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 M in both MnO 4 - (aq) and Mn 2+ (aq). In the second, a zinc electrode is immersed in a M solution of Zn(NO 3 ) 2. Calculate the cell voltage at 25 o C What is the overall cell reaction? Diagram the galvanic cell

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 H 3 O + (aq) concentration, and the other electrode serves as a reference. A calomel electrode has a reduction half reaction Hg 2 Cl 2 (s) + 2 e - -> 2 Hg(l) + 2 Cl - (aq)E o = V When combined with the H + (aq)/H 2 (g) electrode, the overall cell reaction is: Hg 2 Cl 2 (s) + H 2 (g) -> 2 H + (aq) + 2 Hg(l) + 2 Cl - (aq)

Q = [H + (aq)] 2 [Cl - (aq)] 2 / P H 2 If P H 2 is held at 1 atm then Q = [H + (aq)] 2 [Cl - (aq)] 2  E =  E o - (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.  E depends only on [H + (aq)]. Other electrodes are selectively sensitive to ions such as Ca 2+, NH 4 +, Na +, S 2-.

A galvanic cell is constructed of a Cu electrode dipped into a 1 M solution of Cu 2+ and a hydrogen gas electrode (P H2 = 1 atm) in a solution of unknown pH. The potential measured is V at 25 o C. What is the pH of the solution? Cu 2+ (aq) + 2e -  Cu(s)E o = V 2 H + (aq) + 2e -  H 2 (g) E o = 0 V Cell reaction Cu 2+ (aq) + H 2  Cu(s) + 2 H + (aq)  E o = V  E =  E o - (R T / n F) ln Q  E =  E o - ( / n) ln Q =  E o - ( / n) log Q =  E o - ( / n) log Q

Q = [H + (aq) ] 2 / ([Cu 2+ (aq)] P H2 ) V = V - ( / 2) log [H + (aq) ] 2 / ([Cu 2+ (aq)] P H2 ) - log [H + ] 2 = - 2 log [H + ] = 2( )/ log [H + ] = 4.00 = pH

Equilibrium Constants  G r o = - n F  E o  G r o = - R T ln K ln K = (n F  E o )/ (R T) Determine the equilibrium constant at 25 o C 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)E o = V (2) Ag + (aq) + e -  Ag(s)E o = V To get the desired overall reaction, reverse (2) and add to (1)

AgCl(s)  Ag + (aq) + Cl - (aq)  E o = 0.22 V V = V ln K = (n F  E o )/ (R T) n = 1;  E o = V K = 1.6 x

Corrosion

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

When O 2 is dissolved in water, the following half reaction is important: O 2 (g) + 4 H + (aq) + 4 e -  2 H 2 O(l) E o = V At pH = 7, E ≈ V for this reaction; greater than Fe 2+ (aq) + 2e +  Fe(s) E o = V Fe(s) is readily oxidized to Fe 2+ (aq) in O 2 containing water

Rust: Fe 2 O 3. H 2 O(s)

To prevent corrosion: protect surface from exposure to O 2 and H 2 O 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 O 2.

Mg 2+ (aq) + 2e -  Mg(s)E o = V Fe 2+ (aq) + 2e -  Fe (s)E o = V The spontaneous cell reaction: Mg(s) + Fe 2+ (aq)  Fe (s) + Mg 2+ (aq)

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

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 + 1/2O 2 + H +  H 2 O + NAD +  E o = V

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