Half-Cell Potentials

Size: px

Start display at page:

Download "Half-Cell Potentials"

Edwina Ball
5 years ago
Views:

1 Half-Cell Potentials! SHE reduction potential is defined to be exactly 0 v! half-reactions with a stronger tendency toward reduction than the SHE have a + value for E red! half-reactions with a stronger tendency toward oxidation than the SHE have a - value for E red! E cell = E oxidation + E reduction!!e oxidation = -E reduction!!when adding E values for the half-cells, do not multiply the half-cell E values, even if you need to multiply the halfreactions to balance the equation

2 Selected Standard Electrode Potentials (298K) Half-Reaction E 0 (V) F 2 (g) + 2e - 2F - (aq) strength of oxidizing agent Cl 2 (g) + 2e - 2Cl - (aq) MnO 2 (g) + 4H + (aq) + 2e - Mn 2+ (aq) + 2H 2 O(l) NO 3- (aq) + 4H + (aq) + 3e - NO(g) + 2H 2 O(l) Ag + (aq) + e - Ag(s) Fe 3+ (g) + e - Fe 2+ (aq) O 2 (g) + 2H 2 O(l) + 4e - 4OH - (aq) Cu 2+ (aq) + 2e - Cu(s) 2H + (aq) + 2e - H 2 (g) N 2 (g) + 5H + (aq) + 4e - N 2 H 5+ (aq) Fe 2+ (aq) + 2e - Fe(s) 2H 2 O(l) + 2e - H 2 (g) + 2OH - (aq) Na + (aq) + e - Na(s) strength of reducing agent Li + (aq) + e - Li(s) -3.05

3 Writing Spontaneous Redox Reactions!By convention, electrode potentials are written as reductions.!when pairing two half-cells, you must reverse one reduction half-cell to produce an oxidation half-cell. Reverse the sign of the potential.!the reduction half-cell potential and the oxidation half-cell potential are added to obtain the E 0 cell.!when writing a spontaneous redox reaction, the left side (reactants) must contain the stronger oxidizing and reducing agents. Example: Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) stronger reducing agent! stronger oxidizing agent! weaker oxidizing agent! weaker reducing agent!

4 Relative Reactivities (Activities) of Metals 1. Metals that can displace H from acid 2. Metals that cannot displace H from acid 3. Metals that can displace H from water 4. Metals that can displace other metals from solution

5 Predicting Whether a Metal Will Dissolve in an Acid! acids dissolve in metals if the reduction of the metal ion is easier than the reduction of H + (aq)! metals whose ion reduction reaction lies below H + reduction on the table will dissolve in acid

6 E cell,!g and K! for a spontaneous reaction!!one the proceeds in the forward direction with the chemicals in their standard states!!g < 1 (negative)!!e > 1 (positive)!!k > 1!!G =!RTlnK =!nfe cell!!n is the number of electrons!!f = Faraday s Constant = 96,485 C/mol e!

7 The interrelationship of!g 0, E 0, and K.!G 0!G 0 K E 0 cell Reaction at standard-state conditions < 0 > 1 > 0 spontaneous! 0 > 0 1 < 1 0 < 0 at equilibrium! nonspontaneous!!g 0 = -nfe o cell!g 0 = -RT lnk By substituting standard state values into E 0 cell, we get E 0 cell = (0.0592V/n) log K (at 298 K) E 0 cell E 0 cell = -RT lnk nf K

8 Calculating K and!g 0 from E 0 cell PROBLEM: Lead can displace silver from solution: Pb(s) + 2Ag + (aq) Pb 2+ (aq) + 2Ag(s) PLAN: Break the reaction into half-reactions, find the E 0 for each half-reaction and then the E 0 cell. Substitute into the proper equations. SOLUTION: Pb 2+ (aq) + 2e - Ag + (aq) + e - Pb(s) Ag(s) E 0 = -0.13V E 0 = 0.80V E 0 = 0.13V E 0 = 0.80V 2X Pb(s) Pb 2+ (aq) + 2e - Ag + (aq) + e - Ag(s) E 0 cell = 0.93V E 0 cell = 0.592V n log K!G 0 = -nfe 0 cell = -(2)(96.5kJ/mol*V)(0.93V) n x E log K = 0 cell (2)(0.93V) = K = 2.6x10 31!G 0.592V 0.592V 0 = -1.8x10 2 kj

