D. Harvey - Modern Analytical Chemistry (794078), страница 45
Текст из файла (страница 45)
Conversely, the species being reduced is called an oxidizingagent. Thus, in reaction 6.22, Fe3+ is the oxidizing agent and H2C2O4 is the reducingagent.The products of a redox reaction also have redox properties. For example, theFe2+ in reaction 6.22 can be oxidized to Fe3+, while CO2 can be reduced to H2C2O4.Borrowing some terminology from acid–base chemistry, we call Fe2+ the conjugatereducing agent of the oxidizing agent Fe3+ and CO2 the conjugate oxidizing agent ofthe reducing agent H2C2O4.Unlike the reactions that we have already considered, the equilibrium positionof a redox reaction is rarely expressed by an equilibrium constant. Since redox reactions involve the transfer of electrons from a reducing agent to an oxidizing agent,it is convenient to consider the thermodynamics of the reaction in terms of theelectron.The free energy, ∆G, associated with moving a charge, Q, under a potential, E,is given by∆G = EQCharge is proportional to the number of electrons that must be moved.
For a reaction in which one mole of reactant is oxidized or reduced, the charge, in coulombs, isQ = nFwhere n is the number of moles of electrons per mole of reactant, and F is Faraday’sconstant (96,485 C ⋅ mol–1). The change in free energy (in joules per mole; J/mol)for a redox reaction, therefore, is∆G = –nFE6.23where ∆G has units of joules per mole. The appearance of a minus sign in equation6.23 is due to a difference in the conventions for assigning the favored direction forreactions.
In thermodynamics, reactions are favored when ∆G is negative, andredox reactions are favored when E is positive.The relationship between electrochemical potential and the concentrationsof reactants and products can be determined by substituting equation 6.23 intoequation 6.3–nFE = –nFE° + RT ln QNernst equationAn equation relating electrochemicalpotential to the concentrations ofproducts and reactants.where E° is the electrochemical potential under standard-state conditions.
Dividingthrough by –nF leads to the well-known Nernst equation.*Separating a redox reaction into its half-reactions is useful if you need to balance the reaction. One method forbalancing redox reactions is reviewed in Appendix 4.1400-CH06 9/9/99 7:40 AM Page 147Chapter 6 Equilibrium ChemistryE = Eo –RTln QnFSubstituting appropriate values for R and F, assuming a temperature of 25 °C(298 K), and switching from ln to log* gives the potential in volts asE = Eo –0.05916log Qn6.24The standard-state electrochemical potential, E°, provides an alternative way ofexpressing the equilibrium constant for a redox reaction.
Since a reaction at equilibrium has a ∆G of zero, the electrochemical potential, E, also must be zero. Substituting into equation 6.24 and rearranging shows thatEo =RTlog KnF6.25Standard-state potentials are generally not tabulated for chemical reactions, but arecalculated using the standard-state potentials for the oxidation, E°ox, and reductionhalf-reactions, E°red. By convention, standard-state potentials are only listed for reduction half-reactions, and E° for a reaction is calculated asE°reac = E°red – E°oxwhere both E°red and E°ox are standard-state reduction potentials.Since the potential for a single half-reaction cannot be measured, a reference halfreaction is arbitrarily assigned a standard-state potential of zero.
All other reductionpotentials are reported relative to this reference. The standard half-reaction is2H3O+(aq) + 2e–t 2H2O(l) + H2(g)Appendix 3D contains a listing of the standard-state reduction potentials for selected species. The more positive the standard-state reduction potential, the morefavorable the reduction reaction will be under standard-state conditions. Thus,under standard-state conditions, the reduction of Cu2+ to Cu (E° = +0.3419) ismore favorable than the reduction of Zn2+ to Zn (E° = –0.7618).EXAMPLE 6.5Calculate (a) the standard-state potential, (b) the equilibrium constant, and(c) the potential when [Ag + ] = 0.020 M and [Cd 2+ ] = 0.050 M, for thefollowing reaction taking place at 25 °C.Cd(s) + 2Ag+(aq)t Cd2+(aq) + 2Ag(s)SOLUTION(a) In this reaction Cd is undergoing oxidation, and Ag+ is undergoingreduction.
