D. Harvey - Modern Analytical Chemistry (794078), страница 97
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We can, however, calculateCe4+3331400-CH09 9/9/99 2:13 PM Page 334334Modern Analytical ChemistryEeq by combining the two Nernst equations. To do so we recognize that the potentials for the two half-reactions are the same; thus,° 3 + 2+ − 0.05916 logE eq = EFe/ Fe[Fe 2+ ][Fe3+ ]° 4+E eq = ECe/ Ce 3+ − 0.05916 log[Ce3+ ][Ce 4+ ]Adding together these two Nernst equations leaves us with° 3+ 2+ + E ° 4+2E eq = EFe/ FeCe / Ce 3+ − 0.05916 log[Fe 2+ ][Ce3+ ][Fe3+ ][Ce 4+ ]9.18At the equivalence point, the titration reaction’s stoichiometry requires that[Fe2+] = [Ce4+][Fe3+] = [Ce3+]The ratio in the log term of equation 9.18, therefore, equals one and the log term iszero.
Equation 9.18 simplifies toE eq =° 3+ 2+ + E ° 4+EFe/ FeCe / Ce 3+2=0.767 V + 1.70 V= 1.23 V2After the equivalence point, the concentrations of Ce3+ and excess Ce4+ are easyto calculate. The potential, therefore, is best calculated using the Nernst equationfor the titrant’s half-reaction.° 4+E = ECe/ Ce 3+ − 0.05916 log[Ce3+ ][Ce 4+ ]9.19For example, after adding 60.0 mL of titrant, the concentrations of Ce3+ and Ce4+areintial moles Fe2+M FeVFe[Ce3+ ] ==total volumeVFe + VCe=[Ce4+ ] ==(0.100 M)(50.0 mL)= 4.55 × 10 −2 M50.0 mL + 60.0 mLmoles excess Ce4+M V − M FeVFe= Ce Cetotal volumeVFe + VCe(0.100 M)(60.0 mL) − (0.100 M)(50.0 mL)= 9.09 × 10 −3 M50.0 mL + 60.0 mLSubstituting these concentrations into equation 9.19 gives the potential asE = +1.70 V − 0.05916 log4.55 × 10 −2= 1.66 V9.09 × 10 −3Additional results for this titration curve are shown in Table 9.17 and Figure 9.34.1400-CH09 9/9/99 2:13 PM Page 335Chapter 9 Titrimetric Methods of AnalysisTable 9.17Data for Titration of 50.0 mL of0.100 M Fe2+ with 0.100 M Ce4+E (V)Volume Ce4+(mL)E (V)5.0010.0015.0020.0025.0030.0035.0040.0045.0050.000.7110.7310.7450.7570.7670.7770.7890.8030.8231.2355.0060.0065.0070.0075.0080.0085.0090.0095.00100.001.641.661.671.681.681.691.691.691.701.70EVolume Ce4+(mL)Sketching a Redox Titration Curve As we have done for acid–base and complexometric titrations, we now show how to quickly sketch a redox titration curve using aminimum number of calculations.EXAMPLE 9.11Sketch a titration curve for the titration of 50.0 mL of 0.100 M Fe2+ with 0.100M Ce4+ in a matrix of 1 M HClO4.
This is the same titration for which wepreviously calculated the titration curve (Table 9.17 and Figure 9.34).SOLUTIONWe begin by drawing axes for the titration curve (Figure 9.35a). Having shownthat the equivalence point volume is 50.0 mL, we draw a vertical lineintersecting the x-axis at this volume (Figure 9.35b).Before the equivalence point, the solution’s electrochemical potential isdetermined by the concentration of excess Fe2+ and the concentration of Fe3+produced by the titration reaction. Using values from Table 9.17, we plot E for5.0 mL and 45.0 mL of titrant (Figure 9.35c).After the equivalence point, the solution’s electrochemical potential isdetermined by the concentration of excess Ce4+ and the concentration of Ce3+.Using values from Table 9.17, we plot points for 60.0 mL and 80.0 mL of titrant(Figure 9.35d).To complete an approximate sketch of the titration curve, we drawseparate straight lines through the two points before and after the equivalencepoint (Figure 9.35e).
Finally, a smooth curve is drawn to connect the threestraight-line segments (Figure 9.35f).3351.81.61.41.210.80.6020406080Volume of Ce4+100Figure 9.34Redox titration curve for 50.0 mL of 0.100 MFe2+ with 0.100 M Ce4+ in 1 M HClO4.1400-CH09 9/9/99 2:13 PM Page 336336Modern Analytical ChemistryPercent titrated050100Percent titrated15020001.6001.6001.4001.4001.2001.0000.8000.6000.4002001.2001.0000.8000.6000.2000.2000.0000.00020406080Volume of titrant (mL)100(a)020406080Volume of titrant (mL)100(b)Percent titrated050100Percent titrated15020001.800501001502001.8001.6001.6001.4001.400Potential (V)Potential (V)1500.40001.2001.0000.8000.6000.4001.2001.0000.8000.6000.4000.2000.2000.0000.000020406080Volume of titrant (mL)100(c)020406080Volume of titrant (mL)100(d)Percent titrated050100Percent titrated15020001.8001.8001.6001.6001.4001.400Potential (V)Potential (V)1001.800Potential (V)Potential (V)1.800501.2001.0000.8000.6000.400501001502001.2001.0000.8000.6000.4000.2000.2000.0000.000020406080Volume of titrant (mL)(e)Figure 9.35How to sketch a redox titration curve; see text for explanation.1000(f)20406080Volume of titrant (mL)1001400-CH09 9/9/99 2:13 PM Page 337Chapter 9 Titrimetric Methods of Analysis9D.2 Selecting and Evaluating the End PointThe equivalence point of a redox titration occurs when stoichiometrically equivalent amounts of analyte and titrant react.
