Consider the half-reaction for the reduction of Cd2+to Cd(s). Ecd² Cd Cd²+ (aq) + 2 e -Cd(s) Calculate the potential of the cadmium electrode at 25 °C when immersed in a 0.0450 M solution of CdBr₂. Ecd = = -0.403 V V

Consider the half-reaction for the reduction of Cd2+to Cd(s). Ecd² Cd Cd²+ (aq) + 2 e -Cd(s) Calculate the potential of the cadmium electrode at 25 °C when immersed in a 0.0450 M solution of CdBr₂. Ecd = = -0.403 V V

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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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 [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?
A galvanic cell is constructed as follows. One half-cell consists of a platinum wire immersed in a solution containing 1.00 M Sn2+ and 1.00 M Sn4+. The other half-cell has a thallium rod immersed in a solution of 1.00 M Tl+. What is the equilibrium constant at 25° C?(Eo= Tl+ / Tl = −0.34 V.)? What is the cell voltage if the Tl+ concentration is increased 6−fold?
Determine E°, AG°, and K for the overall reaction from the balanced half-reactions and their standard reduction potentials. 4 Fe3+ + 2 H,0 4 Fe²+ + O, + 4H+ Fe3+ + e-= Fe²+ E° = 0.771 V 0, + 2 H* + 2 e¯2 H,0 E° = 1.229 V E° = V K = Is the overall reaction spontaneous as written? AG° = spontaneous not spontaneous
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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).
Use the standard reduction potentials located in the "Tables' linked above to calculate the standard free energy change in kJ for the reaction: 3Hg"(aq) + 2Cr(s)3Hg(1) + 2Cr*(aq) Answer: |kJ K for this reaction would be than one.
0 Consider the half-reaction for the reduction of Cd²+ to Cd(s). Cd²+(aq) + 2e¯ → Cd(s) Ecd² +Cd Calculate the potential of the cadmium electrode at 25 °C when immersed in a 0.0760 M solution of CdBr₂. Eca = Ecd= Calculate the potential of the cadmium electrode at 25 °C when immersed in a 0.0520 M NaOH solution that is saturated with Cd(OH)₂. The Ksp for Cd(OH)₂ is 7.2 × 10-¹5. Eca -0.4383 = -0.4383 Incorrect Calculate the potential of the cadmium electrode at 25 °C when immersed in a 0.0350 M Cd(NH3)² + and 0.105 M NH₂ solution. The 4 for Cd(NH3)²+is 3.63 × 106. = -0.403 V -0.4383 Incorrect V V V
The free energy change for the following reaction at 25 °C, when [Cu2+] = 2.57x10-³ M and [I₂] = 1.10 M, is 88.8 kJ: 2Cu²+(2.57x10-3 M) + 2.001 (aq)—2Cu+ (aq) + 1₂(1.10 M) AG= 88.8 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?
Which of the following reactions occurs at the anode of an electrochemical cell composed of Zn│Zn2+(aq) and Fe│Fe2+(aq) half-cells? Their reduction potentials are as follows: Zn2+ + 2e– --> Zn; Eo = -0.763 V Fe2+ + 2e– --> Fe; Eo = -0.447 V (A) Zn2+ + 2e– --> Cr; (B) Zn --> Zn2+ + 2e–; (C) Fe --> Fe2+ + 2e–; (D) Fe2+ + 2e– --> Fe;