OxidizingReducingElectrodeAgentAgentPotential (V)Au³+ (aq) + 3 eAu(s)+1.50Br₂(!) + 2 e2 Br (aq)+1.08Hg(t)+0.855Hg²+ (aq) + 2 eAg+ (aq) + eHg₂²+ (aq) + 2 eAg(s)+0.802 Hg(1)+0.789Cu²+ (aq) + 2 eCu(s)+0.337Sn4+ (aq) + 2 eSn²+(s)+0.152 H30+ (aq) + 2 eH₂(g) + 2 H₂O(s)0.00Sn²+ (aq) + 2 eSn(s)-0.14Ni²+ (aq) + 2 eNi(s)-0.25Cd²+ (aq) + 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 Agent and a Reducing Agent.Choose Au³+ as the oxidizing agent and Ni as the reducing agent. What is the value of Eºcell?0.70X V→→→→↑Eº cathodeGibbsFreeEnergyEº anode = Eºcell=CathodeAnode

Oxidising agent is the species which oxidises other and itself reduced Reducing agent is the species…

Oxidizing Reducing Electrode Agent Agent Potential (V) Au(s) +1.50 Au³+ (aq) + 3 e Br₂() + 2 e 2 Br²(aq) +1.08 Hg(1) +0.855 Hg2+ (aq) + 2 e Ag+ (aq) + e* Ag(s) +0.80 Hg₂2+ (aq) + 2 e 2 Hg(1) +0.789 Cu²+ (aq) + 2 e Cu(s) +0.337 Sn²+(s) +0.15 Sn4+ (aq) + 2 e 2 H30+ (aq) + 2 e H₂(g) + 2 H₂O(s) 0.00 Sn(s) -0.14 Sn²+ (aq) + 2 e 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+ 0.70 - → → → → → ↑ Eº cathode Gibbs Free Energy Eº anode Cathode = Eºcell Anode as the oxidizing agent and Ni as the reducing agent. What is the value of Eºcell? X V

Oxidizing Reducing Electrode Agent Agent Potential (V) Au(s) +1.50 Au³+ (aq) + 3 e Br₂() + 2 e 2 Br²(aq) +1.08 Hg(1) +0.855 Hg2+ (aq) + 2 e Ag+ (aq) + e* Ag(s) +0.80 Hg₂2+ (aq) + 2 e 2 Hg(1) +0.789 Cu²+ (aq) + 2 e Cu(s) +0.337 Sn²+(s) +0.15 Sn4+ (aq) + 2 e 2 H30+ (aq) + 2 e H₂(g) + 2 H₂O(s) 0.00 Sn(s) -0.14 Sn²+ (aq) + 2 e 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+ 0.70 - → → → → → ↑ Eº cathode Gibbs Free Energy Eº anode Cathode = Eºcell Anode as the oxidizing agent and Ni as the reducing agent. What is the value of Eºcell? X V

