cell:
i) [Cu2+] = 1.0 M; [Zn2+] = 10-5M
ii) [Cu2+] = 10-5M; [Zn2+] = 1.0 M
Example calculation 1
)log(0592.00 Qn
EEcell
Firstly, work out the value of n :
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
Cu2+ + 2e- Cu
Zn Zn2+ + 2e-
2 mol e- transferred per mole of reaction: n = 2
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) E0 = 1.1 V
Lecture 26 - 14
[Zn2+] (M) [Cu2+] (M) Q log(Q) Ecell (V)
1.0 1.0
10-5 1.0
1.0 10-5
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
)log(0592.00 Qn
EEcell
Example calculation
(n=2)
1.0
10-5
105
0.0
-5.0
5.0
1.10
1.25
0.95
][
][2
2
Cu
ZnQ
Lecture 26 - 15
Demo: The effect of concentration
What happens when the concentration of one cell is changed?
Cu|Cu2+||Cu2+|Cu
Both compartments of the voltaic cell are identical.
E0cell = E0
copper – E0copper = 0 (in standard conditions, 1M concentrations)
0.00 V
Lecture 26 - 16
Demo: Cu Concentration Cell
Can we explain this?…
Cu2+ + 2e- Cu High [Cu2+]Low [Cu2+]
Cu Cu2++ 2e-
Cu |Cu2+||Cu2+|Cu
E0 same for both half-reactions, so E0cell= 0.
However, we have reduced the concentration of Cu2+ in one cell = non-standard conditions.
Electrical energy is generated until the concentrations in each half-cell become equal (equilibrium is attained).
Add Na2S –precipitate forms.
Lecture 26 - 17
Cu Concentration Cell
Cu|Cu2+||Cu2+|Cu
Low [Cu2+] High [Cu2+]
Cu Cu2++ 2e- Cu2+ + 2e- Cu
E0cell is same on both sides, but the Cu concentrations are different.
More charge carriers in one half-cell. If we poured the two solutions together, we would expect
spontaneous mixing of two solutions of different concentrations to give one of equal concentration.
The electrical connection allows electrons to pour from one half-cell to the other.
Lecture 26 - 18
Concentration cells The measured cell potential in our experiment was “Voltage”.
Let’s work out what the voltage should be:
Cu Cu2+ (0.01 M) + 2e-
Cu2+ (0.1M) + 2e- Cu
Cu2+ (0.1M) Cu2+ (0.01 M) Ecell =
)log(0592.00 Qn
EEcell
1.0log0296.00.0
y= “Voltage”
Solve for “Voltage”:
“Voltage”
“Voltage” = 0.0296 V
“Voltage”
0.01
Lecture 26 - 19
Concentration cells
The cell potential depends on the concentration of reactants.
Implication: We need to specify concentration when referring to the cell potential.
The overall potential for the Cu/Cu2+ concentration cell is:
E = E0cell – 0.0592/2 · log [Cu2+]dil / [Cu2+]conc
Corollary: It is useful to define a standard concentration, which is 1 M.
Lecture 26 - 20
1. Standard Hydrogen Electrode (SHE)
2. Metal-Insoluble Salt Electrode: Standard Calomel Electrode (SCE) and Silver Electrode
3. Ion-Specific: pH electrode
Reference Electrodes
Lecture 26 - 21
Standard Hydrogen Electrode (SHE)
Platinum – gas electrode
H2 electrode
H2 2H+ + 2 e- Eo = 0.00 V
Metal – Metal ion Electrode
Cu2+ + 2e- Cu Eo = 0.34 V
Pt|H2(g)|H+(aq)||Cu2+(aq)|Cu(s) Eocell = 0.34 – 0 = 0.34 V
anode cathode
Finely divided surface Pt electrode.
HCl solution with [H+] =1,
H2 p= 1 atm bubbling over the electrode.
H2 absorbs on the Pt, forming the equivalent of a 'solid hydrogen‘ electrode in equilibrium with H+.
Lecture 26 - 22
The Standard Hydrogen Electrode (SHE) isn't convenient to use in practice (can be contaminated easily by O2 or organic substances).
There are more practical choices, like metal - insoluble salt electrodes. The potential of these electrodes depends on the concentration of the
anion X- in solution.
