SIMULATION Voltaic Cells Under Standard Conditions Oxidizing Reducing Electrode EO °cathode-E°anode = Eocell Agent Agent Potential (V) Au3+(aq) + 3 e Au(s) +1.50 Brz(t) + 2 e 2 Br (aq) +1.08 Cathode Anode Hg2 (aq) + 2 e Hg(t) +0.855 Ag*(aq) + e Ag(s) +0.80 Hg22 (aq) + 2 e 2 Hg(t) +0.789 Cu2*(aq) + 2 e Cu(s) +0.337 Sn (aq) + 2 e Sn2*(s) +0.15 2 H30*(aq) + 2 e H2(g) + 2 H20(s) 0.00 Sn2*(aq) + 2 e Sn(s) -0.14 Ni2*(aq) + 2 e Ni(s) -0.25 Cd2 (aq) + 2 e Cd(s) -0.40 Al3+ (aq) + 3 e Al(s) -1.66 Mg2+(aq) + 2 e Mg(s) -2.37 Li (aq) + e Li(s) -3.045 Click on an Oxidizing Agent and a Reducing Agent. 1 1 1 Gibbs Free Energy

SIMULATION Voltaic Cells Under Standard Conditions Oxidizing Reducing Electrode EO °cathode-E°anode = Eocell Agent Agent Potential (V) Au3+(aq) + 3 e Au(s) +1.50 Brz(t) + 2 e 2 Br (aq) +1.08 Cathode Anode Hg2 (aq) + 2 e Hg(t) +0.855 Ag(aq) + e Ag(s) +0.80 Hg22 (aq) + 2 e 2 Hg(t) +0.789 Cu2(aq) + 2 e Cu(s) +0.337 Sn (aq) + 2 e Sn2(s) +0.15 2 H30(aq) + 2 e H2(g) + 2 H20(s) 0.00 Sn2(aq) + 2 e Sn(s) -0.14 Ni2(aq) + 2 e Ni(s) -0.25 Cd2 (aq) + 2 e Cd(s) -0.40 Al3+ (aq) + 3 e Al(s) -1.66 Mg2+(aq) + 2 e Mg(s) -2.37 Li (aq) + e Li(s) -3.045 Click on an Oxidizing Agent and a Reducing Agent. 1 1 1 Gibbs Free Energy

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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Calculate the potential developed by a voltaic cell using the following reaction if all dissolved species are 0.012 M at 25 °C. 2 Fe²+ (aq) + H₂O₂(aq) + 2 H+(aq) → 2 Fe³+ (aq) + 2 H₂0 (1) Acidic Solution Standard Electrode Potential, E (volts) +1.77 2 H₂O(l) H₂O₂(aq) + 2 H*(aq) + 2 e Au*(aq) + e —»Au(s) Au*'(aq)+3e — Au(s) +1.68 +1.50 +1.08 NO(g) + 2 H₂O +0.96 Br₂(l) + 2 e2 Br(aq) NO3(aq) + 4 H*(aq) + 3 e Ag (aq) + e→→→→→→ Ag(s) Hg₂²(aq) + 2 e→→→→→2 Hg(l) +0.80 +0.789 Fe³(aq) + e-→→→→→→ Fe²¹(aq) +0.77 Cu²+ (aq) + 2 e → Cu(s) +0.337 HgzClz(s) + 2 e2 Hg() + 2 Cl(aq) +0.27 Sn¹+ (aq) + 2e → Sn²+ (aq) +0.15 2 H*(aq) + 2 e→→→→→→ H₂(g) 0.00 Pb²+ (aq) + 2 e Pb(s) -0.126 Sn²+ (aq) + 2 e → Sn(s) -0.14 Ni²+ (aq) + 2 e → Ni(s) -0.25 Cd²+ (aq) + 2 e → Cd(s) -0.40 Cr³+ (aq) + e →Cr²+ (aq) -0.408 Fe²+ (aq) + 2 e → > Fe(s) -0.44 Zn²+ (aq) + 2 e→→→→→→ Zn(s) -0.763 Cr²+ (aq) + 2 e →→ Cr(s) -0.91 Al³+ (aq) + 3 e →Al(s) -1.66 Mg²+ (aq) + 2 e -2.37 Mg(s) V
Calculate the potential developed by a voltaic cell using the following reaction if all dissolved species are 0.031 M at 25 °C. 2 Fe"(aq) + H,O2(aq) + 2 H*(aq) → 2 Fe³*(aq) + 2 H2O(E) Acidic Solution Standard Electrode Potential, E(volts) H2O2(aq) + 2 H'(aq) + 2 e →2 H2O(t) +1.77 Au*(aq) + e → Au(s) +1.68 Au (aq) + 3 e –→ Au(s) +1.50 > Br2(t) + 2 e → 2 Br(aq) NO: (aq) + 4 H'(aq) + 3 e → NO(g) + 2 H20 Ag*(aq) + e Ag(s) Hgz2"(aq) + 2 e –→2 Hg(t) Fe"(aq) + e- Cu2"(aq) + 2 e – Cu(s) +1.08 +0.96 +0.80 +0.789 - Fe?"(aq) +0.77 +0.337 Hg2Clz(s) + 2 e - 2 Hg(t) + 2 C(aq) +0.27 -> Sn* (aq) + 2e –→ Sn²*(aq) +0.15 Hz(g) 2 H*(aq) + 2 e - Pb"(aq) + 2 e –→ Pb(s) Sn²"(aq) + 2 e –→ Sn(s) > 0.00 -0.126 -0.14 Ni? (aq) + 2 e – Ni(s) -0.25 Cd"(aq) + 2 e – Cd(s) -0.40 Cr**(aq) + e - Fe2"(aq) + 2 e - Zn (aq) + 2 e →Zn(s) Cre"( aq) + 2 e – Cr(s) AB* (aq) + 3 e – Al(s) Mg (aq) + 2 e – Mg(s) Cr**(aq) -0.408 Fe(s) -0.44 -0.763 -0.91 -1.66 -2.37 V Submit Show Approach Show Tutor Steps
