Chapter 21 Electrochemistry - electrochemistry and electrochemical processes are some of the most important sources of power that we have - batteries - much publicized hydrogen fuel cells - photosynthesis - neuron conduction - mitachondria - electrochemistry is pushing the economy forward.. so it is probably worth spending so time on!
21.1 Electrode Potentials and their Measurements - the criteria for spontaneous change (that )G is negative) applies to all reactions, period - for redox reactions, we have: )GE = -nfe where: n is the number of electrons involved F is a Faraday - 96486 coulombs/mole E is the cell potential measured in volts
- clearly this means that if we are going to have a spontaneous reaction, then our cell potential must be positive - that is, if we are going to get some work out of our electrochemical process, then )GE will have to be negative requiring E to be positive (note that both 'n' and 'F' are, by definition, positive numbers) - important to keep this underlying picture in mind - that electrical work can be converted into free energy
- what makes up an electrochemical cell? four components: two electrodes - an anode and a cathode an electrolyte - something that engages in ion transport (not necessarily a liquid) an external circuit - completes the electron path - an electrode is nothing more than a strip of metal in contact with the electrolyte: - two types: oxidation reduction
- for a metal ion in solution, an equilibrium is established at the surface of the electrode called a "half cell": M (s) W M n+ (solv) + ne - where the forward reaction is oxidation and the reverse reaction is reduction. The subscript "solv" indicates that the resulting cation is dissolved in a solvent - most commonly water. - to complete the circuit, we need a second half cell that can receive electrons - and our measurements are based on the flow of electrons from one half cell to the other
- that is, M (s) W M n+ (solv) + ne - X (s) + ne - W X - (solv) - the "cell potential" (E) is the difference in energy between these two equilbria - in a discharging electrochemical cell, oxidation takes place at the anode and reduction takes place at the cathode (note - a way to remember which is which is that "o" and "a" are both vowels while "c" and "r" are both consonants)
- the electrolyte provides a mechanism for the conduction of ions from one electrode to the other - the external circuit provides a route for the conduction of electrons from one electrode to the other
i.e. Zn (s) * Zn 2+ (aq) 2 Cu 2+ (aq) * Cu (s) E cell = 1.103V Cell diagrams and terminology - a cell diagram shows the components of an electrochemical cell - anode is always on the left - cathode is always on the right - a phase boundary is denoted with a single line - a junction between half-cell compartments is a double vertical line anode half-cell: Zn (s) Zn 2+ (aq) + 2e - cathode half-cell: Cu 2+ (aq) + 2e - Cu (s)
note: electrochemical cells that produce electricity are called "voltaic" or "galvanic" cells (most commonly, the latter) electrolytic cells employ electricity to drive a chemical reaction against a free energy gradient ("electrolysis" literally means "breaking with electricity") anodes and cathodes are labeled differently in "galvanic" and "electrolytic" cells
21.2 Standard Electrode Potentials - cell voltages can be measured with a high degree of precision - they are the difference between two reactions - half cell voltages can't be measured at all as there is no zero point - they are the absolute voltage of a reaction - to solve this problem, in aqueous solution, we define zero as the voltage of the "standard hydrogen electrode" - and all electrode potentials are defined based on their reduction potential under standard conditions (activity equal to 1.0M; pressure equal to 1.0 atm)
