Lecture 19 Metal In Indicatr - a cmpund whse clur changes when it binds t a metal in - t be useful, it must bind the metal less strngly than EDTA e.g. titratin f Mg 2+ with EDTA using erichrme black T (In) - very small amunt f In MgIn + EDTA 6 MgEDTA + In red blue - EDTA reacts with free Mg 2+ first, then MgIn - if the metal binds the indicatr mre strngly than EDTA, the metal is said t blck the indicatr and a back titratin must be used e.g. erichrme black T is blcked by Cu 2+, Ni 2+, C 2+, Cr 3+, Fe 3+, Al 3+ -add a knwn excess f EDTA - back titrate excess EDTA with standard Mg 2+ - here Mg 2+ des nt displace Cu 2+ frm EDTA K f (CuEDTA) = 10 19, K f (MgEDTA) = 10 9 ˆ the clur change is frm blue t red. Week 10-1
Metal In Indicatrs Equilibrium Equatins f Erichrme Black T and a Metal In M 2+ ( e.g. Ca 2+ ) HIn 2- º In 3- + H + K 3 = 2.8 x 10-12 M 2+ + In 3- º MIn - K MIn =? M 2+ + HIn 2- º MIn - + H + K =? and K CaIn = 2.5 x 10 5 K 3 K CaIn = 2.8 x 10-12 x 2.5 x 10 5 = 7.0 x 10-7 Week 10-2
Let s lk at the determinatin f hardness f cld tap water frm Guelph. As yu have seen in the lab, impurities (e.g. Cu 2+, Fe 2+, Fe 3+ ) in the tap water frm cmplexes with the erichrme black T and blck the indicatr. In ther wrds, they bind strngly with the indicatr and the metal ins are nt released fr cmplexatin with the EDTA e.g. i. T 100.00 ml f Guelph cld tap water, buffered t ph 10, was added 25.00 ml f 0.01 M EDTA. The excess EDTA was titrated with 17.11 ml 0.01 Mg 2+ slutin ii. T anther 100.00 ml, als buffered t ph 10, was added an excess f CN - slutin. This slutin was then titrated t the end pint with 7.35 ml f the 0.01M EDTA slutin. Calculate the cncentratins f Ca 2+ (including Mg 2+ ) and f the impurities Week 10-3
Titratin Methds with EDTA 1. Direct titratin with EDTA. 2. Back titratin - a knwn excess EDTA added and the excess titrated with anther metal. 3. Displacement Titratin - fr metal ins with n suitable In M n+ + MgY 2-6 MY n-4 + Mg 2+ (+ excess MgY 2- ) 8 8 excess titrated with EDTA K f (MY n-4 ) K f (MgY 2- ) - used fr Hg 2+ e.g. Ag + has n suitable indicatr 2Ag + + Ni(CN) 4 2-6 2Ag(CN) 2! + Ni 2+ 8 titrated with EDTA 4. Masking - a masking agent is a reagent that prtects a cmpnent f the slutin frm frming a cmplex with EDTA. Week 10-4
ELECTROCHEMISTRY In analytical chemistry: 1. electrical wrk using a galvanic cell is nt f interest 2. f interest is the MAXIMUM vltage that can be measured frm a cell Galvanic r Vltaic cells ˆ ˆ ˆ spntaneus chemical reactin t generate electricity half cell ptentials d nt have abslute values instead, they are relative t the standard hydrgen electrde, S.H.E. Week 10-5
The Sign f the Electrde Ptential de Reductin at cathde - lss f electrns - gain f electrns Oxi dat in at an Vltage measurements - ptentimeter/electrnic vltmeter - negligible current draw - fr the diagram abve, the reactin is spntaneus. Week 10-6
A Cmplete Cell Week 10-7
Cmplete Cell and Line Diagram Line Diagram Pt (s) * H 2 (g, " = 1), H + (aq, " = 1) ** Ag + (aq, " = 1)* Ag(s) r S.H.E. ** Ag + (aq, " = 1)* Ag(s) Week 10-8
Ptential f a Cell Nrmal = E + E cell cathde E ande 8 8 Reductin Ptential Oxidatin Ptential Cathde - Reductin: 2Ag + + 2e - º 2Ag (s) ADD Ande - Oxidatin: H 2 (g) º 2H + + 2e - 2Ag + + H 2 (g) º 2Ag (s) + 2H + Week 10-9
