V. Everything’s RelatedV. Everything’s Related

• Positive E°cell means spontaneous.

• Negative ΔG° means spontaneous.• K > 0 means spontaneous.• Thus, all of these must be related somehow.

V. Potential and WorkV. 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°V. Eqn. Relating E°cellcell and ΔG° and ΔG°

V. Sample ProblemV. 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 V. Potential and Equilibrium ConstantsConstants

• 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°V. Eqn. Relating E°cellcell and K and K

V. Sample ProblemV. 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°V. Relating ΔG°, K, and E°cellcell

• With the last equation derived, we summarize how we can convert between ΔG°, K, and E°cell.

V. Nonstandard PotentialsV. Nonstandard Potentials• Just like any other system, electrochemical

cells may not be under standard conditions.

Why is the potential higher?

V. Calculating Nonstandard V. Calculating Nonstandard Cell PotentialsCell 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 EquationV. Deriving the Nernst Equation

V. Nernst Eqn. GeneralizationsV. 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 ProblemV. 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 CellsV. Concentration Cells

V. Concentration CellsV. 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 ElectrochemistryVI. 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 BatteriesVI. 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 BatteriesVI. 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 CellsVI. Fuel Cells

VII. Nonspontaneous Redox VII. Nonspontaneous Redox ReactionsReactions

• 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 CellsVII. Voltaic vs. Electrolytic Cells

VII. Predicting ProductsVII. 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. OvervoltageVII. 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 ElectrolysisVII. Example Electrolysis

• What happens when a solution of NaI undergoes electrolysis?

• Oxidation 2I-

(aq) I2(aq) + 2e- E°ox = -0.54 V

2H2O(l) O2(g) + 4H+(aq) + 4e- E°ox = -1.4 V

• Reduction Na+

(aq) + e- Na(s) E°red = -2.71 V

2H2O(l) + 2e- H2(g) + 2OH-(aq) E°red = -0.41 V

VII. Electrolysis of NaI SolutionVII. Electrolysis of NaI Solution

VII. Electrolysis StoichiometryVII. 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 ProblemVII. Sample Problem

• Copper can be plated out of a solution containing Cu2+ according to the half-reaction: 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?