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.

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.

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

A galvanic cell is powered by the following redox reaction: 2+ + 3 Cu²+ (aq) + Cr(OH)3(s) + 5OH¯(aq) → 3 Cut (aq) + Cro (aq) + 4H₂O(1) Answer the following questions about this cell. If you need any electrochemical data, be sure you get it from the ALEKS Data tab. Write a balanced equation for the half-reaction that takes place at the cathode. Write a balanced equation for the half-reaction that takes place at the anode. Calculate the cell voltage under standard conditions. Round your answer to 2 decimal places. 0 0 0 E = v V ロ→ロ X 5
eq req *3 Use the References to access important values if needed for this question. Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: 2H+ (aq) + Fe(s)→ H₂ (s) + Fe²+ (aq) H₂ (8) E Fe(s) E E 2H+ (aq) +2e Fe2+ (aq) +2e -> AGO kj K for this reaction would be Submit Answer $ 4 R % 5 Retry Entire Group 9 more group attempts remaining = 0.000 V red=-0.440 V T red e than one. Cengage Learning Cengage Technical Support 6 MacBook Pro Y & U * 00 8 1 ( 9 K 0 0 Previous Email Instructo P
Use the standard reduction potentials given below to predict if a reaction will occur when Al metal is put into a 1 M Fe²+ solution. aqueous Fe2+ (aq) + 2e → Fe(s) E red Al³+ (aq) + 3e →→ Al(s) E + red = If a reaction will occur, write a balanced net ionic equation for the reaction. If no reaction will occur, leave all boxes blank. ― -0.440 V -1.660 V (Use the lowest possible coefficients. Be sure to specify states such as (aq) or (s). If a box is not needed, leave it blank.) +
Consider a cell that uses the redox reaction: 3 Na (s) + La3+ (aq) → 3 Na+ (aq) + La (s). Calculate the equilibrium constant (K) for this reaction at 25 ºC, given the following standard reduction potentials: La3+ (aq) + 3 e- → La (s) E° = -2.37 V Na+ (aq) + e- → Na (s) E° = -2.71 V 1.8 x 10-6 5.7 x 10-18 5.6 x 105 1.7 x 1017
Use the standard reduction potentials located in the 'Tables' linked above to calculate the standard free energy change in kJ for the reaction: Cd2+ (aq) + Zn(s)→→→→→→ Cd(s) + Zn²+ (aq) Answer: kJ K for this reaction would be than one.
Calculate the standard free‑energy change for this reaction at 25 °C. A list of standard reduction potentials can be found in this table. 2Au3+(aq)+3Ni(s)↽−−⇀2Au(s)+3Ni2+(aq) Δ?∘= kJ/mol
Enter electrons as e". Use smallest possible integer coefficients. States are not required. If a box is not needed, leave it blank. Standard Reduction Potentials (Volts) at 25 °C Cl₂ (g) + 2 e →2 Cl(aq) O2(g) + 4 H3O+ (aq) + 4 e 6 H₂O(1) 1.360 1.229 2 H2O(l) + 2e Al3+ (aq) + 3 e → Al (s) H2(g) + 2 OH(aq) -0.828 -1.660 A 0.114 M neutral aqueous AlCl3 solution is electrolyzed under 1 atm pressure using platinum electrodes. (a) Write the half-reactions expected to occur, taking into account the effect of the non-standard conditions on the electrode potentials. Half-reaction at anode: + + + + Half-reaction at cathode: + + + + (b) What is the expected decomposition potential?
The standard free energy change for a reaction can be calculated using the equation ΔG∘′=−nFΔE∘′ where n is the number of electrons transferred, F is Faraday's constant, 96.5 kJ·mol−1·V−1, and ΔE∘′ is the difference in reduction potential. For each of the given reactions, determine the number of electrons transferred (n) and calculate standard free energy (ΔG∘′) . Consider the half-reactions and overall reaction for reaction 1. half-reactions: fumarate^2− + 2H^+ ↽−−⇀ succinate^2− CoQH2 ↽−−⇀ CoQ + 2H^+ overall reaction: fumarate^2− + CoQH2↽−−⇀succinate^2− + CoQ ΔE∘′= −0.009 V n=____ ΔG∘′= ____ (kJ⋅mol^−1) Consider the half-reactions and overall reaction for reaction 2. half-reactions: cytochrome (c1) (Fe^2+) ↽−−⇀ cytochrome (c1) (Fe^3+) cytochrome (c) (Fe^3+) ↽−−⇀ cytochrome (c) (Fe^2+) overall reaction: cyt (c1) (Fe^2+)+cyt (c) (Fe^3+)↽−−⇀cyt (c1) (Fe^3+)+cyt (c) (Fe^2+) ΔE∘′=0.034 V n= ____ ΔG∘′=____(kJ⋅mol^−1)
Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy following redox reaction. Be sure your answer has the correct number of significant digits. H₂(g) + 2OH¯(aq) + 2Fe³+ (aq) → 2H₂O (1) + 2Fe²+ (aq) kJ x10 X Ś AGO for the
Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: 2H+ (aq) + Hg(1)→ H₂(g) + Hg²+ (aq) 2H+ (aq) + 2e¯ → H₂(g) Fº red 2+ Hg2 (aq) + 2e¯ → Hg(1) AGO = || kJ K for this reaction would be Eº red = 0.000 V = = 0.855 V than one.