By Cnventin S We use this methd fr ur calculatins: = E E cell cathde E ande cell + E = E E 8 8 Reductin Ptential Reductin Ptential Cathde - Reductin: 2Ag + + 2e - º 2Ag (s) SUBTRACT Ande - Reductin 2H + + 2e - º H 2 (g) 2Ag + + H 2 (g) º 2Ag (s) + 2H + Week 10-10
Lecture 20 NERNST EQUATION expresses the net driving frce fr a reactin E and cncentratin are related t free energy i.e. ) G = -nfe and ) G = -nfe Fr a half reactin, aa + ne - º bb the Nernst equatin is, E = E - RT ln Q nf where Q B b A a = α α = reactins nt at equilibrium and " = activities. Nte: we will use species cncentratins [ ] instead f activities, ", in ur slutins t prblems. The difference between these tw units are negligible fr very dilute slutins. The Usable Frm f the Nernst Equatin at 298.15 K is: E = E - 0.05916 lg Q n Week 10-11
E and E are Independent f Hw the Cell Reactin is Written A cell can be written as: 2H + + 2e - º H 2 (g) E 0.05916 = E lg P 2 [H + ] H 2 2 OR H + + e - º ½ H 2 (g) E 1 2 H 2 0.05916 = E lg P 1 [H + ] withut affecting the values f E r E E = wrk dne per culmb f charge - cnstant fr a specific reactin E = ptential fr a specific cell - will change ONLY if species cncentratins are altered. Week 10-12
Calculatin f Half-Cell Ptentials Let s lk at a half cell cnsisting f a silver rd immersed in AgNO 3 (aq). What is the half cell ptential? Nernst equatin is: Ag + + e - º Ag (s) E + = 0. 799V Ag E = E Ag 0. 05916 1 lg 1 [ Ag ] + + Different descriptin f the same reactin e.g. what happens if an excess f NaCl is added t the half cell? S AgCl(s) is precipitated ut and we have excess Cl - (aq) and [Ag + ] = K sp [Cl - ] S Ag + is still being reduced at the electrde t Ag (s) and the Nernst equatin can nw be written as: E = E + Ag 0. 05916 1 Cl lg [ ] Ksp Week 10-13
But if [Cl - ] = 1.0 M i.e standard state cnditins, then 0. 05916 E = E + + lg K Ag 1 = 0. 799 + ( 0576. ) = E / sp AgCl Ag where E AgCl / Ag refers t: AgCl (s) + e - º Ag (s) + Cl - Thus, the tw electrdes are related via K sp. Week 10-14
E H 2 O / H 2 H 2 O + e - º ½ H 2 (g) + OH - = -0.83 V E H + / H 2 2H + + 2 e - º H 2 (g) = 0.000 V The abve 2 half cells are related via K w E H + / H 2 i.e. E = - 0. 05916 [H 2 ] lg 2 2 [ H + ] H 2 O º H + + OH - and K w = [H + ] [OH - ] [H + ] = K w / [OH - ] ˆ E = 0.000-0. 05916 lg 2 [H 2 ][ OH ] 2 K w 2 at standard state cnditins f 298.15 K and [H 2 ] / [OH - ] / 1.0 E = - 0. 05916 1 lg 2 K 2 w 0. 05916 lg K 2 w E = 2 = E H 2 O / H 2 = 0.02958 lg 10-28 = 0.02958 x (-28) and E H 2 O / H 2 = -0.83 V Week 10-15
E ZnY 2 / Zn ZnY 2- + 2e - º Zn (s) + Y 4- = -1.25 V where Y 4- is the cmpletely deprtnated frm f EDTA E Zn 2+ / Zn Zn 2+ + 2e - º Zn (s) = -0.76 V Shw the abve 2 half cells are related via K f Week 10-16
Nernst Equatin fr a Cmplete Cell Find the ptential f the cell: Cd (s) * CdCl 2 (aq, 0.0167 M) * AgCl (s) * Ag (s) 2AgCl () s + 2e 2Ag () s + 2Cl E = 0222. V 2 + Cd + 2e Cd () s E = 0402. V 2 + 2AgCl () s + Cd () s 2Ag () s + Cd + 2Cl Ecell = + 0624. V + Cathde Ptential 0. 05916 2 E+ = E+ lg [ Cl ] 2 = 0222. 005916. lg[?? ] = 0309. V Ande Ptential E 0. 05916 1 lg 2 [ 2 Cd ] = E + = 0402. 0. 05916 2 1 lg?? = 0. 402 0. 053 = 0. 455V Week 10-17
Cell Ptential E cell =? E cell is psitive ˆ What? Frm the net cell reactin abve, ptential difference can als be calculated i.e. 0. 05916 2+ 2 Ecell = Ecell lg [ Cd ][ Cl ] 2 Week 10-18