3 (5 pts). The solubility product reaction for AgBr is AgBAg Ag+e Ag (s) Eo -0.799...
Consider the following information at 25 ºC: AgBr(s) + e- → Ag(s) + Br-(aq) Eº(AgBr(s),Ag(s)) = 0.07133 V Ag+(aq) + e- → Ag(s) Eº(Ag+(aq),Ag(s)) = 0.7996 V What is the value of the solubility product equilibrium constant Ksp(AgBr) of silver bromide? AgBr(s) ↔ Ag+(aq) + Br-(aq) Ksp(AgBr)
Question 5 Calculate Eº cell for the reaction below, 2 AgBr(aq) + 2 Hg(1) 2 Ag(s) + Hg2Br2(5) given the following standard reduction potentials. Hg2Br2(s) + 2 - 2 Hg(1) + 2 Br"(aq) E = +0.140 V AgBr(aq) + e - Ag(s) + Br" (aq) E = +0.071 V O a. -0.211 V O b. +0.069 V O C. +1.97 v O d. -0.069 V O e. +0.211 V Question 6 Given the following substances and their vapor pressures at...
the solubility product of AgBr(s) is 1.48 ×10^-13. what is the concentration of Ag+ in a saturated solution of AgBr?
Item 10: Question (10 pts.) Given the following reactions, AgBr(s) = Ag+(aq) + Br-(aq) Kap = 5.4E-13 Ag+ (aq) + 2 CN-(aq) — Kp = 1.2E+21 Ag(CN), (aq) determine the equilibrium constant for the reaction below. AgBr(s) + 2 CN- (aq) = Ag(CN)2 (aq) + Br(aq) Submit Submit and Next O: Mark this question for later review. - Skip to Previous
Question 22 (5 points) Given the following half reactions: - AgBr(s) + e- Ag+ (aq) + e- Ag(s) + Br- (aq) E° = 0.07V → Ag (s) E° = 0.80V - Calculate the Ksp for PbSO4 at 25 degrees C. Please enter your answer in scientific notation, such as 1.2E10 or 1.2E-10 and keep two significant figures.
If the following half-reactions are used in an electrolytic cell: Ag+(aq) + e− ⟶ Ag(s) Eo = 0.80 V Ca2+(aq) + 2 e− ⟶ Ca(s) Eo = −2.76 V Click here for a copy of Final Exam cover sheet. 1) The metal solid that is plated out at the cathode is [Ag(s), Ca(s)] 2) It would take [3, 8. 16] hours to deposit 60 g of the solid with a current of 5.0 A.
Question 52 1 pts AgBr has very low solubility in water. Silver ions form a complex ion with thiosulfate, Ag*(aq) + 25,032(aq) =(Ag(S203)213(aq) If sodium thiosulfate is added to a saturated solution of AgBr in equilibrium with solid AgBr, how will the concentrations of free Ag (aq) and Br"(aq) change? What is the equilibrium expression for the reaction below? Fe2O3(s) + 3CO(g) = 2Fe(s) + 3CO2(g) K - [Fe] [co, (Fe,0][co] K - [Fe20][co [Fe] [co,]
Given the following reactions, AgBr(s) = Ag+ (aq) + Br(aq) Ksp = 5.8 x 10-13 Ag+ (aq) + 2 CN"(aq) = Ag(CN)2(aq) Kf = 9.5 x 1021 determine the equilibrium constant for the reaction below. AgBr(s) + 2 CN"(aq) =Ag(CN)2(aq) + Br(aq) Answer: 1.2E21 Check
Consider the following Gibbs energies at 25 "C Substance Ag (aq) Cr(aq) AgCI(s) Br(aq) AgBr(s) 77.1 - 131.2 - 109.8 - 104.0 -96.9 (a) Calculate AG rn for the dissolution of AgCl(s). (b) Calculate the solubility-product constant of AgCl Number Number kJ mol (c) Calculate Δ3rxn for the dissolution of AgBr(s). (d) Calculate the solubility-product constant of AgBr Number Number kJ mol
Consider the two reduction half-reactions: Br2(l) + 2 e− ⟶ 2 Br−(aq) Eo = 1.09 V Ag+(aq) + e− ⟶ Ag(s) Eo = 0.337 V Use the electrode potentials above to calculate Eocell and ∆Gorxn for the reaction below, and determine if it is the reaction for a voltaic cell or an electrolytic cell.