chemistry 2000 lecture 15:...
TRANSCRIPT
Chemistry 2000 Lecture 15: Electrochemistry
Marc R. Roussel
February 28, 2019
Marc R. Roussel Chemistry 2000 Lecture 15: Electrochemistry February 28, 2019 1 / 33
Electrochemical cells
Electrochemical cells
anode= oxidation
K+
−Cl
Ve−
2+
Zn Cu
Cu2+Zn
cathode= reduction
Marc R. Roussel Chemistry 2000 Lecture 15: Electrochemistry February 28, 2019 2 / 33
Electrochemical cells
Key requirements for an electrochemical cell
K+
−Cl
Ve−
2+
Zn Cu
Cu2+Zn
Spatial separation of half-reactions
An external electrical circuit
A means of maintaining electroneutrality
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Electrochemical cells
Cell diagrams
K+
−Cl
Ve−
2+
Zn Cu
Cu2+Zn
There is an abbreviated notation to describe an electrochemical cell.Example:
Zn(s)|Zn2+(aq)‖Cu2+
(aq)|Cu(s)
A single bar represents direct contact between two phases.
The double bar represents an indirect junction between two misciblephases (e.g. a salt bridge).
When we know which is which, we put the anode on the left.
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Electrochemical cells
Reversible emf
K+
−Cl
Ve−
2+
Zn Cu
Cu2+Zn
Recall that reversibility of a process means that the system andsurroundings are in equilibrium during the process.
The voltage produced by the cell can be measured under reversibleconditions by using a voltage source that opposes current flow.
The externally applied voltage that just stops current flow is calledthe electromotive force or emf.
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Electrochemical cells
Some nearly synonymous words
Voltage
(Reversible) emf
Electric potential difference
Marc R. Roussel Chemistry 2000 Lecture 15: Electrochemistry February 28, 2019 6 / 33
Gibbs free energy
So why is G called the Gibbs free energy?
We can show that ∆G is the non-pV reversible work in a process.
If w < 0, work is done by the system. This could be used to, e.g.,raise a weight.
The maximum work that can be extracted from a system is −wrev.
This means that −∆G is the maximum non-pV work that could beextracted from a process.
Marc R. Roussel Chemistry 2000 Lecture 15: Electrochemistry February 28, 2019 7 / 33
Nernst equation
The Nernst equation
For a chemical reaction, ∆rG = wrev,non-pV
Electrical work on a charge q moving through an electric potentialdifference (voltage) ∆φ:
w = q∆φ
If we measure the voltage under reversible conditions, ∆φ is theemf E .
∆rG = qE
The charge carriers are electrons, so q = −nF , where n is the numberof moles of electrons, and F is the charge of a mole of electrons, aquantity called Faraday’s constant:
∆rG = −nFEF = 96 485.3329 C/mol
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Nernst equation
∆rG = −nFE
It is generally more convenient to work in terms of the molar freeenergy ∆rGm. To get this, divide both sides of the equation by thenumber of moles of a reactant or product involved in a particularreaction:
∆rGm = −νeFE
where νe is the stoichiometric coefficient of the electrons in theoverall redox reaction.
We can apply the above equation under any conditions.In particular, under standard conditions,
∆rG◦m = −νeFE ◦
where E ◦ is the emf under standard conditions.
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Nernst equation
∆rGm = −νeFE and ∆rG◦m = −νeFE ◦
Since∆rGm = ∆rG
◦m + RT lnQ,
we get
−νeFE = −νeFE ◦ + RT lnQ,
or E = E ◦ − RT
νeFlnQ.
The highlighted equation is known as the Nernst equation.
Marc R. Roussel Chemistry 2000 Lecture 15: Electrochemistry February 28, 2019 10 / 33
Nernst equation
Some important relationships
K∆rG
◦m = −RT lnK
⇐============⇒ ∆rG◦m
∆rG◦m = −νeFE ◦⇐===========⇒ E ◦
Marc R. Roussel Chemistry 2000 Lecture 15: Electrochemistry February 28, 2019 11 / 33
Nernst equation
The emf is an intensive quantity
Recall that free energy is an extensive quantity.
In deriving the Nernst equation, we divided ∆G by n.
This means that emf is an intensive variable.
The emf won’t change if you multiply a reaction.You never multiply the emf.
Marc R. Roussel Chemistry 2000 Lecture 15: Electrochemistry February 28, 2019 12 / 33
Nernst equation
Thermodynamic feasibility and the emf
∆rGm < 0 for a thermodynamically allowed reaction at constanttemperature and pressure, and ∆rGm = −νeFE .
We now have yet another way to decide if a reaction isthermodynamically allowed:
E > 0 for a thermodynamically allowed reactionat constant temperature and pressure.
When we write a cell diagram, it is assumed that oxidation occurs atthe electrode on the left.If this is wrong, the calculated emf will be negative, indicating thatreduction occurs at the electrode on the left, or equivalently that thereaction occurs in the opposite direction to that implied by thediagram.
