Transcript Ch. 18b Slides

V. Everything’s Related
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Positive E°cell means spontaneous.
Negative ΔG° means spontaneous.
K > 0 means spontaneous.
Thus, all of these must be related somehow.
V. Potential and Work
• Potential difference can be expressed as a
function of work.
• We will use this new view to derive the
relationship between E°cell and ΔG°.
V. Eqn. Relating E°cell and ΔG°
V. Sample Problem
• Using tabulated half-cell potentials,
calculate ΔG° for the reaction 2Na(s) +
2H2O(l)  H2(g) + 2OH-(aq) + 2Na+(aq).
V. Potential and Equilibrium
Constants
• Using our new equation that relates
standard cell potential to standard free
energy, we can derive an equation
between E°cell and K.
• We start with the equation that relates
ΔG° to K.
V. Eqn. Relating E°cell and K
V. Sample Problem
• Use tabulated half-cell potentials to
calculate the equilibrium constant for
the two electron oxidation of copper
metal by H+.
V. Relating ΔG°, K, and E°cell
• With the last equation derived, we summarize
how we can convert between ΔG°, K, and E°cell.
V. Nonstandard Potentials
• Just like any other system, electrochemical
cells may not be under standard conditions.
Why is the potential higher?
V. Calculating Nonstandard
Cell Potentials
• Although we can qualitatively predict
whether Ecell is higher or lower than
E°cell, we’d like to calculate an exact
value.
• We can derive an equation starting with
the nonstandard ΔG equation.
V. Deriving the Nernst Equation
V. Nernst Eqn. Generalizations
• Under standard conditions, log Q = log
1 = 0, so Ecell will equal E°cell.
• If Q < 1, there are more reactants than
products, so redox reaction shifts right;
Ecell will be > than E°cell.
• If Q > 1, there are more products than
reactants, so redox reaction shifts left;
Ecell will be < than E°cell.
• If Q = K, then Ecell = 0.
V. Sample Problem
• Calculate the cell potential for the
electrochemical cell represented by
Ni(s)|Ni2+(aq, 2.0 M)||VO2+(aq, 0.010 M),
H+(aq, 1.0 M), VO2+(aq, 2.0 M)|Pt(s).
V. Concentration Cells
V. Concentration Cells
• If the two half-cells are the same, there
is no reason to reduce something on
one side and oxidize the same thing on
the other side.
• However, if there are [ ] differences,
there is a push to get to equilibrium.
• Electrons flow in order to increase the [ ]
of the dilute cell and decrease the [ ] of
the concentrated cell.
VI. Using Electrochemistry
• If constructed correctly, electrochemical
cells can be used to store and deliver
electricity.
• We briefly look at dry-cell batteries,
lithium ion batteries, and fuel cells.
VI. Dry-cell Batteries
• Called dry-cell because
there’s very little water.
• Voltage derived from the
oxidation of Zn(s) and the
reduction of MnO2(s).
• In acidic battery, MnO2(s)
reduced to Mn2O3(s).
• In alkaline battery, MnO2(s)
reduced to MnO(OH)(s).
VI. Lithium Ion Batteries
• This rechargeable
battery works a bit
differently.
• Motion of Li+ from
anode to cathode
causes e-’s to flow
externally and reduce
the transition metal.
• Battery is recharged by
oxidizing the transition
metal in the cathode.
VI. Fuel Cells
VII. Nonspontaneous Redox
Reactions
• Nonspontaneous reactions can be
forced to occur by inputting energy.
• Nonspontaneous redox reactions can
be forced to occur by inputting energy in
the form of electrical current.
• This type of electrochemical cell is
called an electrolytic cell.
VII. Voltaic vs. Electrolytic Cells
VII. Predicting Products
• In an aqueous electrolytic cell, it’s
possible that H2O can be reduced or
oxidized.
• To predict products, must consider
potentials of all processes that could
occur.
• Processes that are easiest (least
negative or most positive half-cell
potential) will occur.
VII. Overvoltage
• There is one problem in predicting
electrolysis products.
• Some half-cell reactions don’t occur at
their expected voltage potentials!
• e.g. 2H2O(l)  O2(g) + 4H+(aq) + 4e- has
Eox = -0.82 V when [H+] = 1 x 10-7 M.
 However, kinetic factors require a voltage
of 1.4 V for this to occur.
VII. Example Electrolysis
• What happens when a solution of NaI
undergoes electrolysis?
• Oxidation
 2I-(aq)  I2(aq) + 2e 2H2O(l)  O2(g) + 4H+(aq) + 4e-
E°ox = -0.54 V
E°ox = -1.4 V
• Reduction
 Na+(aq) + e-  Na(s)
 2H2O(l) + 2e-  H2(g) + 2OH-(aq)
E°red = -2.71 V
E°red = -0.41 V
VII. Electrolysis of NaI Solution
VII. Electrolysis Stoichiometry
• The # of electrons in any redox reaction
can be used as a stoichiometric ratio.
• If we know how long an electrolysis
takes place, and the magnitude of
current that flowed, we can do
stoichiometric calculations.
• Important relationships:
 1 A = 1 C/s
 F = 96,485 C/mole e-
VII. Sample Problem
• Copper can be plated out of a solution
containing Cu2+ according to the halfreaction: Cu2+(aq) + 2e-  Cu(s). How
long (in minutes) will it take to plate 10.0
g of copper using a current of 2.0 A?