CEM334: ELECTROCHEMISTRY Tutorial / Tutoriaal Summary of selected formulas from CEM334 electrochemistry. / Opsomming van geselekteerde formules van CEM334 elektrochemie
Ecell = Ereduction potential of reduction halfreaction – Ereduction potential of oxidation halfreaction
RT a (products) RT ln = E 0 (T ) − ln Q nF a (reactants) nF

E (T ) = E 0 (T ) −

∆G = -nFEcell
E 0 cell = RT ln K a nF  ∂∆G   ∂E  ∆S = −  = nF    ∂T  P  ∂T  P

 E − E1  ∆S = nF  2  T −T   1 P  2
  E − E1     − E1  ∆H (T1 ) = ∆G + T1∆S = nF T1  2    T2 − T1  P   

  ∂E   ∆H = ∆G + T∆S = nF T   − E   ∂T  P 
NOTE from above: R = gas constant, 8.314 J mol-1 K-1 F = Faraday constant, 96485 C ekw-1

 ∂E  ∆G = ∆H − T∆S = ∆H − nFT    ∂T  P

YOU MUST KNOW THEM AND HOW TO USE THEM – NO FORMULA SHEET WILL BE PROVIDED IN TESTS OR EXAMS! / U MOET HULLE KEN EN KAN GEBRUIK. GEEN FORMULEBLAD WORD IN TOETSE OF EKSAMENS VOORSIEN NIE!

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Selected problems on Electrochemistry
1) Nernst equation
Determine the following cell potential of the cell at 25°C (298K): CuCu2+(2.00 M) Ag+(0.500 M) Ag From the table of standard electrode potentials at 25°C

Cathode (Reduction) Half-Reaction Cu2+(aq) + 2e- -> Cu(s) Ag+(aq) + e- -> Ag(s)

Standard Potential E° (volts) 0.337 0.7991

ANSWER:
Cell reaction: Cu(s) + 2 Ag+(0.500 M) → Cu2+(2.00 M) + 2Ag(s)

E0cell(298K) = E0reduction(298K) - E0oxidation(298K) = 0.799 V - 0.337 V = +0.46 V (standard cell potential) 2+ (8.314J / mol / K )(298K ) ln 2.00 RT Cu E ( 298K ) = E 0 ( 298K ) − ln = 0.46V − + 2 nF 2(96485C / mol ) 0.500 2 Ag

= 0.46V − 0.027V = 0.43V

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2) Equilibrium Constant from Cell Potential
Give the cell notation, cathode and anode reactions of ½ Sn2+ + Tl → ½ Sn + Tl+ Determine the cell potential and Ka for the above reaction at T = 25°C. Given: From the table of