Download Electrochemistry

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Electrochemistry
The kinds of reactions we need to consider are oxidation-reduction reactions. To extract
electrical energy, we need to have electrodes.
Let’s consider a lemon cell: by putting two electrodes in a lemon, simple galvanic cell can be
created. The voltage one measures depends on the metal electrodes chosen, not the lemon. In
class, we chose silver and iron electrodes.
We measured a positive voltage when the positive (red) lead of the meter is attached to the Ag
electrode, the negative (black) lead to the nail. The electrons flow toward the positive lead when
a positive voltage is measured.
Thus, in this simple cell, electrons flow away from Fe, toward Ag. Silver is reduced and iron is
oxidized.
Loss of electrons – oxidation
Gain of electrons – reduction
Mnemonic devices to remember: Oxidation is Loss – Reduction is Gain (OIL RIG)
Or (my personal favorite):
Lose Electrons Oxidation – Gain Electrons Reduction (Leo says ‘Ger’)
We measured 0.608 v for the (Ag, Fe) pair, and the
voltage shown is about right for a (Cu, Zn) pair.
The diagram below shows a more general way to
construct galvanic cells. Note that it is impossible
to observe oxidation without the occurrence of a
corresponding reduction reaction. That is, single
electrode potentials cannot be measured.
The diagram below indicates that the two half cells
can either be connected by a salt bridge or by a
porous plug. Without some such construction to
complete the circuit, current cannot flow.
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Let’s take a closer look at the cell in operation: you can see that electrons flow toward the
cathode. The definition of the cathode is that it is the electrode where reduction occurs.
The idea of a cathode in “cathode ray tubes” is a litle confusing. Here, the cathode supplies
electrons, just like in a galvanic cell, but in addition, the CRT applies a high negative voltage to
the cathode to drive electrons away from it. So, unlike the cathode in a galvanic cell, which is
generally at positive potential, the cathode in a CRT is negative because of an additional voltage
applied to it to make the electrons leave.
The role of a salt bridge in maintaining electroical netruality is important. In the cathode
compartment, reduction is occurring. So, positive ions are leaving the solution. Electrical
neutrality must be maintained, so positive ions must enter the cathode compartment. In the anode
compartment, positive ions are entering the solution, so negative ions must arrive from the salt
bridge. Sometimes, the salt bridge is just rolled up filter paper soaked with a concentrated KNO3
solution. K+ ions can enter the cathode compartment solution. In the anode compartment, the
negative ions required can come from NO3- from the salt bridge.
Now, let’s look at a cell in operation. Let’s consider the Zn, Cu cell.
The notation we use for the cell is:
Zn(s) ⏐Zn2+(aq)⏐⏐ Cu2+ (aq) ⏐Cu(s)
A single vertical line is a phase boundary i.e., between solution and solid, and the doubel line
denotes a salt bridge.
Current flows toward the cathode. We observe that copper is the cathode, zinc is the anode.
Zinc electrode
Copper electrode
Zn 2+ ions in
solution
Cu 2+ ions in solution
The oxidizing
agent is Cu2+
The reducing
agent is Zn
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The cell is not at equilibrium when current flows.
Here are the half reactions:
Cu 2+ + 2e- → Cu(s) the cathode reaction is reduction
Zn(s) → Zn 2+ + 2e- the anode reaction is oxidation
Cu 2+ is an oxidizing agent . The oxidizing agent is reduced. Good oxidizing agents are easily
reduced.
Zn(s) is a reducing agent. The reducing agent is oxidized. Good reducing agents are easily
oxidized.
All we can measure in a galvanic cell is the difference in the reduction potentials of the species
involved in the reaction.
Net reaction: Zn(s) + Cu 2+ (aq) → Zn2+ (aq) + Cu(s) E°cell = +1.10 volts
We know that E°cell represents the helf-cell voltage for the cathode reaction as written
plus the negative of the anode half-cell voltage
E°cell = E°cathode + (-E°anode), or simply
E°cell = E°cathode - E°anode
We need some standard way to determine each half-cell potential independently, and the way
this is done is to use a standard, readily reproducible reference cell, the standard hydrogen
electrode.
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The figure shown above indicates how such a standard electron is constructed. We say
2H3O+ + 2e- → 2H2O + H2(g)
E° is defined to be 0.000.
