ADVANCED PLACEMENT CHEMISTRY ELECTROCHEMISTRY

Galvanic cell- chemical energy is changed into electrical energy (also called a voltaic cell) (spontaneous) Electrolytic cell- electrical energy is changed into chemical energy (nonspontaneous)

Oxidation-reduction reaction (redox)- involves the transfer of one of more electrons. Oxidation can not occur without reduction!!!

Oxidation- *loss of electrons *increase in oxidation number Reduction- *gain of electrons *decrease in oxidation number

OIL RIG Oxidation is loss (of e-), Reduction is gain (of e-) LEO the lion goes GER Lose electrons = oxidation, Gain electrons = reduction

Oxidizing Agent- Substance which is reduced and causes another substance to be oxidized. Reducing Agent- Substance which is oxidized and causes another substance to be reduced.

Half-reaction- Equation written to show either the loss or the gain of electrons (shows number of electrons transferred) Ex. Fe +2  Fe +3 + e - Cu e -  Cu

Voltaic Cell The transfer of electrons can be used as an energy source if it is harnessed. One way that this may be done is by separating the two half-reactions and requiring the electron transfer to occur through a wire. The electron flow (electricity) can be used to do useful work.

The solutions must also be connected in another way or the charge will become unbalanced and the electrical flow will stop. This connection is often by a salt bridge (tube filled with conducting solution)or porous disk. Ions flow from one compartment to the other to keep the net charge zero.

Anode- The electrode at which oxidation occurs. After a period of time, the anode may appear to become smaller as it falls into solution.

Cathode- Electrode at which reduction occurs. After a period of time, it may appear larger, due to ions from solution plating out on it.

An Ox and a Red Cat Oxidation occurs at the anode, Reduction occurs at the cathode.

FAT CAT Electrons travel from anode to cathode (from reducing agent to oxidizing agent) in a galvanic cell.

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Cell potential (E cell ) or electromotive force (emf): *the voltage measured across the electrodes of the half cells *the driving force of the reaction *cells with positive E cell values are always spontaneous *units are volts (1 joule of work per coulomb of charge transferred) 1V = 1J/C

E o cell = E o ox + E o red Standard cell potential is the sum of the standard oxidation potential for the oxidation half-reaction and the standard reduction potential for the reduction half-reaction. Standard conditions are 1 atm, 25 o C,and 1 M *at conditions other than standard, E cell = E ox + E red

Standard potentials for half-reactions: *usually written as reduction reactions (to change to oxidation, reverse reaction and change the sign of the potential) *can't be determined directly *are determined by the use of a standard hydrogen electrode which is arbitrarily assigned a value of 0.00 *all other half reactions are expressed in reference to the standard hydrogen electrode

*elements at the top of the chart are more easily reduced. (usual arrangement) *elements at the bottom of the chart are easily oxidized—in general, metals. (usual arrangement) *Substances with more positive reduction potentials are more easily reduced. They act as oxidizing agents. *The more positive the reduction potential, the stronger the oxidizing agent.

*Substances with negative reduction potentials (and thus positive oxidation potentials) are more easily oxidized. They act as reducing agents. *The more negative the reduction potential, the stronger the reducing agent.

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Standard hydrogen electrode- (SHE)- a platinum electrode in contact with 1M H + and bathed by hydrogen gas at 1 atm 2H + + 2e -  H 2 E o = 0.00 V

When combining half reactions to obtain balanced redox reactions and calculating cell potential, you do not multiply the cell potential by an integer even if you multiply the half-reaction by an integer when writing the redox reaction.

If E cell is positive the reaction is spontaneous. If E cell is negative the reaction is nonspontaneous. Electrons will flow in the direction that will make the E cell positive. Voltaic experiment Voltaic experiment

Ex. What will be the spontaneous reaction between the following set of half-reactions? What is the value of E 0 cell? Cr 3 + (aq) + 3e —  Cr(s) E 0 = MnO 2 (s) + 4H + (aq) + 2e —  Mn 2 + (aq) + 2H 2 O(l) E 0 = One rxn must be oxidation and the other must be reduction (we must reverse one). The sum of the two potentials must be positive. 2(Cr  Cr e - ) E o = (MnO 2 + 4H + + 2e -  Mn H 2 O) E o = Cr + 3MnO H +  2Cr Mn H 2 O E o cell = 2.02 V

A voltaic cell may be described in an abbreviated manner called line-notation. In this form the anode is on the left, followed by its aqueous solution. The right side consists of the aqueous solution for the cathode and finally the cathode. A single vertical line separates the electrodes from their solutions and a double vertical line between the anode solution and the cathode solution represents a salt bridge.