In practice 2 interfaces:
1. M / MX insoluble salt: MX (s) + e- M (s) + X-(s)
2. coating/solution: X- (s) X- (aq)
Overall: MX (s) + e-(metal) M (s) + X- (aq)
Metal-insoluble salt electrodes
M
X-
X-
MXC+
C+
C+
C+
X-
X-
Lecture 26 - 23
The concentration of anions in solution is controlled
by the salt's solubility:
[Ag+] [X-] = Ksp
Normal calomel electrode, Pt | Hg | Hg2Cl2 | KCl (1M) E 0 = 0.28 V
Saturated calomel electrode Pt | Hg | Hg2Cl2 | KCl(sat.) E0 = 0.24 V
Silver/Silver chloride, Ag | AgCl | Cl- (1M) E 0 = 0.22 V
(used in pH meters)
Metal-insoluble salt electrodes
M
X-
X-
MXC+
C+
C+
C+
X-
X-
Lecture 26 - 24
2
The ‘saturated calomel electrode’ (SCE) features the reduction half-reaction:
Hg+ + e– Hg
Hg2Cl2 2Hg+ + 2Cl– Overall:
Hg2Cl2 (s) + 2e– 2 Hg (s) + 2Cl– (sat)
Pt | Hg | Hg2Cl2 | KCl ||
Standard cell potential of
E0 = 0.24 V.
Saturated Calomel Electrode
5 M
Lecture 26 - 25
Calomel: Hg2Cl2(s) + 2e- 2Hg(l) + 2Cl-(aq) E0 = 0.24V Zn Zn2+ + 2e-E0 = + 0.76V
(reversed because it is written as an oxidation)
Q: The standard reduction potential of Zn2+/Zn is - 0.76 V. What would be the observed cell potential for the Zn/Zn2+ couple when measured using the SCE as a reference?
Ans: The Zn will be oxidised (lower reduction potential), soE (cell) = 0.76 + 0.24 = 1.00 V respect to the SCE
So to get the oxidation half-reaction E0 using the SCE as cathode, subtract 0.24 V from the volt meter reading.
Cell Potentials 1
0.24 Hg+/Hg
0.0V H2/H+
-0.76 Zn2+/Zn
Lecture 26 - 26
Summary
CONCEPTS Concentration cells Nernst equation
CALCULATIONS Work out cell potential from reduction potentials; Work out cell potential for any concentration (Nernst equation)
Lecture 26 - 27
Ag+ + e– AgAgCl Ag+ + Cl–
Overall: AgCl + e– Ag + Cl– E0 = 0.22 V
AgCl (s) + e– Ag (s) + Cl– (sat)
A thin coating of AgCl is deposited
on the pure metal surface.
Silver electrode
Ag | AgCl | Cl–
Lecture 26 - 28
Cell Potentials 2 A Fe3+/Fe2+ cell with [Fe3+]=[Fe2+] =1 M has a potential of
0.55V respect to the Ag/AgCl electrode (E0= 0.22 V). What is the potential of this electrode with respect to the SHE?
Answer.
The reactions that occur in the two half-cells are:
Fe3+ + e- → Fe2+ at the cathode E = 0.55 V; E0 = ?
Ag + Cl- +e-→ AgCl at the anode E0= 0.22 V
The potential of this electrode with respect to the SHE is the difference of the two electrode potentials:
E = E0 - 0.22 E0Fe3+/Fe2+ = 0.55 + 0.22 = 0.77 V
0.22 Ag+/Ag
0.0V H2/H+
?
Lecture 26 - 29
Measurement of pH
We could construct a concentration cell, using the standard hydrogen electrode (SHE) and a hydrogen electrode:
H2(g, 1 atm) 2H+(aq, unknown) + 2e- anode
2H+ (aq, 1M) + 2e- H2(g, 1 atm) cathode
2H+ (1M) 2H+ (unknown) Ecell = ?
Using Nernst equation:
i.e. the measurement of the cell potential provides pH directly!
pH 0.0592 ][H ln 22F
RT-
][H
][H ln
nF
RT E E un2
ref
2un0
cell
at 25°C,
unknown
Lecture 26 - 30
pH electrodeThe pH electrode potential is typically measured versus a fixed reference calomel electrode.
E’ is the sum of the constant offset potentials of the inner glass surface/solution, the Ag/AgCl electrode, and the calomel electrode.
Eglass electrode= E’ + RT/2.303F log [H+]
• Based on a thin glass membrane: a modified glass enriched in H+ and resulting in a hydration layer a few micrometers thick.
• Inside the membrane is a 'reference solution' of known [H+] (1M HCl).
• The potential difference relevant to pH measurement builds up across the outside glass/solution interface marked ||.
Lecture 27 - 31
Nernst Equation
]reactants[
]products[
)log(0592.00
Q
Qn
EEcell
Zn(s) + Cu2+(aq) Zn2+
(aq) + Cu(s)
The Nernst equation describes the effect of concentration on cell potential.
When Q < 1, [reactants] >[products] and the cell can do more work.
When Q = 1, Ecell = E0cell (standard conditions [x] = 1 M).
When Q > 1 , [products] > [reactants] and Ecell is lower.
