ELECTROCHEMISTRY: Given the line notation: Pt(s) |X* (aq). X2* (aq) || Y (aq) | Y(s) where: x2+ +e - X* E° = 0.45 V y3+ +3e - Y E° = -2.03 V P4* + 4e - 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). X2* (aq) || Y (aq) | Y(s) where: x2+ +e - X* E° = 0.45 V y3+ +3e - Y E° = -2.03 V P4* + 4e - 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

Give the standard line notation for each cell. (а) 2 103 (ад) + 10 Cu* (aq) + 10 H* (аq) — 10 Cu* (ag) + 12(aq) + 5 H20(1) O Pt(s) | Cu* (aq), Cu²+ (aq) || IO3 (aq), Ht (ag), I2 (aq) | Pt(s) Pt(s) | Cu* (aq), IО; (ад), н* (аq) || Cu?* (ag), I; (ag) | Рt(s) O Pt(s) |1O3 (ag), H* (ag), I2 (aq) || Cu* (aq), Cu²+ (ag) | Pt(s) О P{(6) | Cu?* (ag), I3 (aq) || Cu" (aq), 10; (аg), H (ад) | Pt(5) (b) Zn(s) +2 Ag* (aq) = Zn²+ (aq) + 2 Ag(s) O Zn(8) | Zn²+ (aq) || Ag+(aq) | Ag(s) O Zn(s) | Ag* (ag) || Zn²+ (aq) | Ag(s) Ag(s) | Ag* (aq) || Zn²+ (aq) | Zn(s) Ag(s) | Zn?+ (aq) || Ag* (aq) | Zn(s) (c) H2O2 (ag) + 2 H† (ag) + 2 e¯ –→ 2 H20(1) E° = 1.78 V Ог(9) + 2 H* (аg) + 2 е + H,О, (aq) E° = 0.68 V O Pt(s) | O2 (9) | H2O2 (aq), H* (ag) || H2 O2 (ag), H* (ag) | Pt(s) O Pt(s) | H2O2 (ag), H* (aq) || O2(9) | H2O2 (aq), H* (aq) | Pt(s) O Pt(8) | H2O2(aq) | O2 (9) || H* (ag) | Pt(s) O Pt(s) | H2O2(aq) | Pt(8) (d) Mn2+ (ag) + 2 e¯ → Mn(s) E° = - 1.18 V Fe+ (ag) + 3 e → Fe(s) E° = -0.036 V O Mn(s) | Mn²+…
What is the molar concentration of H', [H*], for following cell:| Zn| Zn2*(1M) || H* (aq) | H2) (1atm)| With a potential of 0.45 V?
6
For the cell Ag | AgCl | Cl (a = 0.0001), Ce4+ (a (a = 0.01), Ce³+ 0.1) | Pt, solve for in V, where the standard half-cell potentials are 0.2225 V for AgCl|CI and 1.45 V for Ce4+|Ce³+ F = 96,485 C mol-¹
Pt(s) IX*(aq), X2+, ELECTROCHEMISTRY: (aq) I| Y3+ *(aq) | Y(s) Given the line notation: where: x2+ + e - x+ E° = 1.17 V y3+ + 3 e - Y E° = -2.16 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
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.
ELECTROCHEMISTRY: Pt(s) |X*(ag). x2+ (aq) | Y(s) Given the line notation: (aq) || Y3+ where: x2+ + e - x* E° = 0.84 V y3+ + 3 e - Y E° = -1.56 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: Piease include the negative (-) sign
An electrochemical cell similar to the cell in Experiment 3 is constructed, but instead of using Cu electrodes and a Cu²+ solution, this cell uses Pb electrodes and a Pb²+ solution. Pb|Pb+²(1 M)||AsO(~ 1 M)|Pb₂ (AsO₂)₂(s) Pb(s) The measured potential, Ecell of this cell is -0.349 V. Calculate Ksp for Pb, (ASO₂)₂. Ksp = 1.479 x10¹2 Incorrect
The potential of electrodes Cu2+ (aq) | Cu(s) is ϕ1 = 0.337 V , and Cu2+ (aq), Cu+ (aq) | Pt is ϕ2 = 0.153 V , respectively. a) Calculate the value of ϕ0 for electrode Cu+(aq) | Cu(s). [Hint: Write half reactions for the above two electrodes, then derive the target electrode.]b) Calculate the equilibrium constant Ka0 for the reaction Cu(s) + Cu2+ (aq) = 2Cu+ (aq) at 298.15K. [Hint: Based on the reaction, design a cell with two electrodes and verify that the two half reactions will give you the target reaction. Then calculate the cell potential.]
ELECTROCHEMISTRY: Given the line notation: Pt(s) | X+(aq), X2+(aq) || Y3+(aq) | Y(s) where: X2+ + e- X+ Eo = 0.44 V Y3+ + 3 e- Y Eo = -1.99 V Pt4+ + 4 e- Pt Eo = 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 show sol'n please.
AG = AG° + RT · In(Q) ΔΕ ΔΕ-). In(Q) Half Reaction (Note: All given as reduction) E° (V) 02 (g) + 4 H*(aq) + 4 e¯ → 2 H20 (I) 1.229 |Z2 (s) + 2 e 2z (aq) 0.426 3+ |A°™ (aq) + 3 e A (s) 0.292 2 H20 (1) + 2 e H2 (g) + 2 OH¯ (aq) - 0.828 > 2+ G (aq) + 2 e¯ → G (s) - 1.245 M2+ (aq) + 2 e → M (s) - 1.893 A student constructs a voltaic electrochemical cell with two metal electrodes [metal G and metal A] in their respective aqueous nitrate solutions [G(NO3)2 and A(NO3)3]. Use this information, as well as the reduction potentials in the table above to complete each statement below. The metal electrode A is... o not changing in mass. o increasing in mass. o decreasing in mass. Question 4 The metal solution A(NO3)3 is... o increasing in concentration. o not changing in concentration. o decreasing in concentration.
Use standard reduction potentials to calculate the equilibrium constant for the reaction: Co²+ (aq) + 2Ag(s) → Co(s) + 2Ag+ (aq) From the table of standard reduction potentials: Eo²+/Co = -0.280 V, E Ag+ /Ag = 0.799 V Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. Equilibrium constant at 298 K: AGO for this reaction would be than zero.