ELECTROCHEMISTRY: Pt(s) |X*(aq), X2* (aq) || Y3+ (aq) | Y(s) Given the line notation: where: x2+ . x* E° = 1.3 V Y3+ + 3 e → Y E° = -2.15 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: Pt(s) |X(aq), X2 (aq) || Y3+ (aq) | Y(s) Given the line notation: where: x2+ . x* E° = 1.3 V Y3+ + 3 e → Y E° = -2.15 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...

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+ Complete and balance the following half-reaction in basic solution. Be sure to include the proper phases for all species within the reaction. Reset 2 Question 3 of 37 03 02 0 3 02 03 04 05 H₂O* OH™ - SO3² (aq) → SO4² (aq) 4 ( ) 1 O 5 02 03+04+ 6 7 8 06 07 0 H* H 9 Do t (s) (1) (g) (aq) S H₂O • H₂O Delete Submit
Give the standard line notation for each cell. (a) 2 IO3(aq) + 10 Fe²+ (aq) + 10 H+ (aq) 10 Fe³+ (aq) + I₂ (aq) + 5 H₂O(1) Pt(s) | Fe³+ (aq), I₂ (aq) || Fe²+ (aq), IO3¯(aq), H+ (aq) | Pt(s) Pt(s) | Fe²+ (aq), Fe³+ (aq) || IO3(aq), H+ (aq), I₂ (aq) | Pt(s) Pt(s) | IO3(aq), H+ (aq), I₂ (aq) || Fe²+ (aq), Fe³+ (aq) | Pt(s) Pt(s) | Fe²+ (aq), IO3¯¯(aq), H+ (aq) || Fe³+ (aq), I₂ (aq) | Pt(s) (b) (c) (d) Zn(s) + 2 Ag+ (aq) = Zn²+ (aq) + 2 Ag(s) Ag(s) | Zn²+ (aq) || Ag+ (aq) | Zn(s) Zn(s) | Zn²+ (aq) || Ag+ (aq) | Ag(s) Ag(s) | Ag+ (aq) || Zn²+ (aq) | Zn(s) Zn(s) | Ag+ (aq) || Zn²+ (aq) | Ag(s) H₂O₂ (aq) + 2 H+ (aq) + 2e → 2 H₂O(l) O2(g) + 2 H+ (aq) + 2 e¯ → H₂O₂ (aq) Pt(s) | H₂O₂ (aq) | Pt(s) ○ Pt(s) | O₂(g) | H₂O₂ (aq), H+ (aq) || H₂O₂ (aq), H+ (aq) | Pt(s) Pt(s) | H₂O₂ (aq), H+ (aq) || O₂(g) | H₂O₂ (aq), H+ (aq) | Pt(s) ○ Pt(s) | H₂O₂ (aq) | O₂(g) || H+ (aq) | Pt(s) Mn²+ (aq) + 2 e → Mn(s) Fe³+ (aq) + 3e → Fe(s) Fe(s) | Mn²+ (aq) || Fe³+ (aq) | Mn(s) Mn(s) | Mn²+ (aq) || Fe³+ (aq) | Fe(s)…
ELECTROCHEMISTRY Write the redox reactions and calculate the standard e.m.f.:a) Ni(s) | Ni2+(aq) || Sn2+(aq), Sn4+(aq) | Ptb) Pt | I2(s), I-(aq) || Ag+(aq) | Ag(s)
Given the cell notation, write the anode half-reaction, the cathode half-reaction and the overall balanced equation. Pt » | H2g | H*(aq)) || Ag*(aq) |Ag) (aq)) || Ag"(aq) |Ag)