Ex. Mg(s) | Mg 2+ (aq) || Al 3+ (aq) | Al(s) This cell has solid magnesium as the anode immersed in a solution of magnesium ions. The cathode is solid aluminum immersed in a solution of aluminum ions. The half reactions would be: Anode: Mg  Mg e - E 0 =2.37 V Cathode: Al e -  Al E 0 = E 0 cell = (-1.66) = 0.71 V This reaction would be spontaneous as written.

Ex. Zn | Zn 2+ (1.0 M) || H + (1.0 M); H 2 (1 atm) | Pt This cell has solid zinc as the anode immersed in a solution of 1.0 M zinc ions. The cathode is an inert platinum electrode immersed in a 1.0 M hydrogen ion solution with hydrogen gas bubbled in at 1 atm pressure. The half reactions would be: anode (oxidation): Zn  Zn e - E o = 0.76 V cathode (reduction): 2H + + 2e -  H 2 E o = 0.00 V E cell = = 0.76 V The reaction is spontaneous as written.

When diagramming an electrochemical cell, the anode is traditionally drawn on the left and the cathode on the right. (Don’t count on this on a test question!)

The greater the positive potential of a cell, the greater the spontaneity of the reaction. Since  G, K, and E 0 are all measurements of the spontaneity of a reaction, we can write equations to interrelate them.

 G 0 = -nFE 0 *  G 0 is the standard free energy change (units are J) *n is the number of moles of electrons transferred *F is a Faraday = 96,500 Coulombs/mol *E 0 is standard cell potential If E 0 is positive,  G is negative

To find K:  G o = -RT lnK

Faraday -amount of electricity that reduces one equivalent mass of a substance at the cathode and oxidizes one equivalent mass of a substance at the anode (equivalent mass is the mass of a species that will yield or consume one mole of electron *amount of electrical charge carried by 1 mol of electrons = 96,500 coulombs

If cell concentrations are not standard (1 M) and 25 0 C, the cell potential is affected. This effect can be determined by using the Nernst equation. This equation is derived from the equation used for free energy when pressures were not all 1 atm :  G=  G 0 + RT lnQ

Nernst equation: E = E 0 -RT lnQ nF R = J/Kmol T = temp in Kelvin Q = ratio of concentrations or pressures of products over reactants each raised to the power of their coefficients n = number of moles of electrons transferred F = 96,500 Coulombs/mol Another form of the Nernst equation is only valid at 25 0 C : E= E 0 -(0.0592/n)log Q

The E value calculated is the maximum potential at the very beginning of the reaction. As the reaction progresses, E decreases until it reaches zero and the battery is at equilibrium (dead).

Ex. Calculate the emf of the Zn/Cu cell at 25 0 C under the following conditions: Zn(s) | Zn 2+ (0.40 M) || Cu +2 (0.020 M) | Cu(s) E cell o = 1.10V The line diagram tells us that the reaction is: Zn(s) + Cu 2+ (0.020 M)  Cu(s) + Zn 2+ (0.40 M) E=1.10 V - ( /2) log (0.40/0.020) E=1.06 V Therefore, the reaction is less spontaneous when the ratio of concentration of products to concentration of reactants is greater than one. This could also be predicted using LeChatelier's principle.

Equilibrium constants can also be calculated using cell potentials. At equilibrium E = 0 and Q = K. 0 = E 0 - (0.0592/n) log K (at 298 K) so E 0 = (0.0592/n) log K

Ex. Determine whether the following reaction is spontaneous and calculate its equilibrium constant at 25 0 C. Sn(s) + Ni 2+  Sn 2+ + Ni(s) The half-reactions would be as follows: Sn(s)  Sn e - E 0 = 0.14 V Ni e -  Ni(s) E 0 = V E 0 cell = (-0.25) = nonspontaneous E 0 = (0.0592/n) log K -0.11= (0.0592/2) log K K = 2 x 10 -4

Galvanic cells are used to power many types of equipment. The common dry cell battery is one example of this. In the original acid version of the battery, the cathode is a graphite rod (inert) immersed in a paste of MnO 2, NH 4 Cl and carbon. The anode is the zinc case. The alkaline dry cell battery has the NH 4 Cl replaced with KOH or NaOH. It lasts longer because the zinc corrodes more slowly under basic conditions.