1:18 Question 14 of 17 The standard cell potential E°(cell) is +0.34 V for the redox reaction 2 In(s) + 6 H*(aq) → 2 In³+ (aq) + 3 H₂(g). 1 4 7 +/- Determine Eº for the half-reaction In³+ (aq) + 3 e → In(s). 2 5 8 V 3 60 Submit 9 O Tap here or pull up for additional resources XU x 100
Calculate the potential developed by a voltaic cell using the following reaction if all dissolved species are 0.049 M at 25 °C. 2 Fe2*(aq) + H2O2(aq) + 2 H*(aq) → 2 Fe³*(aq) + 2 H2O(t) Acidic Solution Standard Electrode Potential, E(volts) H2O2(aq) + 2 H*(aq) + 2 e 2 H2O(t) +1.77 → Au(s) Aus (aq) + 3 e → Au(s) Br2(e) + 2 e –→ 2 Br(aq) NO: (aq) + 4 H*(aq) + 3 e" Au*(aq) + e - +1.68 > +1.50 +1.08 NO(g) + 2 H20 +0.96 Ag*(aq) + e → Ag(s) +0.80 Hg2 (aq) + 2 e Fe"(aq) + e- →2 Hg(t) Fe"(aq) > +0.789 > +0.77 Cu2*(aq) + 2 e Cu(s) +0.337 > Hg2Clz(s) + 2 e →2 Hg(t) + 2 Cr(aq) Sn**(aq) + 2e → 2 H*(aq) + 2 e Pb2"(aq) + 2 e +0.27 Sn2"(aq) +0.15 » H2(g) 0.00 -0.126 → Pb(s) Sn(s) Sn2*(aq) + 2 e Ni"(aq) + 2 e Cd2"(aq) + 2 e → Cd(s) Cr**(aq) + e -0.14 > Ni(s) -0.25 -0.40 » Cr**(aq) -0.408 Fe2"(aq) + 2 e Zn2*(aq) + 2 e → Cr2*( aq) + 2 e A* (aq) + 3 e - » Fe(s) -0.44 Zn(s) -0.763 → Cr(s) Al(s) > -0.91 > -1.66 Mg2"(aq) + 2 e - → Mg(s) -2.37 >
Oxidizing Reducing Electrode Agent Agent Potential (V) Au³+ (aq) + 3 e Au(s) +1.50 Br₂(!) + 2 e 2 Br (aq) +1.08 Hg(t) +0.855 Hg²+ (aq) + 2 e Ag+ (aq) + e Hg₂²+ (aq) + 2 e Ag(s) +0.80 2 Hg(1) +0.789 Cu²+ (aq) + 2 e Cu(s) +0.337 Sn4+ (aq) + 2 e Sn²+(s) +0.15 2 H30+ (aq) + 2 e H₂(g) + 2 H₂O(s) 0.00 Sn²+ (aq) + 2 e Sn(s) -0.14 Ni²+ (aq) + 2 e Ni(s) -0.25 Cd²+ (aq) + 2 e Cd(s) -0.40 Al³+ (aq) + 3 e Al(s) -1.66 Mg²+ (aq) + 2 e Mg(s) -2.37 Li+ (aq) + e Li(s) -3.045 Click on an Oxidizing Agent and a Reducing Agent. Choose Au³+ as the oxidizing agent and Ni as the reducing agent. What is the value of Eºcell? 0.70 X V → → → → ↑ Eº cathode Gibbs Free Energy Eº anode = Eºcell = Cathode Anode
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The electrochemical reaction which powers a lead-acid storage battery is as follows: Pb(s) + PbO2(s) + 4H*(aq) + 2SO,²-(aq) → 2PBSO4(s) + 2H20(1) A single cell of this battery consists of a Pb electrode and a PbO2 electrode, each submerged in sulfuric acid. What reaction occurs at the cathode during discharge? O PbO2(s) is oxidized to PbSO4(s). O PbO2(s) is reduced to PbSO4(s). H* is oxidized to H20(1) . O Pb(s) is oxidized to PbSO4(s). O Pb(s) is reduced to PbSO4(s).
