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Ion-Selective Electrodes

A galvanic cell is made from two half cells. In the first, a platinum electrode is immersed in a solution at pH = 2.00 that is 0.100 M in both MnO 4 - (aq) and Mn 2+ (aq). In the second, a zinc electrode is immersed in a 0.0100 M solution of Zn(NO 3 ) 2 . Calculate the cell voltage at 25 o C

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Ion-Selective Electrodes

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  1. A galvanic cell is made from two half cells. In the first, a platinum electrode is immersed in a solution at pH = 2.00 that is 0.100 M in both MnO4-(aq) and Mn2+(aq). In the second, a zinc electrode is immersed in a 0.0100 M solution of Zn(NO3)2. Calculate the cell voltage at 25oC What is the overall cell reaction? Diagram the galvanic cell

  2. Ion-Selective Electrodes pH or concentration of ions can be measured by using an electrode that responds selectively to only one species of ion. In a pH meter, one electrode is sensitive to the H3O+(aq) concentration, and the other electrode serves as a reference. A calomel electrode has a reduction half reaction Hg2Cl2 (s) + 2 e- -> 2 Hg(l) + 2 Cl- (aq) Eo = +0.27 V When combined with the H+(aq)/H2(g) electrode, the overall cell reaction is: Hg2Cl2 (s) + H2 (g) -> 2 H+ (aq) + 2 Hg(l) + 2 Cl- (aq)

  3. Q = [H+(aq)]2 [Cl- (aq)]2 / PH2 If PH2 is held at 1 atm then Q = [H+(aq)]2 [Cl- (aq)]2 DE = DEo - (RT/ n F ) ln [H+(aq)]2 [Cl- (aq)]2 The [Cl- (aq)] is held constant since the calomel electrode consists of a saturated solution of KCl. DE depends only on [H+(aq)]. Other electrodes are selectively sensitive to ions such as Ca2+, NH4+, Na+, S2-.

  4. A galvanic cell is constructed of a Cu electrode dipped into a 1 M solution of Cu2+ and a hydrogen gas electrode (PH2 = 1 atm) in a solution of unknown pH. The potential measured is 0.573 V at 25oC. What is the pH of the solution? Cu2+(aq) + 2e- Cu(s) Eo = 0.337 V 2 H+(aq) + 2e- H2 (g) Eo = 0 V Cell reaction Cu2+(aq) + H2 Cu(s) + 2 H+(aq) DEo = 0.337 V DE = DEo - (R T / n F) ln Q DE = DEo - (0.025693 / n) ln Q = DEo - (0.025693 / n) 2.303 log Q = DEo - (0.0.05916 / n) log Q

  5. Q = [H+(aq) ]2 / ([Cu2+(aq)] PH2 ) 0.573 V = 0.337 V - (0.0591 / 2) log [H+(aq) ]2 / ([Cu2+(aq)] PH2 ) - log [H+]2 = - 2 log [H+] = 2(0.573 - 0.337)/0.0591 - log [H+] = 4.00 = pH

  6. Equilibrium Constants DGro = - n FDEo DGro = - R T ln K ln K = (n FDEo)/ (R T) Determine the equilibrium constant at 25oC for AgCl(s) -> Ag+(aq) + Cl- (aq) (not a redox reaction itself) Write this reaction in terms of two half reactions; one for oxidation and one for reduction (1) AgCl(s) + e- Ag(s) + Cl-(aq) Eo = +0.22 V (2) Ag+(aq) + e- Ag(s) Eo = +0.80 V To get the desired overall reaction, reverse (2) and add to (1)

  7. AgCl(s)  Ag+(aq) + Cl-(aq) DEo = 0.22 V - 0.80 V = - 0.58 V ln K = (n F DEo)/ (R T) n = 1; DEo = -0.58 V K = 1.6 x 10-10

  8. Corrosion

  9. 2 H2O(l ) + 2 e- H2 (g) + 2 OH- (aq) Eo = - 0.83 V The standard reduction potential is when [OH-(aq)] = 1 M; a pH of 14. At pH = 7, the reduction cell potential is E ≈ - 0.42 V Any metal with a standard reduction potential more negative than - 0.42 V can reduce water at pH = 7; at pH = 7 the metal will be oxidized by water. Fe2+(aq) + 2e+ Fe(s) Eo = - 0.44 V Fe has a very slight tendency to be oxidized by water at pH = 7 Fe(s)  Fe2+(aq) + 2e+

  10. When O2 is dissolved in water, the following half reaction is important: O2 (g) + 4 H+(aq) + 4 e- 2 H2O(l) Eo = + 1.23 V At pH = 7, E ≈ + 0.82 V for this reaction; greater than Fe2+(aq) + 2e+ Fe(s) Eo = - 0.44 V Fe(s) is readily oxidized to Fe2+(aq) in O2 containing water

  11. Rust: Fe2O3. H2O(s)

  12. To prevent corrosion: protect surface from exposure to O2 and H2O Galvanize metal by coating with an unbroken film of Zn. Electro-deposit Zn on metal; Zn lies below Fe and so is more readily oxidized than Fe. Another way is to use a “sacrificial” anode Magnesium is oxidized preferentially, supplying electrons to the iron for the reduction of O2.

  13. Mg2+(aq) + 2e- Mg(s) Eo = -2.36 V Fe2+(aq) + 2e- Fe (s) Eo = -0.44 V The spontaneous cell reaction: Mg(s) + Fe2+(aq)  Fe (s) + Mg2+(aq)

  14. Metabolism Metabolism – process by which living systems acquire and use free energy to carry out vital processes Oxidation-reduction reactions are responsible (directly or indirectly) for all work done in living organisms “Biological circuitry” - spontaneous, enzyme-catalyzed, redox reactions to do biological work break down of complex molecules building complex molecules

  15. A number of steps in metabolism use NAD+ (nicotinamide adenine dinucleotide) as an electron acceptor or oxidizing agent Reduced metabolite + NAD+  oxidized metabolite + NADH + H+ The reduced NADH is then oxidized by oxygen to convert it back to NAD+ NADH + 1/2O2 + H+ H2O + NAD+DEo = 1.135 V

  16. In the reduction of pyruvate by NADH: Pyruvate + NADH + H+ lactate + NAD+ The reduction potentials for the half-reactions are Pyruvate + 2H+ + 2e- lactateE° = -.185 V NAD+ + H+ + 2e- NADHE° = -.315 V For the overall reaction involving the two half-reactions:DE° =E°(cathode) –E°(anode) So for the reduction of pyruvate by NADH: DE° = -.185 V – (-.315 V) = +.13 V

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