Analytical Chem istry - DePauw University

Chapter 6 Equilibrium Chemistry225+ 2+Cd() s + 2Ag ( aq) 2Ag() s + Cd ( aq)So l u t i o n(a) In this reaction Cd is undergoing oxidation and Ag + is undergoingreduction. The standard cell potential, therefore, isoE = ooE − E = − − =+ 2+0. 7996 ( 0. 4030) 1.2026 VAg/ Ag Cd / Cd(b) To calculate the equilibrium constant we substitute appropriate valuesinto equation 6.25.Eo= 1.2026 V =0.059162Vlog KSolving for K gives the equilibrium constant aslog K = 40.6558K = 4.527×10 40(c) To calculate the potential when [Ag + ] is 0.020 M and [Cd 2+ ] is0.050 M, we use the appropriate relationship for the reaction quotient,Q, in equation 6.24.2+o 0.05916 V CdE = E − log [ ]+ 2n [ Ag ]0.05916 V 0 050E = 1.2606 − log ( . )V2 ( 0. 020)2E = 114 . VPractice Exercise 6.4For the following reaction at 25 o C2+ − +5Fe ( aq) + MnO ( aq) + 8H( aq)45Fe( aq) + Mn ( aq) + 4H O( l )3+ 2+calculate (a) the standard potential, (b) the equilibrium constant, and (c)the potential under these conditions: [Fe 2+ ] = 0.50 M, [Fe 3+ ] = 0.10 M,[MnO 4 – ] = 0.025 M, [Mn 2+ ] = 0.015 M, and a pH of 7.00. See Appendix13 for standard state reduction potentials.Click here to review your answer to this exercise.2When writing precipitation, acid–base,and metal–ligand complexation reaction,we represent acidity as H 3 O + . Redox reactionsare more commonly written usingH + instead of H 3 O + . For the reaction inPractice Exercise 6.4, we could replace H +with H 3 O + and increase the stoichiometriccoefficient for H 2 O from 4 to 12.