The standard-state cell potential, therefore, isEo = EoAg + / Ag – EoCd 2 + /Cd = 0.7996 V – (–0.4030 V) = 1.2026 V(b) To calculate the equilibrium constant, we substitute the values for thestandard-state potential and number of electrons into equation 6.25.1.2026 =*ln(x) = 2.303 log(x)0.05916log K21471400-CH06 9/9/99 7:40 AM Page 148148Modern Analytical ChemistrySolving for K gives the equilibrium constant aslog K = 40.6558K = 4.527 × 1040(c) The potential when the [Ag+] is 0.020 M and the [Cd2+] is 0.050 M iscalculated using equation 6.24 employing the appropriate relationship forthe reaction quotient Q.E = E° –0.05916[Cd2 + ]logn[Ag + ]2= 1.2026 –0.05916(0.050)log2(0.020)2= 1.14 V6E Le Châtelier’s PrincipleThe equilibrium position for any reaction is defined by a fixed equilibrium constant, not by a fixed combination of concentrations for the reactants and products.This is easily appreciated by examining the equilibrium constant expression for thedissociation of acetic acid.Ka =Le Châtelier’s principleWhen stressed, a system that was atequilibrium returns to its equilibriumstate by reacting in a manner thatrelieves the stress.[H 3O + ][CH3COO – ]= 1.75 × 10 –5[CH 3COOH]6.26As a single equation with three variables, equation 6.26 does not have a unique solution for the concentrations of CH3COOH, CH3COO–, and H3O+.
At constanttemperature, different solutions of acetic acid may have different values for [H3O+],[CH3COO–] and [CH3COOH], but will always have the same value of Ka.If a solution of acetic acid at equilibrium is disturbed by adding sodium acetate,the [CH3COO–] increases, suggesting an apparent increase in the value of Ka.
SinceKa must remain constant, however, the concentration of all three species in equation 6.26 must change in a fashion that restores Ka to its original value. In this case,equilibrium is reestablished by the partial reaction of CH3COO– and H3O+ to produce additional CH3COOH.The observation that a system at equilibrium responds to a stress by reequilibrating in a manner that diminishes the stress, is formalized as Le Châtelier’s principle. One of the most common stresses that we can apply to a reaction at equilibrium is to change the concentration of a reactant or product.
We already have seen,in the case of sodium acetate and acetic acid, that adding a product to a reactionmixture at equilibrium converts a portion of the products to reactants. In this instance, we disturb the equilibrium by adding a product, and the stress is diminishedby partially reacting the excess product. Adding acetic acid has the opposite effect,partially converting the excess acetic acid to acetate.In our first example, the stress to the equilibrium was applied directly. It is alsopossible to apply a concentration stress indirectly.
Consider, for example, the following solubility equilibrium involving AgClAgCl(s)t Ag+(aq) + Cl–(aq)6.271400-CH06 9/9/99 7:40 AM Page 149Chapter 6 Equilibrium ChemistryThe effect on the solubility of AgCl of adding AgNO3 is obvious,* but what is the effect of adding a ligand that forms a stable, soluble complex with Ag+? Ammonia, forexample, reacts with Ag+ as followsAg+(aq) + 2NH3(aq)t Ag(NH3)2+(aq)6.28Adding ammonia decreases the concentration of Ag+ as the Ag(NH3)2+ complexforms. In turn, decreasing the concentration of Ag+ increases the solubility of AgClas reaction 6.27 reestablishes its equilibrium position. Adding together reactions6.27 and 6.28 clarifies the effect of ammonia on the solubility of AgCl, by showingthat ammonia is a reactant.AgCl(s) + 2NH3(aq)t Ag(NH3)2+(aq) + Cl–(aq)6.29EXAMPLE 6.6What is the effect on the solubility of AgCl if HNO3 is added to the equilibriumsolution defined by reaction 6.29?SOLUTIONNitric acid is a strong acid that reacts with ammonia as shown hereHNO3(aq) + NH3(aq)t NH4+(aq) + NO3–(aq)Adding nitric acid lowers the concentration of ammonia.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
