Chem 222 #11 Ch 16 Feb 15, 2005

Transcription

1 Chem 222 #11 Ch 16 Feb 15, 2005

2 Announcement Download your handout and answer sheet for Exp KH 3-14 starting in the next week at

3 Exp 7 Plot a graph of ph versus titrant volume (V). Plot ( ph/ V vs V) & ( ( ph/ V)/ V vs V) Read Text P 237 and Try Example in P238

4 Find the maximum from the first derivative ( ph/ V) Lab Handout P21

5 Using second derivatives (p237)

6 14-1 Basic Concept of Oxidization and Reduction A species is said to be [Q1] when it loses electrons. It is reduced when it gains electrons. Fe 3+ + V 2+ Fe 2+ + V 3+ Reduced Oxidized A reducing agent, also called a reductant, gives electrons to another substances. An oxidizing agent takes electrons from another substances. Fe 3+ + V 2+ Fe 2+ + V 3+ Q2. Which of Fe 3+ and V 2+ is oxidizing agent?

7 Redox Titration Redox titration is based on an oxidation-reduction reaction.

8 Nernst Equation (p ) aa(oxidized) + ne - bb(reduced) E = E 0 (RT/nF)Log([B] b /[A] a ) = E 0 ( /n)Log([B] b /[A] a ) (14-13) Please memorize the equation (constants should be provided)

9 Higher E 0 means the reaction goes to more easily. (Stronger Oxidizing Agent) Q. Which of the two reactions is preferred?

10 16-1 Shape of a redox titration curve Ce 4+ + Fe 2+ Ce 3+ + Fe 3+ Ceric Ferrous Cerous Ferric Titrant Titration Reaction: Analyte K = in 1M HClO 4 The reaction be almost 100 % Ce 4+ Potentiometer (measure voltage) e - Calomel Reference Electrode Fe 2+ in 1M HClO 4 Pt Wire Stirring bar At the Calomel Reference Electrode At the Pt Electrode (Note: only small amount reacts)

11 So how to get the concentration from voltage? (p350) E =E + -E - G ~ - qe = -nee E + E - Measuring voltage E leads to [Fe 2+ ]/[Fe 3+ ] This is also correct: E [Ce 3+ ]/[Ce 4+ ] See also Nernst equation p293 in Ch 14

12 Cell reactions Cell reactions for Fe V + 2 ( ) Cell reactions for Ce + 2 ( ) Trick: These reactions happen for a small portion of molecules Concentration of Ce 4+ & Fe 3+ are not altered by these cell reactions

13 Potential (V versus S. C. E)

14 How we calculate the titration? Ce 4+ + Fe 2+ Ce 3+ + Fe 3+ Titrant Analyte Before equivalence Excess Fe 2+ & Fe 3+ Little Ce 4+ After equivalence Excess Ce 4+ &Ce 3+ Little Fe 2+ Potential (V versus S. C. E)

15 Case1) Before equivalence point Difficult to calculate [Ce 4+ ] We use [Fe 2+ ] and [Fe 3+ ]. There is a special point at V = V e /2 [Fe 2+ ] = [Fe 3+ ] Namely, Log[Fe 2+ ]/[Fe 3+ ] = 0 E = E(calomel) If you need concentration of Ce 3+, use [Ce 3+ ] = [Fe 3+ ]. Moles of Fe 3+ = C 0 Ce4+V Moles of Fe 2+ =C 0 Ce4+(V e -V)

16 = E +e = ( )/2 E e = E +e E(calomel) = = 0.99V Case 2) At equivalence point Ce 4+ + Fe 2+ Ce 3+ + Fe 3+ Titrant Analyte x x a x a x At equivalence point Mostly Ce 3+ & Fe 3+ (x << a) [Ce 4+ ] = [Fe 2+ ] & [Ce 3+ ] = [Fe 3+ ]

17 Case 3) After equivalence point Excess [Ce 4+ ] and [Ce 3+ ] (= [Fe 3+ ]) Use [Ce 4+ ] & [Ce 3+ ] Again when [Ce 3+ ] = [Ce 4+ ], E = [ Q1 ] E(calomel) When [Ce 3+ ] = [Ce 4+ ], The volume of the added Ce 4+ solution V is V = [Q2] V e Hint: At V= V e, your Ce ions are all in Ce 3+. Ce 3+ moles = Ve C 0 Ce4+ Ce 4+ moles = (V V e ) C 0 Ce4+

