SIMULATION Voltaic Cells Under Standard ConditionsOxidizingReducingElectrodeE° cathodeEOanode = EºcellAgentAgentPotential (V)Au3+(aq) + 3 eAu(s)+1.50Br2(t) + 2 e2 Br (aq)+1.08CathodeAnodeHg2 (aq) + 2 eHg(t)+0.855Ag (aq) + eAg(s)+0.80Hg2 (aq) + 2 e2 Hg(t)+0,789Cu2*(aq) + 2 eCu(s)+0.337Sn1+(aq) + 2 eSn²*(s)+0.152 H30 (aq) + 2 e H2(g) + 2 H20(s)0.00Sn2"(aq) + 2 eSn(s)-0.14Ni2*(aq) + 2 eNi(s)-0.25Cd2+(ag) + 2 eCd(s)-0.40Al3+(aq) + 3 eAl(s)-1.66Mg2 (aq) + 2 eMg(s)-2.37Li"(aq) + eLi(s)-3.045Click on an Oxidizing Aget and a Reducing Agent.Choose Au3+ as the oxidizing agent and Ni as the reducing agent. What is the value of E° cell?VGibbsFreeEnergy

In this question, we want to determine the value of standard potential for cell between Au3+ and Ni.…

Oxidizing Reducing Electrode E°cathode EO anode = E°cell Agent Agent Potential (V) Au (aq) + 3 e Au(s) +1.50 Br2(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 Ni(s) -0.25 Ni2* (aq) + 2 e Cd2+(ag) + 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 Ageat and a Reducing Agent. Choose Aut as the oxidizing agent and Ni as the reducing agent. What is the value of E°cell? V. 1 1 1 Gibbs Free Energy

Oxidizing Reducing Electrode E°cathode EO anode = E°cell Agent Agent Potential (V) Au (aq) + 3 e Au(s) +1.50 Br2(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 Ni(s) -0.25 Ni2* (aq) + 2 e Cd2+(ag) + 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 Ageat and a Reducing Agent. Choose Aut as the oxidizing agent and Ni as the reducing agent. What is the value of E°cell? V. 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...

See similar textbooks

Similar questions

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
Part A If a voltaic cell is created between a Am"/Am electrode and a Fe*/Fe electrode, what species is being oxidized? O Am"(ag) O Fe(s) O Fe"(ag) O Am(s)
9
Fe (s) +3 Au (aq) ^ + Fe log ^ 3+ +3 Au (s) For the first part, identify the oxidation half-reaction , identify the reduction half- reaction, identify the oxidizing agent, and identify the reducing agent under standard conditions. Next, calculate the standard cell potential, E for the reaction . cell E^ red +1.69 V Half-reaction Au^ + (aq) + e^ - Au (s) Fe^ 3+ (aq) + 3 e^ - Fe (s) -0.040 V Now , calculate the non- standard cell potential , when [Au^ + ]=0.0015 M and the [Fe^ 3+ ]= 0.033 M. After finding the cell potential , calculate the Gibb's free energy , AG Finally, calculate the equilibrium constant , K. Express your answers in two significant digits . Show your work for partial credit . Under non -standard conditions , is this reaction spontaneous or nonspontaneous ? Are reactants or products favored at equilibrium under non -standard conditions ?
Consider the following electrochemical cell at 25 C: Pt|H2(g,p)|HCl(aq), 0.010 mol.kg-1|AgCl(s)|Ag For this cell the cell reaction is as follows: 2AgCl(s) + H2(g)  2Ag(s) + 2HCl(aq) (a) Give the electrode half-reactions and determine the standard cell potential.
Consider the reaction corresponding to a voltaic cell and its standard cell potential. Zn(s)+Cu2+(aq)⟶Cu(s) +Zn2+(aq) Eocell=1.1032 V What is the cell potential for a cell with a 2.214 M solution of Zn2+(aq) and 0.1456 M solution of Cu2+(aq) at 437.7 K?
Calculate the cell potential for the galvanic cell in which the given reaction occurs at 25 °C, given that [Sn²+] = 0.00140 M and [Au³+] = 0.841 M. Standard reduction potentials can be found in this table. 3 Sn(s) + 2 Au³+(aq) = 3 Sn²+(aq) + 2 Au(s) E = V * TOOLS x10
Standard Electrode Potentials at 25 °C Reduction Half-Reaction O Cu (aq) O H,Ozlaq) O Br O 10₂ (aq) Fyl) +20 H₂O₂(aq) + 2 H(aq) +2e" PbOg(s) +4 H(aq) +50, (aq) + 2 e Mn0 (aq) +4 H(aq) + 3 e MnO, (aq) +8H(aq) + 5 e Au³(aq) + 3 e PbOy(s) +4 H(aq) + 2e¯ C+20 00, (aq) +14 H(aq) + 6 e 00+ 4H*(aq) + 4 MnO₂ (s) +4 H(aq) +2e¯ 10₂(aq) + 6H(aq) +5e Big(0+20 VO₂ (aq) + 2 H(aq) + e¯ NO₂ (aq) + 4H(aq) + 3 e CO₂(0)+ Ag (ag)+e Fe(aq) + e 0)+2 H(aq) +2 e Mn0, (aq) + e (8) +26 Cu (aq) + e -2 (aq) -2 H₂O(0) Which of the following is the strongest oxidizing agent? -PbSO4(s) + 2 H₂O(0) →→MnO₂ (s) + 2 H₂O(0) Mn²(aq) + 4H₂O(1) - Au(s) ·Pb²+ (aq) + 2 H₂O(0) - 2 Cr (aq) 2 Cr³+ (aq) + 7 H₂O(1) → 2 H₂O( -2(aq) -Cu(s) Mn²+ (aq) + 2 H₂O(0) ₂(aq) + 3 H₂0(0) -2 Br (aq) - Vo²+ (aq) + H₂O( -NO(g) + 2 H₂O(1) — CIO; (aq) -Ag(s) -Fe²+ (aq) -) HyOz(ag) MnO₂ (aq) 1.00 0.96 0.95 0.80 0.77 0.70 0.56 0.54 0.52 E° (V) 2.87 1.78 1.69 1.68 1.51 1.50 1.46 1.36 1.33 1.23 1.21 1.20 1.09
Consider the same diagram of a galvanic cell as in previous question, conventional current anode X X+ (aq) (C) 0.76 V (D) 0.91 V A e flow Salt bridge Y+ (aq) cathode where the half-cells are Ag|AgNO3(aq) and Ni|Ni(NO3)2(aq). Given the following electrode potentials: Ag (aq) + e-> Ag(s); E° = 0.80 V Ni²+ (aq) + 2e --> Ni(s): E = -0.23 V What is the Ecell at 25°C when the electrolyte concentrations are [Ag*] = 1.3 x 10-5 M and [Ni²+] = 0.10 M. (A) 1.15 V (B) 1.30 V
A fictitious electrochemical cell has anode A and cathode B with a standard cell potential of 1.01 V. Currently the concentration of A + is 0.0650 M and the concentration of B 3+ is 0.500 M. If 3 moles of electrons are transferred, what is the cell potential under these non-standard conditions?3 A (s) + B 3+ (aq) → 3 A + (aq) + B (s) 1.47 V 0.818 V 1.60 V 1.22 V 1.07 V