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

A chemist designs a galvanic cell that uses these two half-reactions: half-reaction standard reduction potential CrO2−4 (aq)+4H2O(l)+3e−→ CrOH3(s)+5OH−(aq) =E0red−0.13V Cl2(g)+2e−→ 2Cl−(aq) =E0red+1.359V Answer the following questions about this cell. Write a balanced equation for the half-reaction that happens at the cathode. Write a balanced equation for the half-reaction that happens at the anode. Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous as written. Do you have enough information to calculate the cell voltage under standard conditions? If you said it was possible to calculate the cell voltage, do so and enter your answer here. Be sure your answer has the correct number of significant digits. V
Given Pt2(aq) + 2 e Pt(s) Ag (aq) + e → Ag(s) 2 H*(aq) + 2e →→ H₂(g) °, = 0.00 V Fe2+ (aq) + 2e → Fe(s) = -0.45 V If the iron-iron ion electrode were the standard for values instead of the hydrogen-hydrogen ion electrode, the standard reduction potentials () for the four half-reactions above, in corresponding order, would be: Select one: O a. O b. +1.20 V - +0.80 V OC. O d. +1.65 V, +1.25 V, +0.45 V, 0.00 V -1.65 V, -1.25 V, -0.45 V, 0.00 V +1.20 V, +0.80 V, +0.45 V, 0.00 V -0.75 V, -0.35 V, -0.45 V, 0.00 V
Using the following standard reduction potentials E° = -2.92 V Cs*(aq) + e –→ Cs(s) Cr2*(aq) + 2 e » Cr(s) E° = -0.91 V calculate the standard cell potential and write the cell reaction for a voltaic cell. E°cell = -2.01 V and Cr2+(aq) + 2 Cs(s) → 2 Cs*(aq) + Cr(s) E°c cell = +2.01 V and Cr2*(aq) + 2 Cs(s) → 2 Cs*(ag) + Cr(s) None of these E°cell = +4.91 V and %3D Cr(s) + 2 Cs*(aq) →2 Cs(s) + Cr2+(aq) E°cell = -4.91 V and Cr(s) + 2 Cs*(aq) → 2 Cs(s) + Cr2+(aq)
Consider the following unbalanced oxidation-reductionreactions in aqueous solution:Ag+(aq) + Li(s)----> Ag(s) + Li+(aq)Fe(s) + Na+(aq)----> Fe2+ (aq) + Na(s)K(s) + H2O(l)---->KOH(aq) + H2(g)(a) Balance each of the reactions. (b) By using data inAppendix C, calculate ΔH° for each of the reactions.(c) Based on the values you obtain for ΔH°, which of the reactionswould you expect to be thermodynamically favored?(d) Use the activity series to predict which of these reactionsshould occur. Are these results in accordwith your conclusion in part (c) of this problem?
For a galvanic cell based on the following half-reactions: Fe3+(aq) + e- → Fe2+(aq)Ag+(aq) + e- → Ag(s)
Balance the following oxidation-reduction occurring in acidic solution. MnO 4 ^ - (aq)+Co^ 2+ (aq) Mn^ 2+ (aq)+Co^ 3+ (aq) Identify the oxidation half-reaction, reduction half-reactionoxidizing agent and reducing agent as well as balancing in acidic conditions.
A galvanic cell at a temperature of 25.0 °C is powered by the following redox reaction: Cu²* (ag)+ Zn (s) –→ Cu (s) + Zn²* (ag) 2+ 2+ Suppose the cell is prepared with 7.28 M Cu²" in one half-cell and 7.67 M Zn²™ in the other. Calculate the cell voltage under these conditions. Round your answer to 3 significant digits. x10 믐
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 ?
Show complete solution
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 galvanic cell at a temperature of 25.0 °C is powered by the following redox reaction: 3Cu* (aq)+ 2Al (s) → 3Cu (s) + 2A1³* (aq) 2+ in one half-cell and 1.05 M Al' in the other. .3+ Suppose the cell is prepared with 5.97 M Cu Calculate the cell voltage under these conditions. Round your answer to 3 significant digits.
Give the standard line notation for each cell. (a) 2 IO3 (aq) + 10 Fe2+ (ag) + 10 H* (aq) = 10 Fet (ag) + I2 (aq) + 5 H2O(1) O Pt(s) | Fe?+ (aq), Fe+ (ag) || IO3 (ag), Ht (ag), I2 (ag) | Pt(s) O Pt(s) | Fet (aq), I2 (ag) || Fe2+ (ag), IO3 (ag), H* (ag) | Pt(s) Pt(s) | IO3 (ag), H* (ag), I2 (aq) || Fe2+ (ag), Fet (ag) | Pt(s) Pt(s) | Fe2+ (ag), IO3 (ag), H* (aq) || Fet (aq), I2 (ag) | Pt(s) (b) Zn(s) + 2 Ag* (ag) = Zn?+ (aq) + 2 Ag(s) O Zn(s) | Zn?+ (aq) || Ag* (aq) | Ag(s) O Zn(s) | Ag* (aq) || Zn?+ (ag) | Ag(s) Ag(s) | Ag* (ag) || Zn2+ (aq) | Zn(s) O Ag(s) | Zn?+ (aq) || Ag* (aq) | Zn(s) (c) H2O2 (aq) + 2 H* (ag) + 2 e → 2 H2O(1) E = 1.78 V 02 (9) + 2 H* (ag) + 2 e + H2O2 (ag) E = 0.68 V O Pt(s) | O2(9) | H2O2 (aq), H* (ag) || H2O2 (ag), H* (ag) | Pt(s) Pt(s) | H2O2 (aq), H* (ag) || O2 (9) | H2O2 (ag), H* (ag) | Pt(s) Pt(s) | H2O2 (aq) | O2(9) || H* (aq) | Pt(s) O Pt(s) | H2O2(aq) | Pt(s) (d) Mn2+ (aq) + 2 e → Mn(s) E° = - 1.18 V Fe+ (aq) + 3 e → Fe(s) E - -0.036 V O Mn(s) | Mn²+ (ag)…