Calculate the equilibrium constant of this redox reaction at 25 degrees celsius. F= 96500 C/molAg+ (aq) + Fe(s) --> Ag(s) + Fe²+(aq)TABLE 18.1 Standard Reduction Potentials in Water at 25°CPotential (V)Reduction Half-ReactionF₂(g) + 2 e2 F (aq)MnO₂ (aq) + 8 H*(aq) + 5eCl₂(g) + 2 e 2 Cl(aq)Cr₂O72 (aq) + 14 H*(aq) + 6 eO₂(g) + 4 H*(aq) + 4 e+2.87+1.51+1.36+1.33+1.23+1.06+0.96+0.80+0.77+0.68+0.59+0.54+0.40+0.340 [defined]-0.28-0.44-0.76-0.83-1.66-2.71-3.05Br₂(1)+2 e2 Br (aq)NO3(aq) + 4 H*(aq) + 3eAg (aq) + eAg(s)Fe³+ (aq) + eFe²+ (aq)O₂(g) + 2H*(aq) + 2eH₂O₂(aq)MnO₂ (aq) + 2 H₂O(l) + 3e MnO₂(s) + 4 OH(aq)1₂(s) + 2 e21 (aq)2 H*(aq) + 2 eNi²+ (aq) + 2 eFe²+ (aq) + 2 e.Zn²+ (aq) + 2 e2 H₂O(l) + 2 eMn²+ (aq) + 4H₂O(1)2 Cr³+ (aq) + 7 H₂O(l)2 H₂O(1)O₂(g) + 2 H₂O(l) + 4 e 4 OH(aq)Cu²+ (aq) + 2eAl³+ (aq) + 3 eNa (aq) + eLit(aq) +eLi(s)- Al(s)Na(s)→→→ NO(g) + 2 H₂O(1)Cu(s)H₂(g)Ni(s)Fe(s)> Zn(s)H₂(g) + 2OH(aq) O 4.37 x 10^418.50 x 10^41O 8.13 x 10^41O 1.50 x 10^41O 1.19 x 10^41O None of the above

Formula used ∆G° = — nFE° —(1) ∆G° = — 2.303 RT log K —(2) from (1) & (2) nFE° = 2.303 RT log K…

Calculate the equilibrium constant of thi s redox reaction at 25 degrees celsius. F = 96500 C/mol Ag+ (aq) + Fe(s) --> Ag(s) + Fe²+(aq) TABLE 18.1 Standard Reduction Potentials in Water at 25°C Potential (V) Reduction Half-Reaction +2.87 F₂(g) + 2 e 2 F (aq) +1.51 MnO₂ (aq) + 8 H*(aq) + 5e Cl₂(g) + 2 e 2 Cl(aq) +1.36 +1.33 Cr₂O72 (aq) + 14 H*(aq) + 6 e +1.23 O₂(g) + 4 H*(aq) + 4 e +1.06 Br₂() +2 e +0.96 +0.80 +0.77 +0.68 +0.59 +0.54 +0.40 +0.34 0 [defined] -0.28 -0.44 -0.76 -0.83 -1.66 -2.71 -3.05 O₂(g) + 2 H₂O(l) Cu²+ (aq) + 2e 2 Br (aq) NO3(aq) + 4 H*(aq) + 3e Ag (aq) + e Ag(s) Fe³+ (aq) + e Fe²+ (aq) O₂(g) + 2H*(aq) + 2e H₂O₂(aq) MnO4 (aq) + 2 H₂O(l) + 3e MnO₂(s) + 4 OH(aq) 1₂(s) + 2 e 21 (aq) + 4 e 4 OH(aq) 2 H*(aq) + 2 e Ni²+ (aq) + 2 e Fe²+ (aq) + 2 e. Mn²+ (aq) + 4H₂O(l) 2 Cr³+ (aq) + 7 H₂O(l) Zn²+ (aq) + 2 e 2 H₂O(l) + 2 e Al³+ (aq) + 3 e Na (aq) + e Lit(aq) +eLi(s) 2 H₂O(1) - Al(s) Na(s) Cu(s) H₂(g) Ni(s) Fe(s) > Zn(s) H₂(g) + 2OH(aq) →→→ NO(g) + 2 H₂O(1)

Calculate the equilibrium constant of thi s redox reaction at 25 degrees celsius. F = 96500 C/mol Ag+ (aq) + Fe(s) --> Ag(s) + Fe²+(aq) TABLE 18.1 Standard Reduction Potentials in Water at 25°C Potential (V) Reduction Half-Reaction +2.87 F₂(g) + 2 e 2 F (aq) +1.51 MnO₂ (aq) + 8 H(aq) + 5e Cl₂(g) + 2 e 2 Cl(aq) +1.36 +1.33 Cr₂O72 (aq) + 14 H(aq) + 6 e +1.23 O₂(g) + 4 H(aq) + 4 e +1.06 Br₂() +2 e +0.96 +0.80 +0.77 +0.68 +0.59 +0.54 +0.40 +0.34 0 [defined] -0.28 -0.44 -0.76 -0.83 -1.66 -2.71 -3.05 O₂(g) + 2 H₂O(l) Cu²+ (aq) + 2e 2 Br (aq) NO3(aq) + 4 H(aq) + 3e Ag (aq) + e Ag(s) Fe³+ (aq) + e Fe²+ (aq) O₂(g) + 2H(aq) + 2e H₂O₂(aq) MnO4 (aq) + 2 H₂O(l) + 3e MnO₂(s) + 4 OH(aq) 1₂(s) + 2 e 21 (aq) + 4 e 4 OH(aq) 2 H(aq) + 2 e Ni²+ (aq) + 2 e Fe²+ (aq) + 2 e. Mn²+ (aq) + 4H₂O(l) 2 Cr³+ (aq) + 7 H₂O(l) Zn²+ (aq) + 2 e 2 H₂O(l) + 2 e Al³+ (aq) + 3 e Na (aq) + e Lit(aq) +eLi(s) 2 H₂O(1) - Al(s) Na(s) Cu(s) H₂(g) Ni(s) Fe(s) > Zn(s) H₂(g) + 2OH(aq) →→→ NO(g) + 2 H₂O(1)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

