Cells use a concentration gradient of various ions to assist in normal biological events. Determine the Ecell if the intracellular concentration of K+ is 140 mM and the extracellular concentration of K+ is 4.0 mM. You may assume a temperature of 25°C. K+(aq) + e →K(s) E° = -2.92 V 3.01 V 0.091 V -0.086 V 0.086 V -0.091 V

Cells use a concentration gradient of various ions to assist in normal biological events. Determine the Ecell if the intracellular concentration of K+ is 140 mM and the extracellular concentration of K+ is 4.0 mM. You may assume a temperature of 25°C. K+(aq) + e →K(s) E° = -2.92 V 3.01 V 0.091 V -0.086 V 0.086 V -0.091 V

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

Use the data in Table 16.1 (or Appendix L) to calculate the equilibrium constant K at 25 °C for the following reaction. Mg(s) + Ni2+ (aq)→ Mg²+ (aq) + Ni(s) 0.0592 V log K E = n A
For the cell: Cu | Cu^2+ || Ag+ | Ag. What is the Standard Gibbs free energy? Choose the correct answer. A. 4.45x10^4 J B. -8.9x10^4 J C. 4.45x10^4 J D. 8.9x10^4 J
In this Chemical equation 2S2O32- (aq) + I2 (aq) —> S4O62- (aq) + 2I- (aq) with an ON of +2 -2 0 +2.5-2 -1 respectively, is this a redox or not?why?
If 50 kJ of energy are absorbed when 25 g of Ca(OH) 2 are dissolved in water. How much energy would be absorbed with 1.5 mol of Ca(OH)2 is dissolved in water? 3 kJ 222.3 kJ 75 kJ 3000 kJ 111.15 kJ
What is the value of K for this aqueous reaction at 298 K? A + B 2C+D AG° = 11.65 kJ/mol K =
For the following redox reaction calculate the Ecell and K at 298 K. 2Na+(aq) + Ca(s)→2Na(s) + Ca2+(aq) Half Reaction E(V) Na+(aq) + e- → Na(s) -2.71 Ca2+(aq) + 2e- → Ca(s) -2.87
What is the value of K for this aqueous reaction at 298 K? A + B 2C+ D AG° = 19.27 kJ/mol K =
Calculate the normality of a solution of 0.45 g/L H2C2O4, both as an acid and as a reducing agent Note H2HC2O4 M.Wt= 90 g/mol N= 0.001 meq/ml O N= 0.0001 meq/ml O N= 0.01 meq/ml O N= 0.1 meq/ml O
Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2+ Cu²+ (aq) + 21 (aq) → Cu(s) + I₂ (s) From the table of standard reduction potentials: Cu²+/Cu 0.337 V, E₁₂/1 Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. Equilibrium constant at 298 K:I AGO for this reaction would be than zero. = = 0.535 V
What is the value of K for this aqueous reaction at 298 K? A +B C+D AG° 19.76 kJ/mol K =
Use the reactions and AG° values below to determine AG° for: 3 CH4(g) + 4 O2(g) → C302(g) + 6 H20(g) For CH4(g) + 2 O2(g) → CO2(g) + 2 H20(g) For 3 CO2(g) → C302(g) + 2 O2(g) AG° = -800.8 kJ/mol AG° = +1073.4 kJ/mol %3D 272.6 kJ/mol 817.8 kJ/mol None of the answers are correct -345.8 kJ/mol -1329.0 kJ/mol
Calculate AG° and K at 25°C for the reactions. 7H2O + 2 Cr+ + 3 Cl2 = Cr2 07²- + 6 Cl¯ + 14H = 0.03 V cel a. AG° = kJ K = Cu²+ (aq) + Ba(s) = Ba2+ (aq) + Cu(s) e° = 3.24 V b. AG° : kJ K = Cl2 (g) + 2 Br (aq) = Br2 (g) + 2 Cl¯ (ag) e° = 0.27 V C. AG° = kJ K = 3H2 O(1) + 5 IO4 (ag) + 2 Mn²+ (aq) = 5 IO3 (aq) + 2 MnO4¯(ag) + 6H† (ag) d. 0.09 V ´cell AG° = kJ K =