Using the standard reduction potentials Au3(aq) +3e Au(s) +1.37volt Ca2+ (aq) +2e= Ca(s) -2.74 volt Calculate the value of Eºcell in volt for the cell reaction: 2Au(s) +3Ca2+ (aq) = 2Au³+ (aq) + 3Ca(s) Report your answer with 2 places past the decimal place. Do not report unit your answer. Type your answer...

Using the standard reduction potentials Au3(aq) +3e Au(s) +1.37volt Ca2+ (aq) +2e= Ca(s) -2.74 volt Calculate the value of Eºcell in volt for the cell reaction: 2Au(s) +3Ca2+ (aq) = 2Au³+ (aq) + 3Ca(s) Report your answer with 2 places past the decimal place. Do not report unit your answer. Type your answer...

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 voltaic cell has the following initial state: Cr (s) | Cr3 (aq, 2.0 M) || Cr* (aq, 9.0 M) | Cr (s) Each half-cell has a volume of 1.00 L. Calculate the number of hours the cell can run at 26 A before reaching equilibrium. Enter your answer to 1 decimal place. Reduction half-reaction E (V) Au (aq)+e + Au(s) 1.69 + Ag(s) + Cu(s) Ag (aq) + e 0.80 Cu (aq) +e 0.52 Cu (aq) + 2e + Cu(s) 0.32 Ph2+ (aq)+2e + Pb(s) -0.13 + Sn(s) -0.14 (aq)+ 2e Ni2+ (aq) + 2e Co+(aq) + 2e Sn2+ + Ni(s) -0.23 + Co(s) -0.28 In+ (aq) + 3e + In(s) -0.34 Ca (aq) + 2e + Cd(s) -0.40 Fe (aq) + 2e + Fe(s) -0.45 Cr (aq) + 3e + Cr(s) -0.73 Zn2+ (aq) + 2e + Zn(s) -0.76
Give the standard cell potential of the following half reaction: Cr2O72-(aq) + 3H2O2(aq) + 8H+(aq) → 2Cr3+(aq) + 3O2(aq) + 7H2O 2.28 V 0.65 V -2.28 V -0.65 V
Using values from the tables of standard reduction potentials, calculate the cell potentials of the cell with the given reaction: I2(s)+Mg(s)⟶Mg2+(aq)+2I−(aq) Mg2+(aq)+2e−⟶Mg(s) E∘red=−2.425 VI2(s)+2e−⟶2I−(aq) E∘red= 0.522 V
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 >
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
Calculate Eº, E, and AG for the following cell reaction: 2+ 2+ Mg(s) + Sn²+ (aq) = Mg²+ (aq) + Sn (s) 2+ 2 where [Mg²+] = [Mg²+] = 0.040 M and [Sn²+] = 0.030 M. Round each of your answers to 4 significant digits. Note: Reference the Standard reduction potentials at 25 °C table for additional information. Note: The Faraday constant is 9.649 × 10ª Part: 0 / 3 Part 1 of 3 Eº = v V x10 J V. mol S I
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
For the galvanic cell: Zn(s) + 2Tl+(aq) 2Tl(s) + Zn2+(aq) Eocell = 0.42 V The standard reduction potential for the half cell Zn2+(aq) + 2e– Zn(s) is Eored = –0.76 V What is the standard reduction potential for Tl+(aq) + e– Tl(s) Eored = ? V a. +0.17 V b. –0.17 V c. –0.34 V d. +0.34 V
An Ag/AgCl electrode dipping into 1.00 M HCI has a standard reduction potential of +0.2223 V. The half-reaction is AgCl(s) + e = Ag(s) +Cl'(aq) A second Ag/AGCI electrode is dipped into a solution containing Cl at an unknown concentration. The cell generates a potential of 0.0402 V, with the electrode in the solution of unknown concentration having a negative charge. What is the molar concentration of CI in the unknown solution? [CI] = M
Using standard potentials given in the appendices, calculate the standard cell potentials for the following reaction: Cu(s) + 2Ag+ (aq) = Cu²+ + 2Ag(s) Report the answer in volts with three significant figures. 0.462 V Submit Previous Answers Correct Correct answer is shown. Your answer 0.46 V was either rounded differently or used a different number of significant figures than required for this part. Part B Determine the equilibrium constant for the above reaction.
Please give the line notation for each cell. Use a small L Or a capital i for the line separator. Give the Eo (standard cell potential) of the following equations. Assume at standard conditions. Show work, which reaction you reverse and the total voltage answer. Cr3+(aq) + Cl2(g) ⇐> Cr2O72- (aq) + Cl- (aq) Cu2+ (aq) + Mg(s) ⇐> Mg2+(aq) + Cu(s)