18 Example Suppose that we titrate ml of M Fe 2+ with M Ce 4+ using cells in Fig Q1) Equivalence point occurs at V e Ce4+ = (100 ml * M)/ [Q1 ] = [Q2] ml. A1)? ml Q3) Calculate the cell voltage at V Ce4+ = 36.0 ml, 50.0 ml, 63.0 ml. A3) At V Ce4+ = 36 ml before equivalence Use [Fe 2+ ] & [Fe 3+ ] Moles of Fe 3+ in the solution is M 36.0 ml Remaining Fe 2+ is ( M ml M 36.0 ml) = M 14.0 ml The ratio of moles between Fe 2+ and Fe 3+ in solution is [Fe 2+ ]/[Fe 3+ ] = [ Q4 ]/(36.0 ml ) = E = Log[ Q5 ]

19 When V = 50 ml, use E + = [Q1]/2 E = E+ - E(Calomel) When V = 63 ml, use [Ce 4+ ] and [Ce 3+ ]. Moles of Ce 3+ = C 0 Ce4+ V e = M [Q2] ml Moles of Ce 4+ = C Ce4+0 (V V e ) [Ce 3+ ]/[Ce 4+ ] = [Q3]

20 Shape of Titration Curve

21 P352 Demonstration MnO Fe H + Titrant Analyte Mn Fe H 2 O Dissolve g of Fe(NH 4 )(SO 4 ) 2 6H 2 O (FM ; 1.00 mmol) in 500 ml of 1 M H 2 SO 4. Titrate the well-stirred solution with M KMnO 4. Q1. How much KMnO 4 solution is required to reach the equivalence point? C Fe : 1.00 mmol/0.5l = 2.0 mm C Fe V Fe /C MnO4 V e = [Q1] (2.0mM * 500mL) = 5 * [Q2] *V e V e V e = 10 ml Q2. Calculate the cell voltage for 5 ml, 10 ml, 20 ml

22 MnO Fe H + Titrant Analyte Mn Fe H 2 O Fe 3+ + e- Fe E 0 = 0.68 V in 1M H 2 SO 4 MnO H + + 5e- Mn H 2 O E 0 = V (1) For Fe : E = 0.68 (2) For Mn: E = [ Fe Log 3 [ Fe For equivalence point (1) + 5 (2) yields [ Mn ] Log [ MnO ][ H 4 ] ] + ] E = [ Mn ][ Fe ] Log [ MnO ][ H ] [ Fe + 4 ] [Mn 2+ ] = 5[Fe 3+ ]; [MnO 4- ] = 5[Fe 2+ ]

23 P352 Demonstration MnO Fe H + Titrant Analyte Mn Fe H 2 O Dissolve g of Fe(NH 4 )(SO 4 ) 2 6H 2 O (FM ; 1.00 mmol) in 500 ml of 1 M H 2 SO 4. Titrate the well-stirred solution with M KMnO 4. Q1. How much KMnO 4 solution is required to reach the equivalence point? 1.00 mmol/5 = M V e V e = 10 ml Q2. Calculate the cell voltage when volume of the titrant, V, is 5 ml, 10 ml, 20 ml At V = 5 ml, the fraction of Fe 2+ oxidized into Fe 3+ is V/V e The fraction of remaining Fe 2+ is (V e -V)/V e At V =10 ml, what is the ratio between [MnO 4- ] and [Fe 2+ ]? At V =20 ml, the fraction of Mn 2+ is V e /V. the fraction of MnO 4- is (V-V e )/V.

24 Streamline calculation 1) Before equivalence point Moles of Fe 3+ 5 C MnO4- V Moles of Fe 2+ = Original # of moles Consumed # of moles = C Fe2+ V Fe2+ -5 C MnO4- V Using C Fe2+ V Fe2+ = 5 C MnO4- V e = 5 C MnO4- V e -5 C MnO4- V =5 C MnO4- (V e -V) 3) After equivalence point Moles of Mn 2+ C MnO4- V e Moles of MnO - 4 = added MnO 4- - Consumed MnO - 4 = C MnO4- V C MnO4- V e = C MnO4- (V-V e ) [Mn 2+ ]/[MnO 4- ] = V e /(V V e )

25 MnO Fe H + Titrant Analyte Mn Fe H 2 O Fe 3+ + e- Fe E 0 = 0.68 V in 1M H 2 SO 4 MnO H + + 5e- Mn H 2 O E 0 = V (1) For Fe : E = 0.68 (2) For Mn: E = [ Fe Log 3 [ Fe [ Mn ] Log [ MnO ][ H ] ] + ] [H + ] = 1 M {500 ml/ (500 ml + V )} see p115 about sulfuric acid (3) For equivalence point (1) + 5 (2) yields 6E = [ Mn ][ Fe ].05916Log + 8 [ MnO ][ H ] [ Fe ] [Mn 2+ ] = 5[Fe 3+ ]; [MnO 4- ] = 5[Fe 2+ ]

26 Weighted average + ph dependent factor