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 SolutionStandard Electrode Potential, E(volts)H2O2(aq) + 2 H'(aq) + 2 e →2 H2O(t)+1.77Au*(aq) + e → Au(s)+1.68Au (aq) + 3 e –→ Au(s)+1.50>Br2(t) + 2 e → 2 Br(aq)NO: (aq) + 4 H'(aq) + 3 e → NO(g) + 2 H20Ag*(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+0.337Hg2Clz(s) + 2 e -2 Hg(t) + 2 C(aq)+0.27->Sn* (aq) + 2e –→ Sn²*(aq)+0.15Hz(g)2 H*(aq) + 2 e -Pb"(aq) + 2 e –→ Pb(s)Sn²"(aq) + 2 e –→ Sn(s)>0.00-0.126-0.14Ni? (aq) + 2 e – Ni(s)-0.25Cd"(aq) + 2 e – Cd(s)-0.40Cr**(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)-0.408Fe(s)-0.44-0.763-0.91-1.66-2.37VSubmitShow Approach Show Tutor Steps

Answered: Image /qna-images/answer/ccb90563-5e11-4715-a148-b3f99f615de9.jpg

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 Fe*(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) Au (aq) + 3 e - Au(s) +1.68 +1.50 Br2(t) + 2 e → 2 Br(aq) +1.08 NO: (aq) + 4 H'(aq) + 3 e NO(g) + 2 H20 +0.96 Ag*(aq) + e Ag(s) +0.80 Hg.2"(aq) + 2 e –→2 Hg(t) Fe"(aq) + e-→ Fe2 (aq) Cu2"(aq) + 2 e – Cu(s) +0.789 +0.77 +0.337 HgzClz(s) + 2 e - 2 Hg(t) + 2 C(aq) +0.27 Sn* (aq) + 2e –→ Sn²*(aq) Hz(g) Pb"(aq) + 2 e – Pb(s) Sn2"(aq) + 2 e - Sn(s) Ni? (aq) + 2 e – Ni(s) Cd"(aq) + 2 e – Cd(s) → Cr*(aq) +0.15 2 H*(aq) + 2 e 0.00 > -0.126 -0.14 -0.25 -0.40 Cr**(aq) + e"- Fe2"(aq) + 2 e - Zn (aq) + 2 e →Zn(s) Cr"( aq) + 2 e – Cr(s) A* (aq) + 3 e – Al(s) -0.408 Fe(s) -0.44 > -0.763 -0.91 -1.66 Mg2"(aq) + 2 e » Mg(s) -2.37 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 Fe(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) Au (aq) + 3 e - Au(s) +1.68 +1.50 Br2(t) + 2 e → 2 Br(aq) +1.08 NO: (aq) + 4 H'(aq) + 3 e NO(g) + 2 H20 +0.96 Ag(aq) + e Ag(s) +0.80 Hg.2"(aq) + 2 e –→2 Hg(t) Fe"(aq) + e-→ Fe2 (aq) Cu2"(aq) + 2 e – Cu(s) +0.789 +0.77 +0.337 HgzClz(s) + 2 e - 2 Hg(t) + 2 C(aq) +0.27 Sn* (aq) + 2e –→ Sn²(aq) Hz(g) Pb"(aq) + 2 e – Pb(s) Sn2"(aq) + 2 e - Sn(s) Ni? (aq) + 2 e – Ni(s) Cd"(aq) + 2 e – Cd(s) → Cr(aq) +0.15 2 H*(aq) + 2 e 0.00 > -0.126 -0.14 -0.25 -0.40 Cr**(aq) + e"- Fe2"(aq) + 2 e - Zn (aq) + 2 e →Zn(s) Cr"( aq) + 2 e – Cr(s) A* (aq) + 3 e – Al(s) -0.408 Fe(s) -0.44 > -0.763 -0.91 -1.66 Mg2"(aq) + 2 e » Mg(s) -2.37 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

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)
What is AG" for the following reaction? Use data given in the table. 2Br (ag) + Cla (g) -→ Bra (1) + 2C1 (ag) Standard Electrode (Reduction) Potentials in Aqueous Solution at 25°C Cathode (Reduction) Standard Half-Reaction Potential, E (V) Lit (ag) + e Li(s) Na+(ag) + e Na(s) Mg+ (ag) + 2e Mg(s) -3.04 -2.71 -2.38 Al* (ag) + 3e - Al(s) -1.66 2H20(1) + 2e H2(9) + 20H (ag) -0.83 Zn+ (ag) + 2e Zn(s) -0.76 Cr** (aq) + 3e Cr(s) --0.74 Fe+ (ag) + 2e Fe(s) -0.41 Ca2+ (aq) + 2e Cd(s) -0.40 Ni* (ag) + 2e Ni(s) -0.23 Sn* (aq) + 2e Sn(s) -0.14 Pb+ (ag) + 2e = Pb(s) --0.13 Fe* (ag) + 3e Fe(s) 2H* (ag) + 2e H2 (9) -0.04 0.00 Sn+ (ag) + 2e Sn+ (ag) 0.15
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 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 AG for the following cell reaction: Tl(s) | Tl+ (aq) || Ni²+ (aq) | Ni(s) From AG, determine the standard potential for the half-reaction Tl+ (aq) + e → Tl(s) AG; for Tl+ (aq) is -32.4 kJ/mol. 2+ AG; for Ni²+ (aq) is -45.60 kJ/mol. Standard Electrode (Reduction) Potentials in Aqueous Solution at 25°C Cathode (Reduction) Half-Reaction Ni²+ (aq) + 2e Ni(s) AG = J Standard Potential, E° (V) -0.23 Standard reduction potential for the half-reaction = V
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.
Balance the REDOX reaction associated with the cell shown below, in acidic medium: 2+ Fe(s) | Fe² || MnO (aq) Mn²) | Pt(s) Μη '(aq) 4(aq)' Enter the number for the coefficient of H₂O in the balanced equation:
A galvanic cell at a temperature of 25.0 °C is powered by the following redox reaction: 3Cu2* (ag) +2A1 (s) –- 3Cu (s) + 2AI³* (aq) 2+ in one half-cell and 4.12 M Al" in the other. 3+ Suppose the cell is prepared with 3.11 M Cu Calculate the cell voltage under these conditions. Round your answer to 3 significant digits.