Lecture 27 - 32
Difference between Q and K
Breakdown of N2O4 to NO2:
N2O4 (g, colourless) → 2 NO2 (g, brown)
][][
42
22
ONNOQ
Q =K
Figure from Silberberg, “Chemistry”,
McGraw Hill, 2006.
Lecture 27 - 33
Potential of an electrochemical cell
]C[
]Z[2
2
u
nQ
0 V
~1037
Zn(s) + Cu2+(aq) Zn2+
(aq) + Cu(s)
Ece
ll (V
olts
)
Ecell decreases as the reaction proceeds, until at equilibrium Ecell =0 and
.)log(0592.00 Kn
E
K
)Qlog(n
.EEcell
059200= -
Lecture 27 - 34
Recap: Examining Q To K Ratios
If Q/K < 1 Ecell is positive for the reaction as written. The smaller the Q/K ratio,
the greater the value of Ecell and the more electrical work the cell can do.
If Q/K = 1, Ecell = 0. The cell is at equilibrium and can no longer do work.
If Q/K > 1, Ecell is negative for the reaction as written. The cell will operate in
reverse – the reverse reaction will take place and do work until Q/K = 1 at equilibrium.
)log(0592.00 Qn
EEcell
Lecture 27 - 35
Large K products favoured large standard cell potential, E0
Link between E 0 and KQ: What happens if the reaction proceeds until equilibrium is reached?
A: The reaction stops, therefore the voltage, or electrical potential, is zero (the battery is flat). In mathematical terms:
0)log(0592.00 Qn
EEcell
So the equilibrium constant determines the cell potential.
)log(0592.00 Kn
E
At equilibrium Q=K
Lecture 27 - 36
Redox reactions are special
For redox reactions there is a direct experimental method to measure K and ΔG°.
Figure from Silberberg, “Chemistry”,
McGraw Hill, 2006.
Lecture 27 - 37
Relation between E 0 and K
)log(0592.00 Kn
E
K is plotted on a logarithmic scale to give a straight line.
Figure from Silberberg, “Chemistry”,
McGraw Hill, 2006.
Lecture 27 - 38
Example question 2
Co(s) + Ni2+(aq) Co2+(aq) + Ni(s) E 0 = 0.03 V
Q: A voltaic cell consisting of a Ni/Ni2+ half-cell and Co/Co2+ half-cell is constructed with the following initial concentrations: [Ni2+] = 0.80 M; [Co2+]=0.2 M.
a) What is the initial Ecell?b) What is the [Ni2+] when the voltage reaches 0.025 V?c) What are the equilibrium concentrations of the ions?
Given: E 0 Ni2+
/Ni = -0.25 V; E 0 Co2+
/Co = -0.28 V
Ni2+ + 2e- → Ni E 0 = -0.25V
Co → Co2+ + 2e- E 0 = +0.28V
Lecture 27 - 39
Example question 2a
Co(s) + Ni2+(aq) Co2+(aq) + Ni(s) E 0 = 0.03 V
25.08.0
2.0
][
][2
2
Ni
CoQ
)log(0592.00 Qn
EEcell )25.0log(2
0592.003.0
)602.0(0296.003.0 V 048.0
a) What is the initial Ecell?
Lecture 27 - 40
Example question 2b
)log(0592.00 Qn
EEcell )log(0296.003.0025.0 Q
169.00296.0
005.0)log( Q
Q = 1.47
)8.0(
)2.0(
][
][47.1
2
2
x
x
Ni
CoQ
Co(s) + Ni2+(aq) Co2+(aq) + Ni(s)
0.80 - x 0.20+x
xx 2.047.1176.1
976.047.2 x x = 0.40
So when Ecell = 0.025 V
[Co2+] = 0.60 M
[Ni2+] = 0.40 M
b) What is the [Ni2+] when the voltage reaches 0.025V?
Lecture 27 - 41
Example question 2c
)log(0592.0
0 0 Kn
E )log(0296.003.0 K
014.10296.0
03.0)log( K
K = 10.24Co(s) + Ni2+(aq) Co2+(aq) + Ni(s)
0.80-x 0.20+x
)8.0(
)2.0(
][
][24.10
2
2
x
x
Ni
CoK
xx 2.024.10192.8
986.724.11 x x = 0.71
So at equilibrium,
[Co2+] = 0.91 M
[Ni2+] = 0.09 M
c) What are the equilibrium concentrations of the ions?
Lecture 27 - 42
Concentration Cells in Nature
Concentration cells are present all around us, e.g.
nerve signalling: concentration gradients produce electrical current
ion pumps across cell membranes: Na+ / K+ pump, Ca2+ pump
energy production and storage in cells: ATP
Lecture 27 - 43
Nerve cells
The membrane potential is more positive outside than inside the cell. On nerve stimulation, Na+ enters cell, the inside cell membrane
becomes > +ive, then K+ ions leave cell to re-equilibrate the outside. These rapid (ms) changes in charge across the membrane stimulate
the neighbouring region and the electrical impulse moves down the length of the cell.
