Eo and the Equilibrium Constant, K
Electrochemical measurements (Ch 14-5 to 6) 1. To state the relation between Eo and K. (Do Ex 14-24b, 14-27) 2. From galvanic cell voltage measurement
(a) Ksp (Do Exercise 14-18, 14-34) (b) Ksp and g (Do Exercise 14-39) (c) Ka (Do Exercise 14-G, 14-36) (d) Kf (Do Exercise 14-H, 14-I) (e) Kf and 2 Ka (Do Exercise 14-37) Potentiometric redox titration (Ch 16-1) To perform calculations for potentiometric redox titration (Do Exercise 16-4)
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Eo and the equilibrium constant, K
- 0 Cathodic reaction: aA + ne cC E+ Anodic reaction: - o dD + ne bB E-
c b 0 0.05916 AC é 0 0.05916 AB ù E = E+ - E- = E+ - log a - êE- - log d ú n AA ë n AD û
c d 0.05916 A A o 0.05916 = (E 0 - E o ) - log C D = E - log Q + - a b n n AA AB When the cell is at equilibrium, Q = K and E = 0 with no current flow.
0.05916 0.05916 o \0 = E o - log K E o = log K or K = 10nE / 0.05916 n n
+ - o Example: 2Ag + 2e 2Ag(s) E+ = 0.799V 2 + - o Cd + 2e Cd (s) E - = - 0.402 V o o o 2(1.201) / 0.05916 40.60 40 E = E+ - E- = 0.799 - (-0.402) = 1.201V \K =10 =10 = 4.0´10 Potentiometric measurements allow us to find K that are too large or too small. Chem215/P.Li/electrochem & redox/P 2
1 Electrochemical cells as chemical probes
1. Equilibrium between 2 half-cells and the net reactions are not redox reactions Solubility product determination
- 2- o FeCO3 (s) + 2e Fe(s) + CO3 E+ = -0.756V 2+ - o Fe + 2e Fe(s) E- = -0.44V
2+ 2- o FeCO3 (s) Fe + CO3 E = -0.756 - (-0.44) = -0.316V
2(-0.316 ) / 0.05916 -11 K = K sp =10 = 10
Eo is just another way of expressing the free energy change or equilibrium constant of a reaction DGo = -nFE o and DG o = -RT ln K
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Electrochemical cells as chemical probes 2. Equilibrium within each half-cell If a galvanic cell has a non-zero voltage, then the net cell reaction is not at equilibrium. However, we do allow the half-cells to stand long enough to come to equilibrium within each half-cell. (A) Acid-base equilibrium constant determination
+ - Pt(s) H2 (1.00atm) H (xM ),CH 3CO2 H (0.050M ),CH3CO2 Na(0.0050M ) Cl (0.10M ) AgCl(s) Ag(s)
- - o E = 0.503V AgCl(s) + e Ag(s) + Cl (aq, 0.10M ) E+ = 0.222V + - o 2H (aq, cM ) + 2e H 2 (g,1 atm) E- = 0V The net cell reaction is not at equilibrium because the measured voltage is 0.503V, not 0V.
é 0.05916 ù é 0.05916 PH ù E = E - E = 0.222- log [Cl - ] - 0 - log 2 + - ê ú ê + 2 ú ë 1 û ë 2 [H ] û 0.05916 1.00 0.503 = 0.222 - 0.05916 log(0.10) + log \[H + ] = 10-3.75 = 1.8´10-4 M 2 [H + ]2 This allows us to evaluate the equilibrium constant for the acetic acid dissociation - + -4 -4 [CH 3 CO2 ][H ] (0.0050 +1.8´10 )´ (1.8´10 ) -5 K a = = -4 = 1.87 ´10 [CH 3CO2H ] (0.050-1.8´10 ) The anodic half-cell (the left one) may be thought of as a probe to measure the unknown + Chem215/P.Li/electrochem & redox/P 4 [H ], leading to the evaluation of Ka.
2 Electrochemical cells as chemical probes (B) Complex formation constant determination S.H.E. HgY 2- (aq,0.00500M ), EDTA(aq,0.0150M ) Hg(l) E = 0.331V
2+ - l o Hg + 2e Hg( ) E+ = 0.852V + - o 2H + 2e H 2 (g) E- = 0V
E = E+ - E- é ù 0.05916 1 2+ -17.61 -18 0.331= ê0.852- log 2+ ú - 0 [Hg ] = 10 = 2.4 ´10 M ë 2 [Hg ]û
2- This value allows us to evaluate Kf [HgY ].
