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finding E° cell at 25 degrees celcius how do i find gibbs free energy value? 2)...
finding E° cell at 25 degrees celcius 3) Zn/Zn e Il Cu 2+ Cu 1,0 M Zn, 1.0 M CU
at 25 degrees celcius E cell Determining 1. Cul Cu2+ 11 Agt (Ag 1.0 M Cu, 1.0 M Ag
To decide whether AgCl should dissolve in water ( spontaneous reaction ) at 25 degrees celcius, the delta H and T delta S must be used in the Gibbs Free Energy equation to calculate delta G AgCl (s) + H2O (I) -> Ag+ + Cl- + H2O Standard Enthalpies of formation and standard entropies of common compounds substance state AgCl s -127010 96.20 Cl- aq -167080. 56.50 Ag+ aq. 105790. 72.70
The free energy change for the following reaction at 25 °C, when [Zn2]- 4.80x103 M and [Cd2+ = 1.17 M, is 83.1 kJ: Zn2 (4.80x103 M)+ Cd(s)Zn(s)+ Ca2 (1.17 M) AG-83.1 kJ What is the cell potential for the reaction as written under these conditions? Answer: Would this reaction be spontaneous in the forward or the reverse direction?
Standard cell potential for Zn Cu2+ cell equal 1.104V Calculate change Gibbs free energy in this cell. Gcell=-n x F x E^0cell We were unable to transcribe this imageWe were unable to transcribe this image
Calculate the cell potential (Ecell) and Gibbs Free Energy (Delta G) for the reaction (2Ag+ (aq) + Fe (s) -> Fe2+ (aq) + 2Ag (s) ) under these conditions: a 250mL beaker of 1.0mol Fe metal electrode and 0.10 moles of Fe 2+; and a second 250mL beaker of 2.0 moles Ag metal electrode ad 0.50 moles of Ag+. E naught of the cell= 1.21 V.
Calculate the Gibbs free energy (in kJ/mol) at 25 degrees C of the following reaction using the data in the table. 2A +3B + 3C - 1D + 5E AH (kJ/mol) A 48.24 B -94.95 C-30.59 D 81.90 E -46.95 sº (J/mol/K) 146.14 188.08 133.51 184.43 19.71
Half-cell Potentials: Half Reaction: E value +0.80 V Agt + e → Ag Fe3+ + € → Fe2+ +0.77 v +0.34 V -0.13 V Cu2+ +2e → Cu Pb2+ + 2e - → Ib Ni2+ + 2e → Ni Cd2+ +2e → Cd -0.25 V -0.40 V Fe2+ + 2e → Fe -0.44 V Zn2+ + 2e → Zn -0.76 V Al3+ +3e → AI - 1.66 V Consider an electrochemical cell constructed from the following half cells, linked by...
how do i do nerst equation 1. Standard Cells Voltaie Cell Data Sheet Experimental Cell Potential, volts Calculated Cell Potential volts =.47 v PbPb (1.0 MIC"(1.0 MC SnSn" (1.0 M) "(1.0 M) = 485 ,465 540 .44v 08v CujCư"(1.0 M)||Ag (1,0 M)/AS - ,460 Sn/Sn?"(1.0 M)||Pb (1.0M)/Pb Olv Pb/Pb2+(1.0 M)||Ag" (1.0 MAS 93 Sn/Sn? (1.0 M)||Ag (1.0 M)/Ag 19/ -99 = 94v 80 II. Nonstandard Cells Voltaic Cell Experimental Cell Potential, volts Calculated Potential, Nernst Equation, volts* Pb/Pb*P(1.0 M)||Cu?(0.0010 M)|Cu...
When doing a Nernst equation, how do you calculate n? I know for something like Zn|Zn2+(1.0 M)||Cu2+(1.0 M)|Cu where you have 2+ and 2+ n would be 2, but what about something like Pb|Pb2+(1.0 M)||Ag+(0.10 M)|Ag where you have 2+ and + I'm confused and can't find an answer anywhere