Item 1 < 1 of 25 Constants | Periodic Table Part A The standard cell potential (E') of the reaction below is -0.55 V. The value of A G for the reaction is J/mol. I2 (5) + 2Br (aq) → 21 (aq) + Brz (1) O 1.1- 105 O 0.54 O 0.55 O 5.5 - 106 O none of the above Submit Request Answer
Standard Reduction (Electrode) Potentials at 25 oC Half-Cell Reaction Eo (volts) Standard Reduction (Electrode) Potentials at 25 oC Half-Cell Reaction Eo (volts) F2(g) + 2 e- 2 F-(aq) 2.87 Ce4+(aq) + e- Ce3+(aq) 1.61 MnO4-(aq) + 8 H+(aq) + 5 e- Mn2+(aq) + 4 H2O(l) 1.51 Cl2(g) + 2 e- 2 Cl-(aq) 1.36 Cr2O72-(aq) + 14 H+(aq) + 6 e- 2 Cr3+(aq) + 7 H2O(l) 1.33 O2(g) + 4 H+(aq) + 4 e- 2 H2O(l) 1.229 Br2(l) + 2 e- 2 Br-(aq) 1.08 NO3-(aq) + 4 H+(aq) + 3 e- NO(g) + 2 H2O(l) 0.96 2 Hg2+(aq) + 2 e- Hg22+(aq) 0.920 Hg2+(aq) + 2 e- Hg(l) 0.855 Ag+(aq) + e- Ag(s) 0.799 Hg22+(aq) + 2 e- 2 Hg(l) 0.789 Fe3+(aq) + e- Fe2+(aq) 0.771 I2(s) + 2 e- 2 I-(aq) 0.535 Fe(CN)63-(aq) + e- Fe(CN)64-(aq) 0.48 Cu2+(aq) + 2 e- Cu(s) 0.337 Cu2+(aq) + e- Cu+(aq) 0.153 S(s) + 2 H+(aq) + 2 e- H2S(aq) 0.14 2 H+(aq) + 2 e- H2(g) 0.0000 Pb2+(aq) + 2 e- Pb(s) -0.126 Sn2+(aq) + 2 e- Sn(s) -0.14 Ni2+(aq) + 2 e- Ni(s) -0.25 Co2+(aq) + 2 e- Co(s)…
eq Preq 1req F2 # 3 LU E D What is the calculated value of the cell potential at 298 K for an electrochemical cell with the following reaction, when the Pb²+ concentration is 8.54 x 10-4 M and the Mg2+ concentration is 1.23 M ? Ecell F3 The cell reaction as written above is Submit Answer $ SA 4 2+ Pb²+ (aq) + Mg(s) → Pb(s) + Mg²+ (aq) R F V V F4 % References] Use the References to access important values if needed for this question. стор T Retry Entire Group G DII Cengage Learning Cengage Technical Support F5 OL Y H F6 9 more group attempts remaining & 7 U ✓for the concentrations given. F7 J OO PrtScn BN M F8 K Home F9 L End F10 Previous PgUp F11 Next Save and Exit 4x10 PgDn 8:28 PM 12/2/2022 F12 1 Ins {
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The following question refers to the following system: 3Ag(s) + NO3 (aq) + 4H*(aq) → 3Ag*(aq) + NO(g) + 2H2O() Ag → Ag*(aq) + e- NO3 (aq) + 4H*(aq) + 3e→ NO(g) + 2H20(1) Anode reaction: E° = -0.7990 V Cathode reaction: ɛ° = 0.9637 V Determine the standard cell potential. a. 0.5876 V b. 2.0921 V с. 3.5254 V d. 0.1647 V е. —1.7627 V