Calculate the cell potential for the galvanic cell in which the reaction Fe(s) + Au³+(aq) = Fe*(aq)+Au(s) occurs at 25 °C, given that [Fe+] = 0.00200 M and [Au³+] : = 0.765 M. Refer to the table of standard reduction potentials. E = V
= 2.26 A voltaio cell employs the following redox reaction: 2 Fe (aq) +3 Mg (s) 2 Fe (s) +3 Mg (aq) Calculate the cell potential at 25 °C under each of the following conditions. In this solution, Ecel 1). Therefore, the reaction will proceed to the left. Part C [Fe"] = 2.85 M ; [Mg?+] = 2.1x10-3 M Express your answer in units of volts. Ecell= V
Consider this electrochemical reaction: Zn (s) +Ni2+(aq)→ Zn2+(aq)+Ni(s) The standard cell potential Eocell= 0.51V. Calculate the cell potential at 25 oC if the Ni2+ concentration is 5.0 M and the Zn2+ concentration is 0.050 M F=96500 C/mol, R= 8.314 J K-1 mol-1
An electrochemical cell is made by placing an iron electrode in 1.00 L of 0.19 M FeSO4 solution and a copper electrode in 1.00 L of 8.80 × 10-² M CuSO4 solution. Standard Electrode (Reduction) Potentials in Aqueous Solution at 25°C Cathode (Reduction) Half-Reaction ¯(aq) + 2e¯ ⇒ Fe(s) Cu²+ (aq) + 2e¯¯ ≈ Cu(s) Fe²+ Standard Potential, E° (V) -0.41 = 0.34 a. What is the initial voltage of this cell when it is properly constructed? Initial voltage 2+ b. Calculate the final concentration of Cu²+ in this cell if it is allowed to produce an average current of 1.78 amp for 241 s. Final concentration of Cu²+ M
For the following redox reaction calculate the standard cell potential of the reaction. Redox reaction: Cu (s) + 2 Fe3+(aq) → Cu2+(aq) + 2 Fe2+(aq)
PLEASE HELP
Calculate the cell potential for the galvanic cell in which the given reaction occurs at 25 °C, given that [Sn²+] = 0.0526 M, [Fe³+] = 0.0221 M, [Sn4] = 0.00628 M, and [Fe²+] = 0.01068 M. Standard reduction potentials can be found in this table. E = 4 Sn² (aq) + 2 Fe³+(aq) = Sn+ (aq) + 2 Fe²+(aq) V
A galvanic cell at a temperature of 25.0°C is powered by the following redox reaction: 2Cr3+(aq)+3Ca(s)--->2Cr(s)+3Ca2+(aq) Suppose the cell is prepared with 1.96 M Cr3+ in one half-cell and 6.08 M Ca2+ in the other. Calculate the cell voltage under these conditions. Round your answer to 3 significant digits.
Use the standard reduction potentials in Appendix L Links to an external site.to find the standard cell potential in volts for the following galvanic cell.Cu(s) | 1 M Cu(NO3)2(aq) || 1 M AuCl3(aq), | Au(s)
Calculate the cell potential for the galvanic cell in which the given reaction occurs at 25 °C, given that [Sn] = 0.0595 M, [Co1 = 0.0452 M, [Sn+] = 0.00261 M, and [Co*] = 0.00954 M. Standard reduction potentials can be found in %3D %3D this table. Sn (aq) + 2 Co+(aq) Sn+(aq) +2 Co*(aq) V E =
Consider the following balanced equation for a redox reaction in an electrochemical cell: 3 Cd (s) + 2 NO3 (aq) + 8 H* (aq) → 3 Cd²+ (aq) + 2 NO(g) + 4H2O (1I) E°cell = 1.36 V Calculate the value of pH of the solution containing H*(aq) when the nonstandard cell potential (Ecell) is measured to be 1.12 V given the following conditions at 25 °C: [NO3] = 0.15 M; [Cd²+] = 0.46 M; PNO = 0.28 bar %3D %3D
Calculate E for the following electrochemical cell at 25 °C Pt(s) | H2(g, 1.00 atm)| H+(aq, 1.00 M) | | Cu2+(aq, 0.315 M)| Cu(s) given the following standard reduction potentials. Cu2+(aq) + 2 e- ⟶ Cu(s) E° = +0.337 V 2 H+(aq) + 2 e- ⟶ H2(g) E° = 0.000 V