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Electrochemistry
• “Marriage” of redox and thermo
• Spontaneous electron-transfer reactions can result in spontaneous electric current if the reactants are separated by a wire– Voltaic (Galvanic) cells [Experiment 32!]
– We can use the “spontaneity” of the reaction to do electrical work
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(Continued)
• We can “push” electrons through a cell in order to make a nonspontaneous redox reaction occur.– Electrolysis cell [Experiment 32!]
– Doing work to “force” chemical reaction to occur [opposite of voltaic cell]
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Balancing Redox Equations
• Deferred until later
• For now just know that:– A half reaction has electrons written as a reactant
or a product• Oxidation half reaction: A reactant gets oxidized
(loses electrons); electrons appear as a “product”• Reduction half reaction: A reactant gets reduced
(gains electrons); electrons appear as a “reactant”
– A balanced redox equation does not show electrons explicitly. #e-’s lost = #e-’s gained (called “n”)
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Voltaic Cells
• Recall lab…
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• The spontaneous rxn occurs in the cell
• e-’s flow from – to + (“get to go where they want to go”)
• Anode = where ox occurs• Cathode = where red occurs• Salt bridge prevents charge
buildup (which would stop flow)
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You used graphite in place of Pt in lab for Fe2+/Fe3+ and I2/I-cells. A cheaper “inert” electrode.
Used when neither redox species in a half reaction (or electrode) is a neutral metal.
Neither is a neutral metal
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Standard Reduction Potentials (E°red)
• Recall lab– Make a bunch of different cells, get different
Ecell values (Eºcell if at standard state).
– Clearly some reductions are more favorable than others
• How do you know? [Which direction did e-’s flow?]• By how much?• Rank them? (Must pick a zero as reference.)
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Quick quiz
• NOTE: Every electrode compartment has one oxidizing agent and one reducing agent (this pair is called the redox “couple”)
Ox agent is ___
Red agent is ___ Ni (b/c it’s “more negative”; it has an electron to give)
Ni2+ (b/c it’s “more positive”; it has “room for an electron”
• If an electrode has Ni(s) and Ni2+ ions in it, which species is the oxidizing agent and which the reducing agent (of the pair)?
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Revisit Earlier Cell—Look at this as a “Competition for the electrons”. Which oxidizing agent “wants
them more”?
Who is the (possible)
oxidizing agent on the right? _____
Hint: The two “players” are Cu and Cu2+.
Cu2+
Who is the (possible)
oxidizing agent on the left? _____
Hint: The two “players” are Zn and Zn2+
Zn2+
Who wins? (Which one “got” the electrons?) ____Cu2+
Cu2+ “pulled harder”
So…which of the half reactions shown at the right is more favorable (greater tendency to happen)?
Cu2+ + 2 e- Cu(s)
Zn2+ + 2 e- Zn(s)
By how much?.....11
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Reducing Cu2+ is more favorable than reducing
Zn2+ …by 1.10 V! (Measure it w/voltmeter!)
We define a “standard reduction potential”, E°red, for every reduction half reaction such that:
E°cell = E°red(cathode) - E°red(anode)
The more positive the “E” (Ecell, Ered, or Eox), the more favorable the process
Where reduction takes place
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Reducing Cu2+ is more favorable than reducing
Zn2+ …by 1.10 V! (Measure it w/voltmeter!)
E°cell = E°red(cathode) - E°red(anode)
If E°red(Zn2+/Zn) were 0 V, E°red(Cu2+/Cu) would be +1.10 V
If E°red(Zn2+/Zn) were -1.0 V, E°red(Cu2+/Cu) would be +0.10 V
NOTE:
1.10 V = E°red(Cu2+/Cu) - E°red(Zn2+/Zn)
If E°red(Zn2+/Zn) were +1.0 V, E°red(Cu2+/Cu) would be +2.10 V
The “zero” is arbitrary, but must be chosen / agreed upon!13
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This electrode (SHE) was ultimately the one chosen by the scientific community to be the “zero” of potential.
2 H+ + 2 e- H2 (g); E°red = 0.0 V
Upshot:One can determine any E°red experimentally by just setting up a cell where one of the half cells is SHE!(next slide → )
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Ecell Ecathode - EanodeBoth as reductions
0.76 V = E°red(SHE) - E°red(Zn2+/Zn)
0.76 V = 0 - E°red(Zn2+/Zn)
E°red(Zn2+/Zn) = - 0.76 V
The reduction of H+ is more favorable than the reduction of Zn2+…. by 0.76 V! H+ (not Zn2+) gets reduced
Determining a Standard Reduction Potential using the SHE
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2 H+(aq) + 2 e- H2(g) 0
Zn2+(aq) + 2 e- Zn(s) -0.76
Could use this info to predict that this direct reaction would occur:
E°cell E°red(cat) - E°red(an)
Use these values to:
predict which reactions are spontaneous at standard state
and to
find any E°cell!
E°cell 0 – (-0.76) = +0.76 V
H+ gets reduced; Zn2+ does not (Zn gets oxidized):
2 H+ + Zn → H2(g) + Zn2+
is spontaneous: E°cell > 0
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Refers only to species on the left side of the arrow. E.g., F2, is a better ox agent than H2O2 which is better ox agent than Au3+ (but all of these species are very good oxidizing agents relative to most!)
Refers only to species on the right side of the arrow. E.g., F-, is a poorer red agent than H2O which is poorer red agent than Au(s) (but all of these are very poor reducing agents relative to most!)