9 The Effect of Concentration on Cell Potential!G =!G 0 + RT ln Q -nf E cell = -nf E cell + RT ln Q E cell = E 0 cell - RT nf ln Q!When Q < 1 and thus [reactant] > [product], lnq < 0, so E cell > E 0 cell!when Q = 1 and thus [reactant] = [product], lnq = 0, so E cell = E 0 cell!when Q >1 and thus [reactant] < [product], lnq > 0, so E cell < E 0 cell E cell = E 0 cell n log Q

10 Nonstandard Conditions - the Nernst Equation!!G =!G + RT ln Q! E = E - (0.0592/n) log Q at 25 C! when Q = K, E = 0! use to calculate E when concentrations not 1 M

11 E at Nonstandard Conditions

12 Concentration Cell when the cell concentrations are equal there is no difference in energy between the half-cells and no electrons flow Cu(s)! Cu 2+ (aq) (0.010 M)!! Cu2+ (aq) (2.0 M)! Cu(s)

13 Concentration Cell the cell concentrations are different, electrons flow from the side with the less concentrated solution (anode) to the side with the more concentrated solution (cathode) Cu(s)! Cu 2+ (aq) (0.010 M)!! Cu2+ (aq) (2.0 M)! Cu(s)

14 Electrolytic Cell! uses electrical energy to overcome the energy barrier and cause a non-spontaneous reaction!!must be DC source! the + terminal of the battery = anode! the - terminal of the battery = cathode! cations attracted to the cathode, anions to the anode! cations pick up electrons from the cathode and are reduced, anions release electrons to the anode and are oxidized! some electrolysis reactions require more voltage than E tot, called the overvoltage

16 Electrochemical Cells! in all electrochemical cells, oxidation occurs at the anode, reduction occurs at the cathode! in voltaic cells,!!anode is the source of electrons and has a (!) charge!!cathode draws electrons and has a (+) charge! in electrolytic cells!!electrons are drawn off the anode, so it must have a place to release the electrons, the + terminal of the battery!!electrons are forced toward the anode, so it must have a source of electrons, the! terminal of the battery

17 Electrolysis! electrolysis is the process of using electricity to break a compound apart! electrolysis is done in an electrolytic cell! electrolytic cells can be used to separate elements from their compounds!!generate H 2 from water for fuel cells!!recover metals from their ores

18 Electrolysis of Water

19 Electrolysis of Pure Compounds! must be in molten (liquid) state! electrodes normally graphite! cations are reduced at the cathode to metal element! anions oxidized at anode to nonmetal element

20 Electrolysis of NaCl(l)

The Downs Cell for Sodium Production Cross-sectional view of a Downs cell for commercial production of sodium metal by electrolysis of molten sodium chloride.

21 The Downs Cell for Sodium Production Cross-sectional view of a Downs cell for commercial production of sodium metal by electrolysis of molten sodium chloride. The cell design keeps the sodium and chlorine apart so that they canʼt react with each other. Sodium metal is produced commercially by the electrolysis of a mixture of molten sodium chloride and calcium chloride in a Downs cell.

22 Mixtures of Ions! when more than one cation is present, the cation that is easiest to reduce will be reduced first at the cathode!!least negative or most positive E red! when more than one anion is present, the anion that is easiest to oxidize will be oxidized first at the anode!!least negative or most positive E ox

Electrolysis of NaI (aq) with Inert Electrodes possible oxidations 2 I -1! I 2 + 2 e -1 E =!0.54 v 2 H 2 O! O 2 + 4e -1 + 4H +1 E =!0.82 v possible reductions Na +1 + 1e -1!

23 Electrolysis of NaI (aq) with Inert Electrodes possible oxidations 2 I -1! I e -1 E =!0.54 v 2 H 2 O! O 2 + 4e H +1 E =!0.82 v possible reductions Na e -1! Na 0 E =!2.71 v 2 H 2 O + 2 e -1! H OH -1 E =!0.41 v overall reaction 2 I! (aq) + 2 H 2 O (l)! I 2(aq) + H 2(g) + 2 OH-1 (aq)

24 Faraday s Law! the amount of metal deposited during electrolysis is directly proportional to the charge on the cation, the current, and the length of time the cell runs!!charge that flows through the cell = current! time

25 What mass of Au that can be plated in 25 min using 5.5 A for the half-reaction Au 3+ (aq) + 3 e! " Au(s) Given: Find: 3 mol e! : 1 mol Au, current = 5.5 amps, time = 25 min Mass Au Relationships:

Electrochemical Cells

Electrochemistry Electrochemical Cells The Voltaic Cell Electrochemical Cell = device that generates electricity through redox rxns 1 Voltaic (Galvanic) Cell An electrochemical cell that produces an electrical