As with other titrations, any difference between the equivalence point and the end point is a determinate source of error.Where Is the Equivalence Point? In discussing acid–base titrations and complexometric titrations, we noted that the equivalence point is almost identicalwith the inflection point located in the sharply rising part of the titration curve.If you look back at Figures 9.8 and 9.28, you will see that for acid–base and complexometric titrations the inflection point is also in the middle of the titrationcurve’s sharp rise (we call this a symmetrical equivalence point).
This makes itrelatively easy to find the equivalence point when you sketch these titrationcurves. When the stoichiometry of a redox titration is symmetrical (one moleanalyte per mole of titrant), then the equivalence point also is symmetrical. If thestoichiometry is not symmetrical, then the equivalence point will lie closer to thetop or bottom of the titration curve’s sharp rise. In this case the equivalencepoint is said to be asymmetrical. Example 9.12 shows how to calculate the equivalence point potential in this situation.EXAMPLE 9.12Derive a general equation for the electrochemical potential at the equivalencepoint for the titration of Fe2+ with MnO4–; the reaction is5Fe2+(aq) + MnO4–(aq) + 8H3O+(aq)t 5Fe3+(aq) + Mn2+(aq) + 12H2O(l)SOLUTIONThe redox half-reactions for the analyte and the titrant aret Fe3+(aq) + e–MnO4–(aq) + 8H3O+(aq) + 5e– t Mn2+(aq) + 12H2O(l)Fe2+(aq)for which the Nernst equations are° 3+ 2+ − 0.05916 logE eq = EFe/ Fe°E eq = E MnO4− / Mn 2+−[Fe 2+ ][Fe3+ ]0.05916[Mn2 + ]log5[MnO4 − ][H3O+ ]8Before adding together these two equations, the second equation must bemultiplied by 5 so that the log terms can be combined; thus° 3+ 2+ + 5E °6E eq = EFe/ FeMnO4− / Mn 2+− 0.05916 logAt the equivalence point, we know that[Fe2+] = 5 × [MnO4–][Fe3+] = 5 × [Mn2+][Fe 2+ ][Mn2+ ][Fe3+ ][MnO4 − ][H3O+ ]83371400-CH09 9/9/99 2:13 PM Page 338Modern Analytical ChemistrySubstituting these equalities into the equation for Eeq and rearranging givesE eq ====° 3+ 2+ +°EFe5E MnO/ Fe4− / Mn 2+−0.059165[MnO4 − ][Mn2+ ]log65[Mn2+ ][MnO4 − ][H3O+ ]84− / Mn 2+−0.059161log6[H3O+ ]84− / Mn 2++(0.05916)(8)log[H3O+ ]66° 3+ 2+ + 5E °EFe/ FeMnO6° 3+ 2+ + 5 °EFeE MnO/ Fe6° 3+ 2+ + 5E °EFe/ FeMnO− / Mn 2+− 0.07888pH6For this titration the electrochemical potential at the equivalence point consistsof two terms.
The first term is a weighted average of the standard state orformal potentials for the analyte and titrant, in which the weighting factors arethe number of electrons in their respective redox half-reactions. The secondterm shows that Eeq is pH-dependent. Figure 9.36 shows a typical titrationcurve for the analysis of Fe 2 + by titration with MnO 4 – , showing theasymmetrical equivalence point. Note that the change in potential near theequivalence point is sharp enough that selecting an end point near the middleof the titration curve’s sharply rising portion does not introduce a significanttitration error.41.61.4E3381.21Figure 9.36Titration curve for Fe2+ with MnO4– in 1 MH2SO4; equivalence point is shown by thesymbol .0.80.6020406080Volume of MnO4–100Finding the End Point with a Visual Indicator Three types of visual indicatorsare used to signal the end point in a redox titration. A few titrants, such asMnO4–, have oxidized and reduced forms whose colors in solution are significantly different.
Solutions of MnO4– are intensely purple. In acidic solutions,however, permanganate’s reduced form, Mn2+, is nearly colorless. When MnO4–is used as an oxidizing titrant, the solution remains colorless until the first dropof excess MnO4– is added. The first permanent tinge of purple signals the endpoint.A few substances indicate the presence of a specific oxidized or reduced species.Starch, for example, forms a dark blue complex with I3– and can be used to signalthe presence of excess I3– (color change: colorless to blue), or the completion of areaction in which I3– is consumed (color change: blue to colorless). Another example of a specific indicator is thiocyanate, which forms a soluble red-colored complex, Fe(SCN)2+, with Fe3+.1400-CH09 9/9/99 2:13 PM Page 339Chapter 9 Titrimetric Methods of AnalysisThe most important class of redox indicators, however, are substances thatdo not participate in the redox titration, but whose oxidized and reduced formsdiffer in color.
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