the "Standard Hydrogen Electrode" is: Pt * H 2 (g) * H + (aq) 2 which is "zero" Hence, Pt * H 2 (g) * H + (aq) 2 Ag + (aq) * Ag (s) EE = +0.800 V means that relative to the oxidation/reduction of hydrogen on the surface of platinum, the reduction of silver ions to silver has a potential of +0.800V. Note: this implies that relative to hydrogen production, silver production is a spontaneous process
similarly, Pt * H 2 (g) * H + (aq) 2 Na + (aq) * Na (s) EE = -2.713V which means that this reaction - as written - is not spontaneous. That is, relative to the formation of hydrogen, sodium metal is oxidized to form sodium ions. By tabulating the results of many of these sorts of cells, we can work out the cell potential for any electrochemical couple: EE cell = EE(right) - EE(left) = EE(cathode) - EE(anode) = EE(reduction half cell) - EE(oxidation half cell) i.e. 2Na (aq) + 2H + (aq) 2Na + (aq) + H 2 (g) EE(0.0V) - EE(-2.713) = 2.713 V
21.3 E cell, )G, and K eq - a discharging electrochemical cell is able to do work: ω elec = nfe cell provided that the cell operates "reversibly" (in the chemical sense) and that work is related to free energy (ω = -)G) hence, )GE = -nfee cell i.e. 2Na (aq) + 2H + (aq) 2Na + (aq) + H 2 (g) EE(0.0V) - EE(-2.713) = 2.713 V )GE = -(2)(96486 coul/mol)(2.713 J/coul) = 535.5 kj/mol
Combining Half-Cells - are you wondering from pg. 835 - reduction: M n+ (aq) + ne - M (s) )GE red = - nfee M oxidation: N (s) N n+ (aq) + ne - )GE ox = - nf(-ee N ) = nfee N overall: M n+ (aq) + N (s) N m+ (aq) + M (s) and: )GE = )GE red + )GE ox = -nfee cell = - nfee M + nfee N dividing through by -nf gives: EE cell = EE M -EE N = EE (reduction) -EE (oxidation) regardless of the number of electrons involved!
Hence, when combining half cells, the cell potential is simply the potential of the reduction reaction minus the potential of the oxidation reaction. This is something that (hopefully) you have seen before in balancing redox reactions. While the number of electrons must be adjusted by multiplying the redox reactions by a constant, the cell potential is invariant. The only thing that "changes" it, with regard to adding half cells, is that reversing the reaction reverses the sign.
Spontaneous Change in Redox Reactions -if EE cell is positive, then the reaction is spontaneous as written -if EE cell is negative, then the reaction is not spontaneous as written (the reverse reaction is spontaneous) - if a redox reaction is reversed, then EE cell changes sign Note: both EE and EE cell are "intensive" properties - they do not depend upon the amount of material involved.
The Behaviour of Metals Towards Acids - in considering the reaction of metals with acid, some readily dissolve while others don't - this can be explained through electrochemistry - when a metal reacts with an acid, it is oxidized: oxidation: M (s) M 2+ (aq) + 2e - reduction: 2H + (aq) + 2e - H 2 (g) overall: M (s) + 2H + (aq) M 2+ (aq) + H 2 (g) EE cell = EE H - EE M = 0.0V - EE M = -EE M hence, any metal with a negative reduction potential will have a positive cell potential and )G will be negative!
The Relationship between EE cell and K eq from Chap. 20, from Chap. 21, )G = - RTlnK eq )G = -nfee cell therefore, or, nfee cell = RTlnK eq EE cell = (RT/nF)lnK eq EE cell = (0.025693V/n)lnK eq or EE cell = (0.059152/n)logK eq there is a relationship between thermodynamic properties
Figure 21-8 illustrates this: and how the respective values are measured.
