Multiple ChoiceCalculate the equilibrium constant for the following reaction at 25ºC. Fe (s) + I2 (s) → Fe2+ (aq) + 2 I – (aq)Given the following reduction potentials: Fe2+(aq) + 2 e– →. Fe (s) E°red = – 0.45 VI2 (s) + 2 e– →. 2 I – (aq) E°red = + 0.54 V220views1rank
Textbook QuestionA cell has a standard cell potential of +0.177 V at 298 K. What is the value of the equilibrium constant for the reaction(a) if n = 1?97views
Textbook QuestionAt 298 K a cell reaction has a standard cell potential of +0.17 V. The equilibrium constant for the reaction is 5.5 × 105. What is the value of n for the reaction?99views
Textbook QuestionUsing the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at 298 K: (c) 10 Br -1aq2 + 2 MnO4-1aq2 + 16 H+1aq2 ¡ 2 Mn2+1aq2 + 8 H2O1l2 + 5 Br21l21153views
Textbook QuestionA cell has a standard cell potential of +0.177 V at 298 K. What is the value of the equilibrium constant for the reaction (c) if n = 3?813views1rank
Textbook QuestionCalculate the equilibrium constant for the reaction between Fe2+(aq) and Zn(s) (at 25 °C).1676views
Textbook QuestionBeginning with the equations that relate E°, ∆G°, and K, show that ∆G° is negative and K 7 1 for a reaction that has a positive value of E°331views
Textbook QuestionIf a reaction has an equilibrium constant K 6 1, is E° posi-tive or negative? What is the value of K when E° = 0 V?309views
Textbook QuestionFor the following half-reaction, E° = 1.103 V: Calculate the formation constant Kf for Cu(CN)2-.496views