Calculate the K and AG for the following reaction at 25 °C:Mg(s) + Pb2+(aq)=Mg2+(aq) + Pb (s)Note: Reference the Standard reduction potentials at 25 °C and Fundamental constants tables for additional information.Part 1 of 2Round your answer to 3 significant digits.КсPart 2 of 2GRound your answer to 4 significant digits.AG°kJmol

Answered: Calculate the K and AG for the…

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 Eº, E, and AG for the following cell reaction: 2+ Mg(s) + Sn²+ (aq) ⇒ Mg²+ (aq) + Sn (s) where [Mg²+] = 0.040 M and [Sn²+] = 0.060 M. Round each of your answers to 4 significant digits. 2+ 2+ Note: Reference the Standard reduction potentials at 25 °C table for additional information. Note: The Faraday constant is 9.649 × 10ª J V.mol
35. Calculate the equilibrium constant at 25 °C using standard cell potential data, Sn(s) + CuSO4(aq) = Cu(s) + SnSO4 (aq)
Calculate the AG° (in kJ) for the reaction of elemental chlorine with the bromide ion at 298K. Refer to | the standard cell potentials on the last page of the note packet. Selected Answer: -54 A Calculate the AG° (in kJ) for the reaction of elemental chlorine with the bromide ion at 298K. Refer to the standard X cell potentials on the last page of the note packet. Selected Answer: 27.0
Calculate Eº, E, and AG for the following cell reaction: 2+ 2+ Mg(s) + Sn²+ (aq) = Mg²+ (aq) + Sn (s) 2+ 2 where [Mg²+] = [Mg²+] = 0.040 M and [Sn²+] = 0.030 M. Round each of your answers to 4 significant digits. Note: Reference the Standard reduction potentials at 25 °C table for additional information. Note: The Faraday constant is 9.649 × 10ª Part: 0 / 3 Part 1 of 3 Eº = v V x10 J V. mol S I
Consider the electrolysis reactions cathode anode whose standard reduction potentials can be found in this table. Calculate the voltage needed to drive the net reaction if the current is negligible. Ecell = Cu²+ (aq, 0.177 M) +2e=Cu(s) H₂O(1) ⇒ 10₂(g, 1 bar) + 2 H+ (aq, 0.195 M) + 2 e¯ The cell has a resistance of 66.0 and a current of 7.17 mA when the cathode overpotential is 0.231 V and the anode overpotential is 0.050 V. Calculate the voltage needed to overcome these effects and drive the net reaction. Ecell = V V
Calculate the K and AG for the following reaction at 25 °C: Zn(s) + CuSO₂(aq) → ZnSO₂(aq) + Cu (s) Note: Reference the Standard reduction potentials at 25 °C and Fundamental constants tables for additional information. Part 1 of 2 Round your answer to 3 significant digits. Bo = Part 2 of 2 AGⓇ ☐x10 = Round your answer to 4 significant digits. X kJ mol 0x10 3 X Ś
The standard state cell potential for the reaction below is 1.030 V. Zn(s) + Hg,Cl,(s)→ ZnCl,(aq) + 2Hg(1) Calculate the standard state Gibb's energy change (in kJ) at 298K for the reaction. O -99.4 kJ O -199 kJ O 199kJ O 99.4 kJ
Determine the cell potential for the following cell Ni(s) | Ni2+(aq), 0.500 M || Ag+(aq), 0.010 M | Ag(s) The standard reduction potentials of Ni2+(aq) and Ag+(aq) are –0.25 V and +0.80 V, respectively. A. 1.05 V B. 1.10 V C. 1.00 V D. 1.16 V E. 0.94 V
Consider the following galvanic cell: Agls) | Ag"(3.0 x 104 M)|| Fe*(1.1 x 10° M), Fe?-(0.040 M) | Pt(s) Choose the correct cell potential. O Ecel = + 0.12V O Ecel = - 0.15 V O Ecell = +0.09 V O Ecell = - 0.12 V Determine the sign of the electrodes in the cell. The left side half-cell is the and charged. The right side half cell is the and charged. Enter the equation for the spontaneous cell reaction. (Enter states of matter in your answer.) |
Consider the following reaction at 298K. Hg²+ (aq) + Ni(s)→→→→→→ Hg({) + Ni²+ (aq) The standard Gibbs free energy change for this reaction is -213.27 kJ. Use the Gibbs free energy change to calculate the standard cell potential for this reaction. Eºcell =
The standard cell potential for the reaction shown here is .653 V. Calculate the cell potential at 25°C with [Hg^2+] = 0.0500 M, [V^2+] = 0.100 M, [Hg2^2+] = 0.00500 M and [V^3+] = %3D 0.0200 M. [Hg?+] do.o500 M [V2+] = 0.100 M H0.0500 M [Hg* = 0.00500 M [V³*] = 0.0200 M 2 Hg²+ (aq) + 2 V²+ (aq) = Hg2²+ (aq) + 2 V³+ (aq)

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