The lead storage battery (common car battery) consists of lead as the anode and lead coated with lead dioxide as the cathode. The electrolyte solution (battery acid) is sulfuric acid. The car's alternator recharges the battery by forcing current to flow in the opposite direction.

Electrolytic cells- When the desired chemical reaction is nonspontaneous (E 0 cell is negative) the reaction can proceed by the addition of electrons from an outside source. This results in the conversion of electrical energy into chemical energy. Examples are electrolysis of water, NaCl solution, silver plating

Faraday's Law- The amount of substance being oxidized or reduced at each electrode during electrolysis is directly proportional to the amount of electricity that passes through the cell.

Coulomb = amp x s : amount of charge that passes a given point when 1 amp of electrical current flows for 1 second. Ampere (amp) : flow rate of 1 coulomb/s (i is used as symbol for current) Joule (J): amount of energy absorbed or evolved when 1 coulomb of electrical charge moves through a potential difference of 1 V (1J = 1 volt-coul and 1F = 1 J/V mol)

current(amps) x time  coulombs  Faradays  mol reactant  g reactant or g reactant  mol reactant  Faradays  coulombs  current(amps) x time

Ex. How many grams of Cr can be plated out by passing 2.05 amps through acidic CrO 3 for 1.0 x 10 4 s? Cr must go from an ox # of +6 to zero, therefore 6 electrons must be gained. Cr e -  Cr 10000s 2.05C 1 mol e - 1 mol Cr 52.0 g Cr = 1.84 g Cr 1 s 96,500 C 6 mol e - 1 mol Cr Chrome plating

Ex. How many hours would it take to produce 25.0 g of Cr from a solution of CrCl 3 by a current of 2.75A? Cr e -  Cr 25.0g Cr 1 mol Cr 3 mol e C 1 s 1 hr =14.0 hr 52.0g Cr 1 mol Cr 1 mol e C 3600s

When electrolysis is carried out in aqueous solutions, there is always the possibility that water will be oxidized and/or reduced instead of the solute. In order to predict the products produced when aqueous solutions are electrolyzed, you must determine the potentials of each possibility.

Ex. The electrolysis of aqueous NaBr Possible anode reactions (oxidation): 2Br - (aq)  Br 2 (g) + 2e - E 0 = H 2 O(l)  O 2 (g) + 4H + (aq) + 4e - E 0 = The first reaction occurs because it is less negative (more spontaneous)

Possible cathode reactions (reduction): Na + (aq) + e -  Na(s) E 0 = H 2 O(l) + 2e -  H 2 (g) + 2OH - (aq) E 0 = The second reaction occurs because it is less negative (more spontaneous). The overall reaction is: 2H 2 O + 2Br -  H 2 + 2OH - + Br 2 Electrolysis of KI Electrolysis of KI

This method of determining whether water is oxidized or reduced does not always work. In the electrolysis of salt water (brine), we would predict that water would be both oxidized and reduced. In fact, the chloride ions are oxidized instead of the water. This is due to some fairly complex factors. This phenomenon is called overpotential or overvoltage.

The electrolysis of molten NaCl: NaCl(l)  Na(l) + ½ Cl 2 (g) anode: Cl -  ½ Cl 2 + e - cathode: Na + + e -  Na

The electrolysis of aqueous CuSO 4 : Anode: H 2 O  ½ O 2 + 2H + + 2e - Cathode: Cu e -  Cu

The electrolysis of water: Anode: 2H 2 O(l)  O 2 (g) + 4H + (aq) + 4e - E 0 = Cathode: 2H 2 O(l) + 2e -  H 2 (g)+ 2OH - (aq) E 0 = E 0 cell = V Overall: 6H 2 O  2H 2 + O 2 + 4H + + 4OH - 2H 2 O  2H 2 + O 2