Question is solved using concepts of cell potential.
znt" + 2e nt 4. Calculate the cell potential, the equilibrium constant, and the free energy...
Calculate the cell potential, the equilibrium constant, and the free energy change for Ca(s) + Mn2+ (aq)(1M)=Ca²+ (aq)(1M) + Mn(s) given the following E' values: Ca2+ (aq) + 2e →Ca(8) Eº = -1.50 V Mn2+ (aq) + 2e →Mn(s) E° = -0.58 V
calculate equilibrium constant and free energy change
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Calculate the cell potential, the equilibrium constant, and the free-energy change for Ca(s) + M12+ (aq)(1M) = Cap+ (aq)(1M) + Mn(3) given the following E' values: Ca2+ (aq) + 2e + Ca(8) E° = -1.50 V Mn2+ (aq) + 2e - →Mn(s) E° = -0.58 V Part A Calculate the cell potential 0.920 V Previou Correct Part B Calculate the equilibrium constant I ALCO ? Submit Previous Answers Request Answer X Incorrect;...
Calculate the cell potential, the equilibrium constant, and the free-energy change for: Ca(s)+Mn2+(aq)(1M)⇌Ca2+(aq)(1M)+Mn(s) given the following Eo values: Ca2+(aq)+2e−→Ca(s) Eo = -1.59 V Mn2+(aq)+2e−→Mn(s) Eo = -0.54 V 1.) Calculate the equilibrium constant. 2.) Free-energy change?
Hi, if someone can explain how to do this for me, I'd appreciate it! Calculate the cell potential, the equilibrium constant, and the free-energy change for: Ca(s)+Mn2+(aq)(1M)⇌Ca2+(aq)(1M)+Mn(s) given the following Eo values: Ca2+(aq)+2e−→Ca(s) Eo = -1.57 V Mn2+(aq)+2e−→Mn(s) Eo = -0.63 V
Calculate the cell potential for Ca(s) + Mn2+(aq) ↔ Ca2+(aq) + Mn(s). Assume any aqueous species has a concentration of 1 M. Ca2+(aq) + 2e- → Ca(s) Eº = -2.87 V Mn2+(aq) + 2e- → Mn(s) Eº = -1.18 V
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Calculate the change in free energy (AG) for the following standard cell between Barium and Copper. Recall AG = -nFEcell, and F = 9.65 x 10 C/mol Ba Ba2 (aq) + 2e, E red -2.90 V 2e+Cu2+ (aq) Cus), Ered = +0.337 V +625 kJ/mol +495 kJ/mol -625 kJ/mol 495 kJ/mol
Free-energy change, AGº, is related to cell potential, Eº, by the equation AG° = -nFE° where n is the number of moles of electrons transferred and F = 96,500 C/(mol e ) is the Faraday constant. When Eº is measured in volts, AGⓇ must be in joules since 1 J =1C.V. Part A Calculate the standard free-energy change at 25°C for the following reaction: Mg(s) + Fe2+ (aq)Mg2+ (aq) + Fe(s) Express your answer to three significant figures and include...
Cell Potential and Equilibrium Standard reduction potentials The equilibrium constant, K, for a redox reaction is related to the standard cell potential, Ecel, by the equation Reduction half-reaction (V) Ag+ (aq) + e-→Ag(s) Cu2+ (aq) + 2e-→Cu(s) 0.34 Sn (a) 4e-Sn(s 0.15 2H' (aq) + 2e-→H2 (g) Ni2+ (aq) + 2e-→Ni(s)-0.26 Fe2+ (aq) + 2e-→Fe(s)-0.45 Zn2+ (aq) + 2e-→Zn(s)-0.76 Al3+ (aq) +3e-→Al(s) -1.66 Mg2+ (aq) + 2e-→Mg(s) -2.37 0.80 n FEcell where n is the number of moles of electrons...
The equilibrium constant, K, for a redox reaction is related to the standard potential, Eº, by the equation In K = nFE° RT where n is the number of moles of electrons transferred, F (the Faraday constant) is equal to 96,500 C/(mol e), R (the gas constant) is equal to 8.314 J/(mol · K), and T is the Kelvin temperature. Standard reduction potentials Reduction half-reaction E° (V) Ag+ (aq) + e +Ag(s) 0.80 Cu²+ (aq) + 2e + Cu(s) 0.34...
Calculate the standard Gibbs free energy changes at 25 °C for each of the reactions shown below using the Eº values given. Select whether each of these reactions is nonspontaneous, at equilibrium, or spontaneous under standard conditions. (a) 2 Lit(aq) + 2 I'(aq) = 2 Li(s) + 1 12(5) ° = -3.586 V AGC= X kJ/mol nonspontaneous at equilibrium spontaneous (b) 2 CO3+ (aq) + 1 Hg(0) = 2 Co2+ (aq) + 1 Hg2+(aq) ° = 0.988 V AGº =...