(8) Calculate 46 for the ions in solution in Problem # 7 above. a) - 61...

  
 

Transcribed Image Text:

(8) Calculate 46° for the ions in solution in Problem # 7 above. a) - 61 kJ/md 6) -71 kJ/mol c) - 81 kJ/mol Hint a) Here are of ions @ 25°C a half reactions for 3+ Fe (ag) Cet te → Ce E cell 14 2x10 = K 2+ te Fe First calculate. 24 b) 2x10 = K a solution. 0.77v E cathole = 1.61V Then, calculate K. 34 anode 34 e) 2x10 = = K (8) Calculate 46° for the ions in solution in Problem # 7 above. a) - 61 kJ/md 6) -71 kJ/mol c) - 81 kJ/mol (8) Calculate 46° for the ions in solution in Problem # 7 above. a) - 61 kJ/md 6) -71 kJ/mol c) - 81 kJ/mol Hint a) Here are of ions @ 25°C a half reactions for 3+ Fe (ag) Cet te → Ce E cell 14 2x10 = K 2+ te Fe First calculate. 24 b) 2x10 = K a solution. 0.77v E cathole = 1.61V Then, calculate K. 34 anode 34 e) 2x10 = = K (8) Calculate 46° for the ions in solution in Problem # 7 above. a) - 61 kJ/md 6) -71 kJ/mol c) - 81 kJ/mol

Answer rating: 100% (QA)

714 Fe 0 3 Ce t C Cell reaction Ecell 2 t Fe aq Eanode 077v Ecathode 16... View the full answer