The standard state condition requires that [H3O+] = 1.00 M. By teaming this standard electrode
with any other half cell, observing which reactions occur, and measuring galvanic cell potentials,
half-cell potentials for all systems can be determined.
In the diagram shown above, when a Zn/Zn2+ half cell is coupled with a normal hydrogen
electrode (HNE), hydrogen is the cathode, and zinceis the anode. The cell voltage is +0.76 volts.
So E°cell = E°cathode + - E°anode = +0.76 v = 0.00 – (-E°anode )
The best oxidizing agents are found at
the upper left of the table of E° values
The best reducing agents are found at
the lower right of the table of E° values
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Here’s another way to think about the meaning of E° values.
Let’s consider a few entries from the table:
Cathode reaction runs like this
O3(g) + 2H+ + 2e-
→
O2(g) + H2O
+2.07 v
Cl2(g) + 2e-
→
2Cl-
+1.36
O2(g) + 4H+ + 4e-
→
2H2O
+1.23
Note: the oxidizing agent
(on the left, with the higher
reduction potential) reacts
with the reducing agent on
the right.
Anode reaction runs like this
Fe3+ + eCu 2+ + 2e-
→
→
Fe2+
Cu(s)
+0.77
+0.34
2H+ + 2e-
→
H2(g)
0.000
Fe 2+ + 2eZn 2+ + 2eAl 3+ + 3eNa+ + e-
→
→
→
→
Fe(s)
Zn(s)
Al(s)
Na(s)
-0.44
-0.76
-1.71
-2.74
So, now you can see how reactions occurs spontaneously. When E°cell is positive, the galvanic
cell runs spontaneously.
So, values of the reduction potential, yielding cell potential values, are a measure of the
spontaneity of the electron transfer process:
Electrical Work
E can make that identification quantitative by remembering that
∆G = reversible non-PV (electrical) work.
The way electrical work is done in an electrochemical cell is through transporting charge across
a potential difference.
If we move charge of magnitude Q across a potential difference of E, we can understand work
from the following diagram:
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+
+
+
+
+
+
+
Q
E
-
Pushing a charge Q against a poential difference means the charge (the system) does work
WBY = QE
This is work done BY the charge. Work done ON the charge (system) is –QE.
Charge is expressed in Coulombs (one fundamental charge = 1.6 × 10-19 Coulombs, so a mole of
electrons has a charge of (1.6 × 10-19 Coulombs/particle) (6.023 × 1023 particles/mole) = 96,485
Coulombs per mole.
We call this constant the Faraday, F = 96,485 Coulombs/equivalent
Thus, Q = nF. Q is measured in Coulombs and is the charge associated with n moles of
electrons.
So our definition of free energy tells us that ∆G = -nFE
If the potential difference conrresponds to a potential arising from a system in its standard
state, then
∆G°= -nFE°
So now we see that positive cell potentials correspond to cells passing current spontaneously.
Let’s think about this definition in the context of a table of standard reduction potentials. The
standard half cell potential is a measure of the driving force of the reaction relative to the
reduction of protons to hydrogen gas.
How does a reduction potential or a half cell potential vary with concentration?
We just use ∆G = ∆G° + RT ln Q
-nFE = -nFE° + RT lnQ
Divide both sides by –nF
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The result is
E = E°- (RT/nF) ln Q = E°- (2.303RT/nF) log Q
Evaluate this at 298 K, and you get
E = E°-
0.0591
log Q
n
This is the famous Nernst equation. Walter Nernst was a famous chemist in the late 19th and
early 20th century, well-known for his work on thermodynamics.
As an example of the Nernst equation, let’s see how the cell potential for the hydrogen electrode
varies with hydrogen ion concentration:
2H+ + 2e- → H2(g)
E° = 0.00 volts
So, according to the Nernst equation
E = 0.00 – (.0591/2) log
P(H 2 )
[H + ]2
Let’s assume that the hydrogen gas pressure is always 1 atm.
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So log + 2 = -log [H+]2 = -2 log [H+] = 2 pH Do you follow this math?
[H ]
Then, . This is an important relationship because it shows that a pH meter can be based
on a hydrogen electrode.
Suppose we want to know the half cell potential for a hydrogen electrode operated at a
pH of 7.00. The Nernst equation tells us that E = -0.0591 (7.00) = -0.41 volts.
We’ll use this result when we discuss the electrolysis of water.