Analytical Chem istry - DePauw University

274 Analytical Chemistry 2.0AgBr() s Ag + ( aq) + Br− ( aq )The remaining two reactions are the stepwise formation of Ag(S 2 O 3 ) 2 3– ,which we characterize by K 1 and K 2 .+ 2− −Ag ( aq) + SO ( aq) Ag( SO)( aq)2 32 3Ag( SO) − 2( ) SO − 3aq + ( aq) Ag( SO)− ( aq)2 3 2 32 3 2Using values for K sp , K 1 , and K 2 from Appendix 10 and Appendix 11, wefind that the equilibrium constant for our reaction is−13 8 4K = K × K × K = ( 50 . × 10 )( 66 . × 10 )( 71 . × 10 ) = 23sp 1 2Click here to return to the chapter.Practice Exercise 6.4The two half-reactions are the oxidation of Fe 2+Feand the reduction of MnO 4 – .( aq) Fe ( aq)+e2+ 3+ −− + − +MnO ( aq) + 8H ( aq) + 5e Mn ( aq) + 4HO()l42From Appendix 13, the standard state reduction potentials for these halfreactionsareEo= 0. 771 V E = 151 . Vo3+ 2+ − 2+Fe / FeMnO4/ Mn(a) The standard state potential for the reaction iso ooE = E− 2+ − E3+ 2+= 151 . V− 0. 771 V = 074 . VMnO4/ MnFe / Fe(b) To calculate the equilibrium constant we substitute appropriate valuesinto equation 6.25.Eo= 074 . V =Solving for K gives its value as0.059165log K = 62.5K = 32 . × 10 62log K(c) To calculate the potential under these non-standard state conditions,we make appropriate substitutions into the Nernst equation.2