Energy from ATP hydrolysis is used by ion pumps, so that across the nerve cell membrane concentration gradients are maintained:
Ion Concentration Gradient: Inside Outside
K+ High LowNa+ Low High
Figure from Silberberg, “Chemistry”,
McGraw Hill, 2006.
Lecture 27 - 44
Nerve cells Nernst Equation gives the membrane potential generated by the
differing extracellular versus intracellular concentrations of each ion:
inside
outsideion
inside
outsideion
ion
ionE
ion
ion
nF
RTE
][
][log5.61
][
][log
303.2
Substituting n = 1, T = 37oC:
( in mV)
1 mV = 10-3 V
Consider K+: [K+]outside = 3 mM, [K+]inside = 135 mM
102
65.15.61
135
3log5.61
][
][log5.61
K
inside
outsideK
E
K
KE
mV
mV
(Eo = 0)
Lecture 27 - 45
Cellular Electrochemistry Biological cells apply the principles
of electrochemical cells to generate energy in a complex multistep process.
Bond energy in food is used to generate an electrochemical potential.
The potential is used to create the bond energy of the high-energy molecule adenosine triphosphate (ATP) (energy currency for the cell).
N
NN
N
NH2
O
OHOH
HH
HH
OPO
O-
O
POP-O
O- O-
OO
adenosinetriphosphate
N
NN
N
NH2
O
OHOH
HH
HH
OPO
O-
O
P-O
O-
O
adenosinediphosphate
ATP
ADP
H2O
+ HPO42- +H+
ATP4- + H2O → ADP3- + HPO42- + H+
ΔG °’ = -30.5 kJ mol-1
ΔGo’ (solution at pH 7 and at human body temperature 37oC.)
Lecture 27 - 46
Bond Energy to Electrochemical Potential
Inside mitochondria, redox reactions are performed by a series of proteins that form the electron-transport chain (ETC) which contain redox couples, such as Fe2+/Fe3+.
Large potential differences provide enough energy to convert ADP into ATP.
Fig
ure
from
Silb
erb
erg
, “Ch
em
istry”,
McG
raw
Hill, 2
00
6.
Lecture 27 - 47
In nature, the most important reducing agent is a complex molecule named nicotinamide adenine dinucleotide, abbreviated NADH, which functions as a hydride donor (H-).
= NADH = biological reducing
agent
NAD+ = biological oxidising agent
Bond Energy to Electrochemical Potential
Lecture 27 - 48
ETC consists of three main steps, each of which has a high enough potential difference to produce one ATP molecule.
The reaction that ultimately powers ETC is the reduction of oxygen in the presence
of NADH:
2H+ + 2e- + ½ O2 → H2O Eo’ = +0.82
NADH + H+ → NAD+ + 2H+ + 2e- Eo’ = -0.32
NADH(aq) + H+(aq) + ½O2(aq) → NAD+
(aq) + H2O(l) Eo’overall = 1.14 V
Substantial energy release!
ΔGo’ = -nFEo’ = -2 · 96485 C mol-1· 1.14 V = - 220 kJ mol-1
Electron Transport Chain (ETC)
Lecture 27 - 49
ATP Synthesis
In short: e- are transported along the chain, while protons are forced into the intermembrane space.
This creates a H+ concentration cell across the membrane.
In this step, the cell acts as an electrolytic cell, i.e. uses a spontaneous process to drive a non-spontaneous process.
When [H+]intermembrane/[H+]matrix ~ 2.5 a trigger allows protons to flow back across membrane, and ATP is formed.
It’s not simple: Noble prize in 1997 to Boyer and Walker for elucidating this.
Figure from Silberberg, “Chemistry”,
McGraw Hill, 2006.
Lecture 27 - 50
SummaryCONCEPTS
Concentration cells Nernst equation E 0 and K Link between E , Q and K Applications of concentration cells
CALCULATIONS Work out cell potential from reduction potentials; Work out cell potential for any concentration (Nernst equation) Work out K from E 0
Work out pH from concentration cell
Lecture 26 - 51
Pop Quiz 1
Balance the following reaction in basic solution:
MnO4- + CN- MnO2 + CNO-
Answer: H2O + 2 MnO4- + 3 CN- --> 2 MnO2 + 3 CNO- + 2 OH-
Lecture 26 - 52
Pop Quiz 2
Balance the following reaction in basic solutions:
NO3- + Zn Zn2+ + NH3
Answer: NO3- + 4 Zn + 6 H2O → 4 Zn2+ + NH3 + 9 OH-
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