Exercise 18.67: Problems by Topic - Cell Potential, Free Energy, and the Equilibrium Constant Calculate the equilibrium constant for each of the reactions at 25 ∘C. Part A Pb2+(aq)+Mg(s)→Pb(s)+Mg2+(aq) Express your answer using one significant figure. K = 5×1075 SubmitMy AnswersGive Up All attempts used; correct answer displayed Part B Br2(l)+2Cl−(aq)→2Br−(aq)+Cl2(g) Express your answer using one significant figure. K = 6•10−10 SubmitMy AnswersGive Up Incorrect; Try Again; 5 attempts remaining Part C MnO2(s)+4H+(aq)+Cu(s)→Mn2+(aq)+2H2O(l)+Cu2+(aq) Express your answer using one significant figure. K = SubmitMy AnswersGive Up
part A
Pb2+(aq)+Mg(s)→Pb(s)+Mg2+(aq)
E0cell = E0cathode - E0anode
= (-0.126)-(-2.372)
= 2.246 v
DG0 = - nFE0cell
= -2*96500*2.246
= -433.478 kjoule
DG0 = - RTlnK
-433478 = -8.314*298lnk
k = 9.65*10^75
part B
Br2(l)+2Cl-(aq) ----> 2Br-(aq)+Cl2(g)
E0cell = E0cathode - E0anode
= (1.0873)-(1.36)
= -0.2727 v
DG0 = - nFE0cell
= -2*96500*-0.2727
= 52631.1 joule
DG0 = -RTlnK
52631.1 = -8.314*298lnk
k = 5.95*10^-10
part C
MnO2(s)+4H+(aq)+Cu(s) ----> Mn2+(aq)+2H2O(l)+Cu2+(aq)
E0cell = E0cathode - E0anode
= (0.337)-(0.95)
= -0.613 v
DG0 = - nFE0cell
= -2*96500*-0.613
= 118309 joule
DG0 = -RTlnK
118309 = -8.314*298lnk
K = 1.83*10^-21
Exercise 18.67: Problems by Topic - Cell Potential, Free Energy, and the Equilibrium Constant Calculate the...
Calculate the equilibrium constant for each of the reactions at 25 ∘C. Standard Electrode Potentials at 25 ∘C Reduction Half-Reaction E∘(V) Pb2+(aq)+2e− →Pb(s) -0.13 Zn2+(aq)+2e− →Zn(s) -0.76 Br2(l)+2e− →2Br−(aq) 1.09 Cl2(g)+2e− →2Cl−(aq) 1.36 MnO2(s)+4H+(aq)+2e− →Mn2+(aq)+2H2O(l) 1.21 Pb2+(aq)+2e− →Pb(s) -0.13 Br2(l)+2Cl−(aq)→2Br−(aq)+Cl2(g) Express your answer using two significant figures.
calculate equilibrium constant and free energy change please 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;...
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Exercise 14.54 with feedback For the following reaction, Kc = 255 at 1000 K. CO (g) + Cl2 (g) ⇌ COCl2 (g) A reaction mixture initially contains a COconcentration of 0.1550 M and a Cl2concentration of 0.171 M at 1000 K. Part A What is the equilibrium concentration of CO at 1000 K? Express your answer in molarity to three significant figures. [CO] = M SubmitMy AnswersGive Up Part B What is the equilibrium concentration of Cl2 at 1000 K?...
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
use the appropriate standard reduction potentials in the appendix of your book to determine the equilibrium constant at work for the following reaction. 4H+(aq) +MnO2(s) +2Fe2+(aq)-->Mn2+(aq)+2Fe3+(aq)+2H2O(l) find K.
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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...