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Nernst equation
Criteria for thermodynamic feasibility
> 0
T, p
∆r G < 0
∆Suniverse > 0
Q < K E
constant
Marc R. Roussel Chemistry 2000 Lecture 15: Electrochemistry February 28, 2019 14 / 33
Nernst equation
Half cells
A half-cell consists of an electrode (possibly including a gas flowing overthe electrode) and the necessary solution for one of thehalf-reactions in an electrochemical process.
Example: a zinc electrode bathing in a zinc nitrate solution
Marc R. Roussel Chemistry 2000 Lecture 15: Electrochemistry February 28, 2019 15 / 33
Nernst equation
Half-cell potentials
Suppose that we have three different half-cells, A, B and C.
We measure the emfs of the following cells:
A‖B B‖C A‖CEAB EBC EAC
Experimentally, we find the following:
EAB + EBC = EAC
This can be explained if each cell emf can be calculated as thedifference of half-cell potentials:
EAB = EB − EA
EBC = EC − EB
EAC = EC − EA
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Nernst equation
By convention, we use reduction potentials as our half-cell potentials.
Since oxidation occurs at the anode and we use half-cell reductionpotentials,
Ecell = Ecathode − Eanode
Conceptually, it is often useful to think of this equation as anaddition:
Ecell = E redcathode + E ox
anode
withE oxanode = −E red
anode
The negative sign is the result of reversing the reduction reaction toobtain the oxidation reaction that occurs at the anode.
Marc R. Roussel Chemistry 2000 Lecture 15: Electrochemistry February 28, 2019 17 / 33
Nernst equation
Standard reduction potentials
Our tables will contain standard reduction potentials, i.e. reductionpotentials under the usual thermodynamic standard conditions (unitactivities of all reactants and products, etc.).
Electric potential is a relative measurement, so there is no way toassign absolute values to the standard reduction potentials.
We arbitrarily assign a standard reduction potential of zero to thehalf-reaction
H+(aq) + e− → 1
2H2(g)
This choice fixes all other reduction potentials, both under standardand nonstandard conditions.
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Nernst equation
The (standard) hydrogen electrode
H2
H+
Pt
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Nernst equation
We can use a hydrogen electrode to measure electrode potentials,then use the Nernst equation to calculate standard reductionpotentials.
We then build tables of standard reduction potentials.
We can also use the hydrogen electrode to measure the reductionpotentials of electrodes for some defined conditions in order to createreference electrodes that are easier to use than the hydrogenelectrode.
Marc R. Roussel Chemistry 2000 Lecture 15: Electrochemistry February 28, 2019 20 / 33
Nernst equation
Example: Calculating an emf with the Nernst equation
Calculate the emf generated by the following cell at 25 ◦C:
Zn(s)|Zn2+(aq)(0.125 M)||Ag+(aq)(0.053 M)|Ag(s)
Data:
Ag+(aq) + e− → Ag(s) E ◦ = +0.7996 V
Zn2+(aq) + 2e− → Zn(s) E ◦ = −0.7618 V
Answer: 1.5126 V
Marc R. Roussel Chemistry 2000 Lecture 15: Electrochemistry February 28, 2019 21 / 33
Nernst equation
Example:Obtaining the standard reduction potential of Al3+
The cell
Pt(s)|H2(g)(1.00 bar)|H+(aq)(pH = 5.00)
‖AlCl3(aq)(0.00100 mol/L)|Al(s)
has an emf of -1.425 V at 25◦C.
We want to get the standard reduction potential of Al3+, i.e. thestandard half-cell potential for
Al3+(aq) + 3e− → Al(s)
Marc R. Roussel Chemistry 2000 Lecture 15: Electrochemistry February 28, 2019 22 / 33
Nernst equation
Example: standard reduction potential of Al3+
(continued)
Pt(s)|H2(g)(1.00 bar)|H+(aq)(pH = 5.00)‖AlCl3(aq)(0.00100 mol/L)|Al(s)
emf = −1.425 V
E = E ◦ − RT
νeFlnQ
Strategy:
We know the cell emf, E in the Nernst equation.We first want to calculate the standard cell potential E ◦
by rearranging the Nernst equation.Then we can relate E ◦ to the standard reductionpotentials of the two half cells, one of which is ourunknown.
Marc R. Roussel Chemistry 2000 Lecture 15: Electrochemistry February 28, 2019 23 / 33
Nernst equation
Example: standard reduction potential of Al3+
(continued)
Pt(s)|H2(g)(1.00 bar)|H+(aq)(pH = 5.00)‖AlCl3(aq)(0.00100 mol/L)|Al(s)
From the cell diagram:
Anode: H2(g) → 2H+(aq) + 2e− E ◦
H+/H2= 0 V
Cathode: Al3+(aq) + 3e− → Al(s)
Marc R. Roussel Chemistry 2000 Lecture 15: Electrochemistry February 28, 2019 24 / 33
Nernst equation
In order to get the electrons to cancel, we multiply the firsthalf-reaction by 3 and the other by 2:
3H2(g) → 6H+(aq) + 6e−
2Al3+(aq) + 6e− → 2Al(s)
3H2(g) + 2Al3+(aq) → 6H+(aq) + 2Al(s)
with νe = 6.
