What is AG° in kJ for the following reaction at 298K?O2(g) + 4H*(aq) +4FE2+(aq) → 4Fe3+(aq) + 2H2O(1)E°cell = 0.04 VAG° = –NFE°,cell =-RT In(K)F = 96,485 C / mol eR = 8.314 J/mol KAnswer with 3 significant figures and do NOT include units.

Since the E0cell is given = 0.04 V and from the reaction we can see that 4 Fe2+ is converting into 4…

Answered: What is AG° in kJ for the following…

Chemistry

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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...

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The following half-reaction °E = -2,07 V. AlF63-(aq) + 3e- à Al(s) + 6F-(aq) calculate the equilibrium constant at 25°C for this reaction: Al3+(aq) + 6F-(aq) ↔ AlF63-(aq) K = ?
7
(Q86) What is the value (in V) of the standard cell potential (E°CELL) for the following reaction that occurs in an electrochemical cell at 25°C? Fe3+ (aq) + Cu* (aq) ---> Fe²+ (aq) + Cu2+ (aq) The reduction potentials for the two half-reactions involved are: Fe3+ + e- --> Fe2+ (E°RED) = 0.77 V Cu2+ + e- --> Cu* (E°RED) = 0.16 V %3D
Part A If a voltaic cell is created between a Am"/Am electrode and a Fe*/Fe electrode, what species is being oxidized? O Am"(ag) O Fe(s) O Fe"(ag) O Am(s)
Given the following information: F1- (aq): ΔGf° = -276.48 kJ/mol Na1+ (aq): ΔGf° = -261.87 kJ/mol 2 Na (s) + F2 (g) → 2 NaF (aq) a) Find ΔG° for this reaction using the above numbers. b) Write the expression for Qeq. (Note that NaF is a strong electrolyte!) c) Find ΔG at 325.0 K for 0.01500 atm F2 and 3.065 M NaF.
(Q89) What is the value (in kJ) of the standard Gibbs free energy change (AG°rxn) for the following reaction: 3 Ni (s) + 2 Au³+ (aq) ----> 3 Ni²+ (aq) + 2 Au (s) The reduction potentials for the two half-reactions involved are: Ni²+ + 2 e---> Ni (s) (E°RED) = -0.23 V Au³+ + 3 e- --> Au (s) (E°RED) = 1.50 V
7>) True or False
Consider the following standard reduction potentials: F2(g) + 2 e - 2 F (aq) E° = +2.87 V Mg2*(aq) + 2 e - Mg(s) E° = -2.37 V Under standard conditions, O Mg(s) is a stronger oxidizing agent than F (aq), and Mg2*(aq) is a stronger reducing agent than F2(g), O F (aq) is a stronger oxidizing agent than Mg(s), and F2(g) is a stronger reducing agent than Mg2*(aq). O F2(g) is a stronger oxidizing agent than Mg-"(aq), and Mg(s) is a stronger reducing agent than F (aq). O Mg2*(aq) is a stronger oxidizing agent than F2(g), and F(aq) is a stronger reducing agent than Mg(s).
Calculațe Eo cell from the tabulated standard reduction potentials for each of the following reactions in aqueous solution. Then calculate AG° and K at 25°C from E° cell' F = 9.65 x 10 J/V•mol e R = 8.314 J/mol•K Acidic solution Standard reduction potential, E (Volts) Cu2" (ag) + e Cu'(ag) + e Cu(s) Mno, (aq) + 8H"(aq) + 5e Mn?"(aq) + 4H,0 Fe (aq) + e – Fe? (aq) Cu*(aq) 0.153 0.521 1.507 0.771 Basic solution Zn(OH) (s) + 2e → Zn(s) + 40H (aq) Mno, (aq) + 2H,O + 3e MnO,(s) + 4OH(aq) -1.22 0.588
The potential, AE°, for an electrochemical cell for the reaction shown below is 0.339 V: Cu2*(aq) + H2(g) = Cu(s) + 2H*(aq) (1) What is AE° for reaction 2? ½Cu2*(aq) + ½H2(g) = ½Cu(s) + H*(aq) (2) Select your response from the list below. i) 0.339 V ii) 0.678 V iii) 0.1695 V
4-ele The free energy change for the following reaction at 25 °C, when [Cr³"]= 4.83×103 M and [Hg²+]= 1.13 M, is 937 kJ: 2Cr*(4.83×103 M) + 3Hg(1)– 2+ →2Cr(s) + 3Hg²*(1.13 M) AG = 937 kJ What is the cell potential for the reaction as written under these conditions? Answer: V Would this reaction be spontaneous in the forward or the reverse direction forward reverse

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What is AG° in kJ for the following reaction at 298K? O2(g) + 4H*(aq) +4FE2+(aq) → 4Fe3+(aq) + 2H2O(1) E°cell = 0.04 V AG° = –NFE°, cell =-RT In(K) F = 96,485 C / mol e R = 8.314 J/mol K Answer with 3 significant figures and do NOT include units.