1. Classical Mechanics (energy, position, momentum (mv), time?) U = U tr + U vib + U rot + U el + U int + U nuc Macroscopic — E = K + V + U  U = q + w Microscopic — F = ma – Fails for small particles moving at high speeds Quantum Mechanics — A theory based on wave/particle duality used to treat the energy, and motion of particles. Effective for all particles, but required for the small/quick. More difficult computationally (which is why CM is still taught).

2. Chapter 8 – Electrochemistry and ionic solutions U = U tr + U vib + U rot + U el + U int + U nuc Microscopic — Quantum Mechanics The energy of a single particle can still be thought of as the sum of its kinetic and potential energy, E = K + V. The treatment of kinetic energy is mathematically distinct from CM, whereas the treatment of potential energy uses the same equations as CM. Coulomb’s Law – governs forces between charges Debye-Hückel Theory – predicts nonideal behavior of ions in solution

3. Coulomb’s Law – defining the force between two charges, and the potential energy of a single charged particle. Electric potential of a charge q 1  = q 2 /(4  o r) (V or J/C) Force between two charges in vacuum – F = q 1 q 2 /(4  o r 2 ) N (J/m) Charges in solution F = q 1 q 2 /(4  o  r r 2 ) Electric field E = F/q 1 = q 2 /(4  o r 2 ) |E| ≡ -d  /dr 8.1 – 8.3 (assign 8.2 and 8.5) Chapter 8 – Electrochemistry and ionic solutions  o = permittivity of a vacuum (C 2 J -1 m -1 )  r = dielectric constant (unitless) how medium shields charge forces

4. Coulomb’s Law – Force between two charges in vacuum F = q 1 q 2 /(4  o r 2 ) N q 1 = (4  o r 2 ) F/q 2 = (4  8.85 x 10 -12 100 2 ) 0.025/1 = 8.1 Charge on sphere#1 attracted to sphere #2 with q 2 = 1.00 C if r = 100.0 m and F = 0.0225 N? q 1 = 2.50 x 10 -8 C

5. Electric potential of a charge q 1 8.6  = q 2 /(4  o r) V (J/C) Charges in solution F = q + q - /(4  o  r r 2 ) 8.4 Electric field E + = F/q + = q - /(4  o r 2 ) |E| ≡ -d  /dr V/m  r (H2O) r (m)F (N)q+ q- (C) 780.060751.55E-064.96E-17 8.3 Two charged metal spheres in water (  r = 78) with r = 6.075 cm and F = 1.55 x 10 -6 N. q(-) = 2q(+) a) What are q(-) and q(+)? b) What are the electric fields of the two bodies? q + q - = F (4  o  r r 2 ) = 2q + 2 q+q- 4.98E-099.96E-09 E+ (V/m)E- 311156 (assign 8.2 and 8.5)

6. 8.2 Gravity F = G m 1 m 2 /r 2 F =G =m(earth)m(sun)rqnm(e) 3.54E+226.67E-115.97E+241.98E+301.494E+112.97E+173.07E+121.69E+06 kg mCvs. 10 26 8.1 – 8.3 (assign 8.2 and 8.5) 5 F = q 1 q 2 /(4  o r 2 ) r (Å)r (m)e (C)F (N) 0.5295.29E-111.60E-198.24E-08

7. E˚ rx = E˚ red - E˚ ox E = Eº - RT/nF lnQ F = 96,485 C mol -1.  Gº = -nFEº Standard reaction potentials can be determined from a table of standard reduction potentials of half reactions. The Nernst equation defines reaction potentials at non-standard concentrations. Standard reaction potentials can be determined from standard thermodynamic tables (assuming redox reaction).  G = -nFE

8. Ions in solution tend to deviate from ideality at low concentrations Debye-Hückel Law: ln  ± = A z + z - I 1/2 A = 1.171 m -½ (H 2 O) I = Ionic strength: I = ½  i c i z i 2  ± = The averaged activity coefficient of ions in solution The ionic strength measures the impact that ions have in a solution for properties (e.g. Conductivity – salting out proteins in biochemistry – etc.) Problem 33 abcd HClNaHCO3Fe(NO3)2Fe(NO3)3 m0.00550.0750.025 I0.00550.075 0.15 ln  ± -0.087-0.321-0.641-1.361 ±± 0.9170.7260.5270.257 Problem 41 NaClmI ln  ± ±± 0.90%0.154 -0.45950.632

9. Use activities rather than [ ]s in Q when applying Nernst Equation however, each ion has to have a separate . Extended Debye-Hückelln  = - (A·z 2 ·I 1/2 )/(1+B · å · I 1/2 ) B = 2.32 x 10 9 m -3/2 Å represents the ionic diameter