21.4 EE cell as a function of Concentration - combining standard electrode potentials gives the standard EE cell but what happens under non-standard conditions? - from Chap. 20, )G = )GE + RTlnQ - if we substitute in E cell and EE cell for )G and )GE, we get -nfe cell = -nfee cell + RTlnQ or: E cell = EE cell - (RT/nF)lnQ
This is called "the Nernst Equation" and by switching from natural logs to common logs and evaluating the constants, we actually get the more commonly used form: E cell = EE cell - (0.059152V/n)log(Q) Using this, we can calculate the potential for any system of reactions at any concentration. for example, consider a 0.10M Cu 2+ (aq) and 0.50M Zn 2+ (aq) cell: Cu 2+ (aq) + Zn (s) ÿ Cu (s) + Zn 2+ (aq) Q = [Zn 2+ (aq)] = 0.50 [Cu 2+ (aq)] 0.10
and E cell = Eº cell - RT log [Zn 2+ (aq)] nf [Cu 2+ (aq)] in this case, Eº cell is given by: Cu 2+ (aq) + 2e - ÿ Cu (s) Eº ½ = +0.340V Zn (s) ÿ Zn 2+ (aq) + 2e - Eº ½ = +0.763V Cu 2+ (aq) + Zn (s) ÿ Cu (s) + Zn 2+ (aq) Eº cell = 1.103V and E cell = 1.103V + 0.59152V log (0.50) = 1.082V 2 (0.10)
Concentration Cell -ifq can be used to get voltage, then we can construct a cell that has nothing more than a difference in concentration of a single ion to generate voltage - that is, we can construct a cell with Eº cell = 0.0V and E cell = -0.059152 log Q n - this is the basis of the ph electrode - the concentration of hydrogen ions in the solution are measured against the Standard Hydrogen Electrode (although, in practice, a secondary electrode is used)
that is, E cell = Eº cell - 0.059152 log [H + (aq)]» unknown n [H + (aq)]º with n = 1 and recognizing that the standard state for hydrogen ions is 1.0M giving log 1.0 = 0, we get for the concentration cell: E cell = -0.059152 log[h + (aq)] = 0.059152V(pH) Hence, measuring the potential of a solution with hydrogen ions present gives a cell potential that is directly related to ph. For example, if ph = 3.92, then E cell = 0.059152V(3.92) = 0.232V
Measurement of K sp - we can use a concentration cell to determine the concentration of sparingly soluble salts for example, Ag (s) * Ag + (aq)(sat'd AgBr) 2 Ag + (aq) (0.100M) * Ag (s) E cell = 0.4230V the two half cells are: Ag + (aq) (0.100M) + 1e - Ag (s) AgBr (s) Ag + (aq) + Br - (aq) (sat'd) overall: Ag + (aq) (0.100M) Ag + (aq)(sat'd)
giving us, Q = [Ag + (aq)(sat'd)] [Ag + (aq)(0.100m)] E cell = EE cell - 0.059152 log [Ag + (aq)(sat'd)] n [Ag + (aq)(0.100m)] = EE cell - 0.059152 log(x/0.100) hence, 0.4230 = 0-0.059152(log x - log 0.100) 0.4230 = -log(x) + log(0.100) 0.059152
and: hence, 7.151 = -log(x) + 1.00 x = 10-6.151 = 7.06 x10-7 (= [Ag + (aq)]) and since [Ag + (aq)] = [Br - (aq)] we have, K sp = [Ag + (aq)][br - (aq)] = (7.06 x10-7 )(7.06 x10-7 ) = 5 x10-13
21.5 Batteries: Producing Electricity Through Chemical Reactions - better might be to say "producing energy from chemical reactions" as batteries harvest the free energy available and turn it from "heat" into "electrical energy" - batteries are storage devices and they come in three forms: primary cells secondary cells flow cells (or fuel cells)
primary cells - these are batteries in which there is an irreversible reaction so that once all of the chemicals in the battery are consumed, it is dead (i.e. Duracell) secondary cells - these are rechargeable batteries. The chemical reaction that results in the electrical energy can be reversed by the application of an external energy source. (i.e. Lead-Acid batteries) flow cells - strictly speaking, these are not batteries as they are not storage devices. Rather, they draw their energy from the continual flow (and reaction) of chemical reactants through the cell.
Most common form of voltaic cell is the Leclanche cell invented in the 1860s. It is a "dry cell" because the electrolyte is a solid.
The Lead-Acid (Storage) Battery - truly a storage battery as it will "take a charge" from an external power source, store it until needed, and then deliver the charge on demand. These are the batteries found in automobiles.
The chemical reactions in the Lead-Acid battery can be written: PbO 2 (s) + 3H + (aq) + HSO 4 - (aq) + 2e - PbSO 4 (s) + 2H 2 O (l) Pb (s) + HSO 4 - (aq) PbSO 4 (s) + H + (aq) + 2e - PbO 2 (s) + Pb (s) + 2H + (aq)+ HSO 4 - (aq) 2PbSO 4 (s) + 2H 2 O (l) E cell = E PbO2/PbSO4 - E PbSO4/Pb = 1.74V - (-0.28V) = 2.02V Hence, it takes six of these cells to create a 12V battery.