27. Use the standard reduction potenital to calculate equilibrium constants for the following reaction. CdS(s) →...
28. Use the standard reduction potenital to calculate equilibrium constants for the following reaction. H2O (1) H (aq) +OH (aq) at 298.15 K H2 (g)2H (aq) +2e 2H2O (aq) +2e - H2+ 20H (aq) A.K=1.0 x10 B. K=1.6 x10-1l C.K=1.0 x10-14 0 Define electrolvsis. E= 0V E -0.8277 V -10 spontaneous nonspontaneous nonspontaneous
19.03 pts Question 10 Use the standard reduction potenital to calculate equilibrium constants for the following reaction at 298.15 K. AgCl (s) Ag+ (aq) +Cl- (aq) Ag+ (aq) +Ag(s) E° = 0.7996 V AgCl(s) + e - Ag(s) + CH- (aq) E° = 0.22233 V O K=5.2 x 105 spontaneous O K = 3.8 x 10"nonspontaneous K - 1.7 x 10-10 nonspontaneous
Question 10 3.03 pts Use the standard reduction potenital to calculate equilibrium constants for the following reaction at 298.15 K. AgCl(s) Ag+ (aq) +CH(aq) Ag+ (aq) +eAg(s) E° =0.7996 VAgCl(s) + e - Ag(s) + CH(aq) E° = 0.22233 V OK-5.2 x 10% spontaneous OK=3.8 x 10nonspontaneous N O K=1.7 x 10-10 nonspontaneous
Given the following standard reduction potentials choose the cell which will work as a voltaic cell. All cells below are written according to the usual cell diagram convention. Cu2+(aq) + 2e → Cu(s) E° = 0.34 V 2H+(aq) + 2e → H2(g) E° = 0.00 V Sn2+ (aq) + 2e → Sn(s) E° = -0.14 V Ni2+(aq) + 2e → Ni(s) E° = -0.26 V Cd2+(aq) + 2e → → Cd(s) E° = -0.40 V Sn(s) | Sn2+(aq) || Ni2+(aq)...
Use the standard half-cell potentials listed below to calculate the standard free energy(K]for the following reaction occurring in an electrochemical cell at 25°C. Pb 2+ (aq) +2e--- Pb(s) E* - -0.13 Volt A13+ (aq) + 3 e-Al(s) E* =-1.66 volt a. 1.53 b. - 886 c-434 d. - 443 e. -1036 What is the standard free energy Gº) in Kilojoules for the reaction below at 298 Kelvin: Farady's constant = 96,485 joules/V. mole e Zn2+ (aq) + 2e ......> Zn(s)...
Consider the half‑reaction for the reduction of Cd 2+ Cd2+ to Cd(s) Cd(s) . Cd 2 +(aq)+2 e − ⟶Cd(s) ? ∘ Cd 2+ /Cd =−0.403 V Cd2+(aq)+2e−⟶Cd(s)ECd2+/Cd∘=−0.403 V Calculate the potential of the cadmium electrode at 25 ∘ C 25 ∘C when immersed in a 0.0560 M 0.0560 M solution of CdBr 2 . CdBr2. ? Cd = ECd= V V Calculate the potential of the cadmium electrode at 25 ∘ C 25 ∘C when immersed in a 0.0330...
A) Use tabulated electrode potentials to calculate ΔG∘ for the reaction. 2K(s)+2H2O(l)→H2(g)+2OH−(aq)+2K+(aq) B) (Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V) An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:Ni(s)VO2+(aq,0.024M)+2H+(aq,1.4M)+e−→→Ni2+(aq,1.8M)+2e−VO2+(aq,1.8M)+H2O(l) Calculate the cell potential under these nonstandard concentrations. C) Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V ) Half-reaction E∘ (V ) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) − 0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) − 0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) − 0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) − 0.76...
Calculate E°cell for the following reaction:2Fe2+(aq) + Cd2+(aq) ? 2Fe3+(aq) + Cd(s)A. -0.37 VB. 0.37 VC. -1.17 VD. 1.17 VE. none of these
use tabulated standard electrode potential to calculate the standard cell potential for the reaction occurring in an electrochemical cell at 25 C. (The equation is balanced.) 3Ni^2+(aq)+2Cr(s)--->3Ni(s)2Cr^3+(aq) Express your answer to two significant figures and include the appropriate units. em 26 E (V) -0.45 -0.50 -0.73 -0.76 -1.18 Standard reduction half-cell potentials at 25°C Half-reaction E° (V) Half-reaction Aul+ (aq) + 3e +Au(s) 1.50 Fe2+ (aq) + 2eFe(s) Ag+ (aq) +e-Ag(s) 0.80 Cr3+ (aq) + Cr²+ (aq) Fe+(aq) + 3e...
Use the data in Appendix L to determine the equilibrium constant for the following reactions. Assume 298.15 K if no temperature is given. (a) AgCl(s) ⇌ Ag+(aq) + Cl−(aq) (b) CdS(s) ⇌ Cd2+(aq) + S2−(aq) at 377 K (c) Hg2+(aq) + 4Br−(aq) ⇌ [HgBr4 ] 2−(aq) (d) H2 O(l) ⇌ H+(aq) + OH−(aq) at 25 °C