The free energy change for the following reaction at 25 °C, when [Cd2*] = 1.17 M and [Cr+] = 8.47x10-3 M, is -207 kJ: 3ca**(1.17 M) + 2Cr(s)3Cd(s) + 2Cr³+(8.47x103 M) AG =-207 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?

The free energy change for the following reaction at 25 °C, when [Cd2*] = 1.17 M and [Cr+] = 8.47x10-3 M, is -207 kJ: 3ca**(1.17 M) + 2Cr(s)3Cd(s) + 2Cr³+(8.47x103 M) AG =-207 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?

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...

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7. Please answer the following questions. (a) Given the following reduction potentials Cd (OH)2 (s) + 2e → Cd (s) + 2 OH (aq) Eºred = -0.76 V NIO(OH) (s) + H₂O (1) + e → Ni(OH)2 (aq) + OH (aq) Eºred = +0.49 V Calculate the standard emf of the nickel-cadmium (nicad) rechargeable battery. *hint: the reaction for the discharge of nicad battery is: Cd (s) + 2 NIO(OH) (s) + 2 H₂O (1)→ 2Ni(OH)2 (aq) + Cd(OH)2 (s) (b) A typical nicad voltaic cell generates an emf of +1.30 V. Why is there a difference between this value and the one you calculated in part (b)? (c) Calculate the equilibrium constant for the overall nicad reaction based on this typical emf value.
Calculate the cell potential, E, for the given reactions at 25.00 "C using the ion concentrations provided. Then, determine if th cells are spontaneous or nonspontaneous as written. Refer to the table of standard reduction potentials (E). Pt(s) + Fe²(aq) = P² (aq) + Fe(s) [Fe²*1 = 0.0064 M E= incorrect -4761 E= Cu(s) + 2 Ag* (aq) = Cu² (aq) + 2 Ag(s) |Cu²+1 = 0.027 M Ag1=0.027 M E [P1²*1 = 0.067 M Co² (aq) + Ti (aq) — Co¹ (aq) + Ti² '(aq) Co²1 = 0.055 M Co= 0.030 M ITP 1 = 0.0070 M T² = 0.0107 M mam V The cell is not spontaneous spontaneous. The cell is spontaneous not spontaneous. incorrect The cell is spontaneous not spontaneoux.
Calculate the cell potential for the reaction as written at 25.00 °C, given that [Cr²+] = 0.768 M and [Fe2+] = 0.0170 M. Use the standard reduction potentials in this table. Cr(s) + Fe²+(aq) Cr² (aq) + Fe(s) E = x10 TOOLS V
The free energy change for the following reaction at 25 °C, when [Pb2+] = 3.82×10-3 M and [Cu2+] = 1.20 M, is 68.1 kJ: Pb2+(3.82×10-3 M) + 2Cu+(aq)Pb(s) + 2Cu2+(1.20 M) ΔG = 68.1 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?
K124.
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