Solve the given problems with complete solution. Calculate the concentration of the chromic ion in the cell which has an electrode potential of 0.688 at 298.15 K for the given reaction: Cr207 2- (aq) + 14 H* (aq) + 61-1 (aq) → 2 Cr 3* (aq) + 3l2(s) + 7 H2O1) When [Cr20,2-] = 2.25 M; [H*] = 1.0 M ; [I-1] = 1.0 M ; [H20] = 1.0 x 10 -5 M Calculate the Gibbs free-energy and the standard Gibbs Free-Energy occur in the system

Solve the given problems with complete solution. Calculate the concentration of the chromic ion in the cell which has an electrode potential of 0.688 at 298.15 K for the given reaction: Cr207 2- (aq) + 14 H* (aq) + 61-1 (aq) → 2 Cr 3* (aq) + 3l2(s) + 7 H2O1) When [Cr20,2-] = 2.25 M; [H*] = 1.0 M ; [I-1] = 1.0 M ; [H20] = 1.0 x 10 -5 M Calculate the Gibbs free-energy and the standard Gibbs Free-Energy occur in the system

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

The free energy change for the following reaction at 25 °C, when [Fe²*] = 1.18 M and [Cr³3+] = 8.62×10-3 M, is -186 kJ: 3FE2+(1.18 M) +2Cr(s)- →3F€(s) + 2C1³*(8.62×10-3 M) AG = -186 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?
Consider the following cell reaction: Fe(s) + 2 H*(? M) –Fe²*(1.00 M) + H2(E)(1.00 atm) If the cell potential at 298 K is 0.261 volts, what is the pH of the hydrogen electrode? pH
When the oxidation-reduction reaction was balanced in acidic solution. If the [H2O2] = 0.352 M, partial pressure of Oz is 2.21 atm, [K2Cr2O7) = 0.642 M, the [Cr3+] = 0.963 M, and the pH was measured to be 8.42 then what is the cell potential at 25°C? O2(g) + Cr3+ *(aq) → H202(aq) + Cr20-2" (aq)
Calculate the reaction temperature of the electrochemical cell described below which measures a potential of 0.313 V. Cu(s) | Cu2+ (0.80 M) || O2(g) (0.98 atm) | H+ (0.047 M), H2O2 (0.92 M) | Pt(s) (Use R = 8.314 J/mol-K; F = 96485 C/mol)
Plese don't provide handwritten solution
Give the standard cell potential of the following half reaction: Cr2O72-(aq) + 3H2O2(aq) + 8H+(aq) → 2Cr3+(aq) + 3O2(aq) + 7H2O 2.28 V 0.65 V -2.28 V -0.65 V
9) At 25°C, what is the potential and the cell reaction for the following cell: Ni | Ni2+ (aq, 0010 M) || Cl(aq, 0.20M) | Cl2 (g, 1 atm) | Pt? What is the Keg at 298K?
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
What is the cell potential of the following electrochemical cell? (Report your answer to 2 significant figures.) Ni(s) | Ni²⁺(aq, 0.10 M || Ni²⁺(aq, 0.65 M) | Ni(s) E⁰red Ni²⁺/Ni(s) = -0.23 V
K124.
1) Calculate the following for the reaction at 298.0 K . The reaction is 2 SmCl3(aq) + 6 LiCl(aq) + 6 e- → 2 Sm(s) + 6 Li+(aq) + 6 Cl-(aq) A) The standard Gibbs free energy change of the reaction @ 298.0 K B) The equilibrium constant for the reaction C) The standard cell potential 2) Calculate the free energy change and the cell potential for the reaction at 298.0 K if all reactants are at 2.00 M and/or 2.00 atm and all products are at 0.0100 M and/or 0.0100 atm
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?