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

Ion-Selective ElectrodespH 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)

Q = [H+(aq)]2 [Cl- (aq)]2 / PH2

If PH2 is held at 1 atm then Q = [H+(aq)]2 [Cl- (aq)]2

E = Eo - (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 Ca2+, NH4

+, Na+, S2-.

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) Eo = 0.337 V

E = Eo - (R T / n F) ln Q

E = Eo - (0.025693 / n) ln Q

= Eo - (0.025693 / n) 2.303 log Q = Eo - (0.0.05916 / n) log Q

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

Equilibrium Constants

Gro = - n F Eo

Gro = - R T ln K

ln K = (n FEo)/ (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)

AgCl(s) Ag+(aq) + Cl-(aq) Eo = 0.22 V - 0.80 V = - 0.58 V

ln K = (n F Eo)/ (R T)

n = 1; Eo = -0.58 V

K = 1.6 x 10-10

Corrosion

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+

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

Rust: Fe2O3. H2O(s)

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.

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)

MetabolismMetabolism – 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 + 1/2O2 + H+ H2O + NAD+ Eo = 1.135 V

In the reduction of pyruvate by NADH:Pyruvate + NADH + H+ lactate + NAD+

The reduction potentials for the half-reactions arePyruvate + 2H+ + 2e- lactate E° = -.185 VNAD+ + H+ + 2e- NADH E° = -.315 V For the overall reaction involving the two half-reactions: E°

= E°(cathode) – E°(anode)

So for the reduction of pyruvate by NADH:E° = -.185 V – (-.315 V) = +.13 V