Select the pair of products formed in the electrolysis of a solution of ZrCl4(aq) usinginert electrodes. Assume that there are no overpotentials involved in any reactions.Standard reduction potentials areCl2(g) +2 e 2 Cl¯(aq)→>E° = +1.36 VO2(g) + 4 H+(aq) + 4 e → 2 H2O(l) E° = +1.23 V2 H2O(l) +2 e→H2(g) + 2 OH(aq) E° = -0.83 VZr+(aq) +4 e → Zr(s)->E° = -1.53 VH2 and Cl2Zr and H2Zr and O2H2 and O2Zr and Cl2

The objective of the question is to find the pair of products formed in the electrolysis of a…

Select the pair of products formed in the electrolysis of a solution of ZrCl4(aq) using inert electrodes. Assume that there are no overpotentials involved in Standard reduction potentials are any reactions. Cl2(g) +2 e 2 Cl(aq) -> E° = +1.36 V O2(g) +4 H+(aq) + 4 e¯ -> 2 H2O(1) E° = +1.23 V 2 H2O(l) +2 e→H2(g) + 2 OH(aq) E° = -0.83 V -> Zr+(aq) + 4 e → Zr(s) H2 and Cl2 Zr and H2 Zr and O2 H2 and O2 Zr and Cl2 E° = -1.53 V

Select the pair of products formed in the electrolysis of a solution of ZrCl4(aq) using inert electrodes. Assume that there are no overpotentials involved in Standard reduction potentials are any reactions. Cl2(g) +2 e 2 Cl(aq) -> E° = +1.36 V O2(g) +4 H+(aq) + 4 e¯ -> 2 H2O(1) E° = +1.23 V 2 H2O(l) +2 e→H2(g) + 2 OH(aq) E° = -0.83 V -> Zr+(aq) + 4 e → Zr(s) H2 and Cl2 Zr and H2 Zr and O2 H2 and O2 Zr and Cl2 E° = -1.53 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...

See similar textbooks

Similar questions

Calculate Ecell for the following cell: Pt(s) | H2(g) | H *(aq) || Pb2*(aq) | PBSO4(s) | Pb(s) given the following standard reduction potentials. 2H*(aq) + 2 e ® H2(g) PbSO4(s) + 2 e ® Pb(s) + SO42-(aq) Oa. +0.712 V Ob. +0.356 V Oc. -0.712 V Od.-0.178 V Oe. -0.356 V
Sketch the galvanic cell based on the following overall reactions. Calculate e°; show the direction of electron flow and the direction of ion migration through the salt bridge; identify the cathode and anode; and give the overall balanced reaction. Assume that all concentrations are 1.0 M and that all partial pressures are 1.0 atm. Refer to the table above for the standard reduction potential a. Cr3+ (aq) + Cl2(g) = Cr,03 (aq) + Cl-
A cell is set up with copper and lead electrodes in contact with CuSO4(aq), and Pb(NO3)2(aq). If sulfuric acid is added to the Pb(NO3)2 solution to form PbSO4(s), what will happen to the potential of the cell? Explain.
Use the following standard reduction potentials to determine which species is the strongest oxidizing agent. Fe2+(aq) + 2 e– → Fe(s); E° = –0.41 V Pt2+(aq) + 2 e– → Pt(s); E° = 1.18 V Cr2O72–(aq) + 14 H+(aq) + 6 e– → 2 Cr3+(aq) + 7 H2O(); E° = 1.33 V Which of the following is the strongest oxidizing agent? Fe Pt Cr^3+ Fe^2+ Cr2O7^2-
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
A student wanted to observe the reaction of Fe²+, (aq) V3+ (aq) Fe³+ (aq) + √2+ (aq) using an electrochemical cell. He did the setup and found out that the cell produced a voltage of -1.03V. Assuming standard conditions (298.15 K, 1 atm), which of the following observations would be correct? O a. The standard cell potential is equal to Eº v³+²+ - E°F 3+ Fe / Fe +2 3+ b. The reaction was spontaneous The reaction will proceed without any external power source O d. Exchanging the anode and cathode will make the voltage more negative
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.
Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: (aq) + 2Cu+ (aq) → Pb(s) + 2Cu²+ (aq) AG° Pb²+ = Pb²+ (aq) + 2e¯ → Pb(s) Ee = -0.126 V red 2+ Cu²+ (aq) + e → Cu+ (aq) Fº red kJ K for this reaction would be = = 0.153 V + than one.
(a) Following reactions occur at cathode during the electrolysis of aqueous sodium chloride solution:Na+(aq) + e-1 ——-> Na(s) E° = – 2.71 VH+(aq) + e-1 ——–> 1/2H2(s) E° = 0.00 VOn the basis of their standard reduction electrode potential (E°) values, which reaction is feasible at the cathode and why?(b) Why does the cell potential of mercury cell remain constant throughout its life?
Examine the following half-reactions and select the strongest oxidizing agent among the substances. v Jersey Departm E° = 0.755 V [PtCl4]²¯(aq) + 2e¯ 2 Pt(s) + 4CI'(aq); RuO4(s) + 8H*(aq) + 8e¯ 2 Ru(s) + 4H2O(1); FeO4²(aq) + 8H*(aq) + 3e¯ 2 Fe3*(aq) + 4H2O(/); H4XEO6(aq) + 2H*(aq) + 2e¯ 2 XeO3(aq) + 3H2O(/); E° = 2.42 V E° = 1.038 V E° = 2.07 V RuO4(s) H4XEO6(aq) [PtCl4]² (aq) O HFEO4¯ (aq) CI(aq)
Enter electrons as e". Use smallest possible integer coefficients. States are not required. If a box is not needed, leave it blank. An aqueous ScCl3 solution is electrolyzed under 1 bar pressure using platinum electrodes. (a) Write the half-reactions predicted to occur at the anode and cathode, based on the standard cell potentials given below. Standard Reduction Potentials (Volts) at 25 °C Ch(g) + 2 e → 2 Cl'(aq) 1.360 02(g) + 4 H30*(aq) + 4 e» 6 H2O() 1.229 2 H2O(1) + 2 e– H2(g) + 2 OH (aq) -0.828 Sc*(aq) + 3 e - Sc(s) -2.080 Half-reaction at anode: Half-reaction at cathode: (b) What is the expected decomposition potential? V +
Consider the following standard reduction potentials: Zn2*(aq) + 2 e → Zn(s) E° = -0.76 V Mg2*(aq) + 2 e → Mg(s) E° = -2.37 V Ag*(aq) + e → Ag(s) E° = +0.80 V Which is the strongest oxidizing agent? Mg2*(aq) Zn2*(aq) Ag*(aq) 5:08 PM 90°F Partly sunny P Type here to search 8/6/2021