We have the measured cell emf E = −1.425 V.We can rearrange the Nernst equation to solve the for standard cellemf E ◦:
E ◦ = E +RT
νeFlnQ
= E +RT
νeFln
((aH+)6
(aH2)3(aAl3+)2
)
Marc R. Roussel Chemistry 2000 Lecture 15: Electrochemistry February 28, 2019 25 / 33
Nernst equation
E ◦ = E +RT
νeFln
((aH+)6
(aH2)3(aAl3+)2
)
aH+ = 10−pH = 10−5.00 = 1.0× 10−5
aH2 = pH2/p◦ = (1.00 bar)/(1 bar) = 1.00
aAl3+ = [Al3+]/c◦ = (0.00100 mol/L)/(1 mol/L) = 0.00100
Therefore
E ◦ = − 1.425 V +(8.314 460 J K−1mol−1)(298.15 K)
6(96 485.3329 C/mol)
× ln
((1.0× 10−5)6
(1.00)3(0.00100)2
)= −1.662 V
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Nernst equation
We now have E ◦ = −1.662 V for the reaction
3H2(g) + 2Al3+(aq) → 6H+(aq) + 2Al(s)
However, E ◦ = E ◦cathode − E ◦
anode, or in our case
E ◦ = E ◦Al3+/Al − E◦
H+/H2
∴ E ◦Al3+/Al = E ◦ + E◦
H+/H2
= −1.662 + 0 V = −1.662 V
Marc R. Roussel Chemistry 2000 Lecture 15: Electrochemistry February 28, 2019 27 / 33
Nernst equation
Example: Standard free energy of formation of Al3+(aq)
Emf measurements are an important source of standard free energiesof formation.
In the last example, we found that E ◦ = −1.662 V and νe = 6 for thereaction
3H2(g) + 2Al3+(aq) → 6H+(aq) + 2Al(s)
∴ ∆rG◦m = −νeFE ◦
= −6(96 485.3329 C/mol)(−1.662 V)
= 962.2 kJ/mol
Marc R. Roussel Chemistry 2000 Lecture 15: Electrochemistry February 28, 2019 28 / 33
Nernst equation
3H2(g) + 2Al3+(aq) → 6H+(aq) + 2Al(s)
∆rG◦m = 6∆f G
◦(H+) + 2∆f G◦(Al)
−[3∆f G
◦(H2) + 2∆f G◦(Al3+)
]= −2∆f G
◦(Al3+)
∴ ∆f G◦(Al3+) = −1
2∆rG
◦m
= −1
2(962.2 kJ/mol) = −481.1 kJ/mol
Marc R. Roussel Chemistry 2000 Lecture 15: Electrochemistry February 28, 2019 29 / 33
Nernst equation
Example: A thiosulfate/chlorine cell
Consider the following cell:
Pt(s)|S2O2−3(aq)(0.0083 mol/L),HSO−
4(aq)(0.044 mol/L),H+(aq)(pH = 1.5)
‖Cl−(aq)(0.038 mol/L)|Cl2(g)(0.35 bar)|Pt(s)
We want to know the emf generated by this cell.
Problem: There are no appropriate half-reactions in ourelectrochemical table.
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Nernst equation
We should start by balancing the reaction.
We need to know which of the materials at the anode will be oxidized.Oxidation state of S in chemical species at anode:
S2O2−3 HSO−
4
+2 +6
=⇒ S2O2−3 will be oxidized.
Now balance those half reactions!
Pt(s)|S2O2−3(aq),HSO−
4(aq),H+(aq)‖Cl−(aq)|Cl2(g)|Pt(s)
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Nernst equation
Half-reactions:S2O2−
3 + 5H2O→ 2HSO−4 + 8H+ + 8e−
Cl2 + 2e− → 2Cl−
Overall reaction:
S2O2−3(aq) + 5H2O(l) + 4Cl2(g) → 2HSO−
4(aq) + 8H+(aq) + 8Cl−(aq)
νe = 8
Standard free energy change:
∆rG◦m = 2∆f G
◦(HSO−4 ) + 8∆f G
◦(Cl−)
−[∆f G
◦(S2O2−3 ) + 5∆f G
◦(H2O)]
= 2(−755.7) + 8(−131.0)
− [−522.5 + 5(−237.1)] kJ/mol
= −851.4 kJ/mol
Marc R. Roussel Chemistry 2000 Lecture 15: Electrochemistry February 28, 2019 32 / 33
Nernst equation
We can now proceed in either of two ways:1 Calculate ∆rGm = ∆rG
◦m + RT lnQ, then use E = −∆rGm/(νeF ).
2 Calculate E◦ = −∆rG◦m/(νeF ), then use the Nernst equation to get E .
Both ways will of course give the same answer.(Try it!)
Answer: E = 1.267 V
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