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Excerpt from Voltaic Cell lab reading
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Table 18.1 (continued)
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Recall earlier slide
We define a “standard reduction potential”, E°red, for every reduction half reaction such that:
E°cell = E°red(cathode) - E°red(anode)
Ecell Ered + Eox
OR (could also write Ecell as…
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Lab interlude
• See overhead / board
• The lab manual initially asks you to pretend that the Ag+/Ag reduction potential is 0.0 V just to show you the “arbitrary-ness” of this.
• Then it tells you that in reality, Ag+/Ag reduction potential is 0.80 V if the H+/H2 potential (SHE) is 0.0 V
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Cu/Cu2+ & Zn/Zn2+
Cu/Cu2+ & Fe3+/Fe2+
Zn/Zn2+ & Ag/Ag+
**Circle the species that is the better oxidizing agent**
1.50 V Zn/Zn2+
1.05 V Zn/Zn2+
0.32 V Cu/Cu2+ Cu Cu2+ + 2e-
Cu2+ + 2e- Cu
Zn Zn2+ + 2e-
Zn Zn2+ + 2e-
Fe3+ + e- Fe2+
Ag+ + e- Ag 0.0 V1.50 V
1.50 V -0.45 V
+0.45 V -0.13 V
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Ag+ + e- Ag0.0 V
Fe3+ + e- Fe2+-0.13 V
Cu2+ + 2e- Cu-0.45 V
Zn2+ + 2e- Zn-1.50 V
0.80 V
0.67 V
0.35 V
-0.70 V
0.80 V
0.77 V
0.34 V
-0.76 V
From Text Table
(See next slide)
Zn2+ + 2e- Zn -1.50 VFlip
Cu2+ + 2e- Cu
Zn Zn2+ + 2e-
Fe3+ + e- Fe2+
Ag+ + e- Ag 0.0 V
1.50 V
-0.45 V
-0.13 V
From Table A.1:
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Table 18.1 (continued)
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Determine the cell reaction, calculate Ecell , identify the cathode and anode, label the (+) and (-) electrodes, and show electron flow (see board)
Ni2+(aq) + 2 e- Ni(s) -0.23 V
Mn2+(aq) + 2 e- Mn(s) -1.18 V
The better oxidizing agent is:___
So ___ actually gets reduced, and thus electrons flow to the _______ side, which must be the ____ode.
So the Ni electrode must be ______ive
E°cell = _____ - _____
= _______
Ni2+
Ni2+
rightcath
posit
-0.23 V -1.18 V
+0.95 V 26
(Because its Ered is more positive)
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Determine the cell reaction, calculate Ecell , identify the cathode and anode, label the (+) and (-) electrodes, and show electron flow (see board)
Fe2+(aq) + 2 e- Fe(s) -0.45 V
Mg2+(aq) + 2 e- Mg(s) -2.37 V
The better oxidizing agent is:___
So ___ actually gets reduced, and thus electrons flow to the _______ side, which must be the ____ode.
So the Fe electrode must be ______ive
E°cell = _____ - _____
= _______
Fe2+
Fe2+
rightcath
posit
-0.45 V -2.37 V
+1.92 V 27
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Saltbridge
1 M Fe2+
1 M Pb2+
e
Fe2+(aq) + 2 e- Fe(s) -0.45 V
Pb2+(aq) + 2 e- Pb(s) -0.13 V
The better oxidizing agent is:___
So ___ actually gets reduced, and thus electrons flow to the _______ side, which must be the ____ode.
So the Pb electrode must be ______ive
E°cell = _____ - _____
= _______
Pb2+
Pb2+
leftcath
posit
-0.13 V -0.45 V
+0.32 V
Pb(s) Fe(s)
Determine the cell reaction, calculate Ecell , identify the cathode and anode, label the (+) and (-) electrodes, and show electron flow (see board)
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• See last week’s pink lab handout (Voltaic Cells), board, and below– Start with G = G + RTlnQ and substitute in
G = -nFEcell and G = -nFEcell
After some algebra (and substituting in values for R, assuming T = 298 K, and converting to a base 10 log):
Nernst Equation
(T = 25C)
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cell cell
0.0592 Vlog E E Q
no o
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Standard vs. Nonstandard Cell
cell cell
0.0592 Vlog E E Q
no o
30**Always write out the balanced redox equation before using the Nernst Equation or predicting whether Ecell should increase or decrease.**
Zn + Cu2+ Zn2+ + Cu; Q = ??
Recall lab—adding NH3 to Cu2+ side!
(T = 25C)
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Explain in detail an in a conceptual way why the cell potential goes up when the NH3 is added. Is the driving force for the cell rxn greater or smaller after the NH3 is added?
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Relationship between variables (at any conditions; Mines Fig)
G G RT ln Q
cell cell
0.0592 Vlog E E Q
no o
Q
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(T = 25C)
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Relationship between variables (at standard state conditions; From Tro)
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(T = 25C)
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EXAMPLE 18.8 Calculating Ecell Under Nonstandard ConditionsDetermine the cell potential for an electrochemical cell based on the following two half-reactions:
Oxidation:
Reduction:
E°cell = E°red(cat) - E°red(an)
ORE°cell = E°red(cat) + E°ox(an)
cell cell
0.0592 Vlog E E Q
no o**Need to write balanced equation before
using Nernst! What will “n” be here?**
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Fig. 18.22
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What mass of gold is plated in 25 minutes if the current is 5.5 A?
Au3+ (aq) + 3 e- Au(s)
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