Give the standard line notation for each cell. (a) 2 IO3¯(aq) + 10 Cr²+ (aq) + 10 H+ (aq) ⇒ 10 Cr³+(aq) + I₂ (aq) + 5 H₂O(1) Pt(s) | Cr²+ (aq), IO3¯(aq), H+ (aq) || Cr³+ (aq), I2 (aq) | Pt(s) Pt(s) | Cr³+ (aq), I2 (aq) || Cr²+ (aq), 103¯(aq), H+ (aq) | Pt(s) Pt(s) | IO3¯(aq), H+ (aq), I2 (aq) || Cr²+ (aq), Cr³+ (aq) | Pt(s) Pt(s) | Cr²+ (aq), Cr³+ (aq) || IO3¯¯(aq), H† (aq), I₂ (aq) | Pt(s) (b) (c) (d) Co(s) + 2 Ag+ (aq) ⇒ Co²+ (aq) + 2 Ag(s) Co(s) | Ag+ (aq) || Co²+ (aq) | Ag(s) (aq) || Ag* (aq) | Co(s) Ag(s) | Co²+ 2+ Ag(s) | Ag+ (aq) || Co²+ (aq) | Co(s) 2+ Co(s) | Co²+ (aq) || Ag+ (aq) | Ag(s) H2O2(aq)+2 H*(aq)+2e O₂(g) + 2 H+(aq) + 2 e¯ → H2O2(aq) Pt(s) | H₂O₂ (aq) | 02 (9) || H+ (aq) | Pt(s) Pt(s) | H₂O₂ (aq) | Pt(s) Pt(s) | H₂O2 (aq), H+ (aq) || O2(g) | H₂O₂ (aq), H+ (aq) | Pt(s) Pt(s) | O2(g) | H₂O2 (aq), H† (aq) || H₂O2 (aq), H† (aq) | Pt(s) 00 Mn²+ (aq) + 2 e → Mn(s) → 2H,O(l) 3+ Fe³+ (aq) + 3 e → Fe(s) Eº E° -1.18 V 3+ Fe(s) | Fe³+ (aq) || Mn²+ (aq) | Mn(s) Fe(s) | Mn²+ (aq)…
ELECTROCHEMISTRY: Pt(s)|X(2g)- (aq) IY(s) Given the line notation: (aq) II Y3+ where: x2* +e - x* E° = 0.79 V y3+ +3 e - Y E° = -2.38 V P4+ + 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 inciude the negative (-) sign
Consider the electrolysis reactions cathode anode whose standard reduction potentials can be found in this table. Calculate the voltage needed to drive the net reaction if the current is negligible. Ecell = Cu²+ (aq, 0.177 M) +2e=Cu(s) H₂O(1) ⇒ 10₂(g, 1 bar) + 2 H+ (aq, 0.195 M) + 2 e¯ The cell has a resistance of 66.0 and a current of 7.17 mA when the cathode overpotential is 0.231 V and the anode overpotential is 0.050 V. Calculate the voltage needed to overcome these effects and drive the net reaction. Ecell = V V
Cu(s)||Cu2+(aq, 0.0155 M)‖‖Ag+(aq, 3.50 M)||Ag(s)Cu(s)⁢|Cu2+⁢(aq, 0.0155 M)‖Ag+⁢(aq, 3.50 M)|Ag(s) net cell equation: Cu+2Ag+⟶Cu2++2AgCu+2⁢Ag+⟶Cu2++2Ag Calculate ?∘cell, Δ?∘rxn, Δ?rxn, and ?cell at 25.0⁢ ∘C, using standard potentials as needed. ?∘cell= 0.46 V Δ?∘rxn= -88.76 kJmol Δ?rxn= kJmol ?cell= V I just need help with the last 2
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
Use standard reduction potentials to calculate the equilibrium constant for the reaction: (aq) + Ag(s) → Fe²+ (s) + Ag+ (aq) Fe³+ From the table of standard reduction potentials: = 0.771 V, E E3+ = 0.799 V Ag+ /Ag Fe³+/Fe²+ 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 Submit Answer than zero.
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.
ELECTROCHEMISTRY: x2+ (aq) || Y3+ Given the line notation: Pt(s) |X*(ag). (aq) IY(s) where: x2+ + e - x* E° = 0.47 V y3+ + 3 e - Y E° = -2.65 V Pt4+ + 4 e E° = 1.20 V - Pt %3D 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