2- [HgY ] 0.00500 21 K f = 2+ = -18 -5 = 6´10 Hg a - EDTA (2. ´10 )(2.3´10 )(0.0150) [ ] Y 4 [ ] 4
-5 Re call aY 4- = 2.3´10 at pH 6 Chem215/P.Li/electrochem & redox/P 5
Potentiometric redox titration
Titration of 100.0 mL of 0.0500M Fe2+ with 0.100M Ce4+.
Titration reaction: 17 K » 10 in 1M HClO4 Ce4+ + Fe 2+ ® Ce3+ + Fe 3+ (16-1)
This is a rapid and quantitative reaction. Now, we need to perform some potentiometric calculations in monitoring [Fe2+]/[Fe3+] for the establishment of a titration curve.
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3 Potentiometric redox titration
- - At the S.C.E.: ½Hg2Cl2(s) + e Hg(l) + Cl (aq) E-° = 0.241V At the Pt indicator electrode: o Fe 3+ + e- Fe 2+ E = 0.767V (16- 2) o Ce4+ + e- Ce3+ E =1.70V (16- 3) At the Pt electrode, reactions 16-2 and 16-3 are both in equilibrium. 0.05916 [Fe2+ ] \E = 0.767- log (16-9) + 1 [Fe3+ ] 0.05916 [Ce3+ ] Note that there is only one value of E+. E =1.70- log (16-10) + 1 [Ce4+ ] The line notation of either cell (1) or (2): 1 - 3+ 2+ (1) Hg(l) 2 Hg2Cl2 (s) Cl (aq) Fe (aq),Fe (aq) Pt(s)
1 - 4+ 3+ (2) Hg(l) 2 Hg 2Cl2 (s) Cl (aq) Ce (aq),Ce (aq) Pt(s)
There are 3 regions in the titration curve. Chem215/P.Li/electrochem & redox/P 7
Potentiometric redox titration 1) Before EP (e.g. v = 36.0 mL) Since there is excess Fe2+, and both [Fe2+] and [Fe3+] are known, it is convenient to use equation 16-9 to calculate E+.
[Fe 2+ ] E = E - E(S.C.E.) = 0.767- 0.05916log - 0.241 + [Fe3+ ]
æ 0.0500M ´ 0.1000L - 0.100M ´ 0.0360L ö = 0.526 - 0.05916logç ÷ è 0.100M ´ 0.0360L ø
æ14.0 ö = 0.526 - 0.05916logç ÷ = 0.550V è 36.0 ø
Chem215/P.Li/electrochem & redox/P 8
4 Potentiometric redox titration
2) At EP (e.g. ve = 50.0 mL) ve ´ 0.100M = 100.0mL ´ 0.0500M After the complete reaction of Ce4+ and Fe2+, equal amount of Ce3+ and Fe3+ should be produced. \[Fe3+ ] = [Ce3+ ] (16 - 7) There will be tiny amount of Ce4+ and Fe2+ present because of the following equilibrium. Ce3+ + Fe3+ Ce 4+ + Fe 2+ \[Fe 2+ ] = [Ce 4+ ] (16 - 8) Since [Fe2+], [Fe3+], [Ce3+] and [Ce4+] are not known, we must use both equations 16-9 and 16-10 to calculate E+. We add the two equations, æ [Fe2+ ] [Ce3+ ] ö 2E = 0.767+1.70- 0.05916çlog + log ÷ + ç 3+ 4+ ÷ è [Fe ] [Ce ]ø æ[Fe 2+ ] [Ce3+ ] ö = 2.46 - 0.05916logç ÷ 7 ç 3+ 4+ ÷ = 2.467 - 0.05916log 1 = 2.467 è [Fe ] [Ce ] ø
Note that E is just the average of E of equations 16-2 & 16-3. \E+ =1.23V +
\ E = E+ - E(S.C.E.) = (1.23 - 0.241)V = 0.99V Chem215/P.Li/electrochem & redox/P 9
Potentiometric redox titration 3) After EP (e.g. v = 63.0 mL) Since there is excess Ce4+, and both [Ce4+] and [Ce3+] are known, it is convenient to use equation 16-10 to calculate E+. æ [Ce3+ ]ö E = E - E(S.C.E.) = ç1.70- 0.05916log ÷ - 0.241 + ç 4+ ÷ è [Ce ]ø
æ 0.0500M ´ 0.1000L ö =1.459 - 0.05916logç ÷ è 0.100M ´ 0.0630L - 0.0500M ´ 0.1000L ø
æ 50.0 ö = 1.459 - 0.05916logç ÷ = 1.424V è 13.0 ø
See Fig. 16-2 for a potentiometric redox titration curve for Fe2+ with Ce4+. Note the cell voltage at any point depends only on the ratio, but not absolute values, of the concentrations of the reactants. We should expect to see the same curve if both reactants (e.g. Fe2+ and Ce4+) are diluted by a factor of 10.
Chem215/P.Li/electrochem & redox/P 10
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