ELECTROCHEMISTRY: Given the line notation: Pt(s) |X*(aq). X<" x2+, (aq) || y3+ (aq) IY(s) where: x2+ +e , x* E° = 0.47 V y3+ + 3 e - Y E° = -2.65 V Pt4+ + 4 e - Pt E° = 1.20 V Calculate the EMF. Input values only with 2 decimal places. Do not include the unit. If positive value: DO NOT include the positive (+) sign If negative value: Please include the negative (-) sign

ELECTROCHEMISTRY: Given the line notation: Pt(s) |X(aq). X<" x2+, (aq) || y3+ (aq) IY(s) where: x2+ +e , x E° = 0.47 V y3+ + 3 e - Y E° = -2.65 V Pt4+ + 4 e - Pt E° = 1.20 V Calculate the EMF. Input values only with 2 decimal places. Do not include the unit. If positive value: DO NOT include the positive (+) sign If negative value: Please include the negative (-) sign

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

Pt(s) IX*(ag), X2+, ELECTROCHEMISTRY: (aq) | Y(s) Given the line notation: (aq) || y3+ where: x2+ + e - x+ E° = 0.94 V y3+ + 3 e → Y E° = -2.71 V Pt4+ + 4 e E° = 1.20 V Pt Calculate the EMF.
Calculate the EMF of the following electrochemical cell: Pt | VO2+(0.25 M), V3+(0.10 M), H+ (0.001 M) || TI3+ (0.10 M), TI+(0.05 M), | Pt Indicate whether the cell is galvanic or electrolytic. Evo2+/v3+ = 0.361 V, ET³+/T+ = 1.25 V Ans: 1.23 V, galvanic
Calculate K (both parts of the question)
2. The cell potential of this electrochemical cell depends on the pH of the solution in the anode half-cell. Pt (s) |H2 (g, 1 atm)|H*(aq, ? M) || Cu2+ (aq, 1.10 M)|Cu What is the pH of the solution if Ecell is 375 mV.
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
For the galvanic cell at 298 K 2X(s) + 3Y2*(aq) –→ 2x3*(aq) + 3Y(s) E° cell = 0.39 V What is the measured cell potential (Ecell) when [Y2*] = 1.0 M and [X3+] = 0.01 M? Periodic Table and Datasheet O 0.39 V O 0.35 V O 0.43 V O 0.51 V
Ff.39.
Pt(k)| Fe³+(aq), Fe²+(aq) // I-(aq) | 12(k) | Calculate the standard cell potential for the Pt(k) Cell and interpret whether the reaction taking place in the cell will proceed spontaneously under standard conditions. E° (Pt|Fe³+, Fe²+) = 0.77 V . E°( 12 | I-) = 0.53 V
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)
At a temperature of 25 ° C the galvanic element Pt | MnO4- 0,10M, Mn2 + 0,1M, pH = 1 | Ag + 0.0001M | Ag has the electromotive force 0,80V. Calculate the standard electrode potential of Ag + if E ° (MnO4 -, H + | Mn2 +) = 1.50V
Balance in acid-medium. Thanks
Calculate the potential for each cell as written at 25 °C. Indicate whether the metal electrode in the half-reaction opposite the standard hydrogen electrode, S.H.E., would be the anode or the cathode if the cell was shorted and electrons were allowed to flow freely. S.H.E. || Zn²+ (aq, 0.0110 M)| Zn(s) Ecell = Pt(s) | V3+ (aq, 0.450 M), V²+ (aq, 0.00420 M) || S.H.E. Ecell = EZn²+ Zn Incorrect = -0.762 V V V E√3+√√²+ = -0.255 V