OxidizingReducing- EoElectrodeE°cathodeanode = E°cellAgentAgentPotential (V)Au3*(aq) + 3 eAu(s)+1.50Br2(t) + 2 e2 Br (aq)+1.08CathodeAnodeHg2*(aq) + 2 eHg(1)+0.855Ag*(aq) + eAg(s)+0.80Hg22*(aq) + 2 e2 Hg(t)+0.789Cu2* (aq) + 2 eCu(s)+0.337Sn*(aq) + 2 eSn2*(s)+0.152 H30*(aq) + 2 eH2(g) + 2 H20(s)0.00Sn2* (aq) + 2 eNi2*(aq) + 2 ecd2*(aq) + 2 eSn(s)-0.14Ni(s)-0.25Cd(s)-0.40AI3* (ag) + 3 eAl(s)-1.66Mg2+(aq) + 2 eLi*(aq) + eMg(s)-2.37Li(s)-3.045Click on an Oxidizing Agent and a Reducing Agent.Select Ag+ as the oxidizing agent.What is the value of E°cathode?X VEnergyFreeGibbs111 11

Given: Ag+ as the oxidizing agent. To find: What is the value of Eºcathode ?

Oxidizing Reducing Electrode E°cathode - E°anode = E°cell 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 cd2*(aq) + 2 e Al3* (aq) + 3 e Ni(s) -0.25 Cd(s) -0.40 Al(s) -1.66 Mg2*(aq) + 2 e Li*(aq) + e Mg(s) -2.37 Li(s) -3.045 Click on an Oxidizing Agent and a Reducing Agent. Select Ag+ as the oxidizing agent. What is the value of E°cathode? x V Gibbs Free Energy

Oxidizing Reducing Electrode E°cathode - E°anode = E°cell 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 cd2(aq) + 2 e Al3* (aq) + 3 e Ni(s) -0.25 Cd(s) -0.40 Al(s) -1.66 Mg2(aq) + 2 e Li(aq) + e Mg(s) -2.37 Li(s) -3.045 Click on an Oxidizing Agent and a Reducing Agent. Select Ag+ as the oxidizing agent. What is the value of E°cathode? x V 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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3.2 calculate the standard cell potential of the following cell at 25C Sn(s)|Sn"(aq)||Cu?²*(aq)|Cu(s)
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
9
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 following galvanic cell at 25°C. Pt | Cr²+ (0.25 M), Cr³+ (2.0 M) || Co²+ (0.17 M) | Co The overall reaction and equilibrium constant value are given below. 2 Cr²+ "(aq) + Co2+ Calculate the cell potential & for this galvanic cell and AG for the cell reaction at these conditions. + (aq) → 2 Cr³+ (aq) + Co(s) K = 2.79x107
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
(a) In the electrolysis of aqueous Na2C204, how many liters of H2(g) (at STP) are generated by a current of 82.6 A for a period of 54.0 min? The unbalanced chemical reaction representing this electrolysis is shown below. Na2C204(aq) + H20(1) → CO2(g) + H2(g) + NaOH(aq) liters of H2(g) is generated by this electrolysis. (b) How many moles of NaOH(aq) are formed in the solution in this process? moles of NaOH(aq) are formed.
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