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39. Given: E +0.46 V Eo- +0.34 V Cu(s) + 2 Ag+ → 2 Ag (s) + Cu2. Cu2+ + H2 (g) → Cu (s) + 2 H' Find the standar...
Half-Reaction Fe2+(aq) + 2e Fe(s) Hg2+ (aq) + 2e Hg() Ag+ (aq) + e + Ag(s)...
Half-Reaction Fe2+(aq) + 2e Fe(s) Hg2+ (aq) + 2e Hg() Ag+ (aq) + e + Ag(s) Cu2+ (aq) + 2e + Cu(s) Zn2+ (aq) + 2e → Zn(s) E (V) -0.44 0.86 0.80 0.34 - 0.76 Using the table, calculate Eºcell for the following electrochemical cell under standard conditions voltmeter a) 1.24 V Fe. salt bridge Ag b) -1.24 V c) 2.04 V d) - 2.04 V Ag a b С
Fill in the Blanks 0.34 Half Reaction E. (V) Half Reaction Erd® (V) F + 2e...
Fill in the Blanks 0.34 Half Reaction E. (V) Half Reaction Erd® (V) F + 2e →2F 0+ 2H + 2e →H,02 0.68 Ag* + e → Ag Cu* + e → Cu 0.52 Co? + e Co? O2 + 2H,0 + 4e → 40H 0.40 H,O, + 2H + 2e → 2H,0 Cu2+ + 2e → Cu Cet+e → Ces Cu2+ + e → Cut 0.16 PbO, + 4H+ +50,2 +2e → PbSO, + 2H,0 2H* + 2e →...
The % difference = Standard E° (using table) = (show calculation) Ee=Fe+2/ Fe= -0.44V Eo=Cu2+/ Cu=...
The % difference = Standard E° (using table) = (show calculation) Ee=Fe+2/ Fe= -0.44V Eo=Cu2+/ Cu= +0.34 V Eo cell=Eocathode-Eganode = Ee Cu+2/cu- Eo Fe2+/ Fe = +0.34+ 0.44 V = +0.78 V Note that the list of reduction reactions includes the reduction of ironfil) to elemental iron, and also the reduction of iron(III) to iron(II). Is your data more consistent with the formation of Fe? or Felt? Explain your answer.
Consider the 2 half-reactions: Cu2+(aq) + 2e- --> Cu(s) Eo= 0.34V Ag+(aq) + e- --> Ag(s)...
Consider the 2 half-reactions: Cu2+(aq) + 2e- --> Cu(s) Eo= 0.34V Ag+(aq) + e- --> Ag(s) Eo= 0.8V If you increase the concentration of Cu2+(aq), which of the following would be true about the cell potential E? It would remain constant It would increase It would decrease This cannot be determined
2.87 Ered® (V) 0.68 0.52 0.40 0.34 0.16 Half Reaction F,+ 2e →2F Ag* + e...
2.87 Ered® (V) 0.68 0.52 0.40 0.34 0.16 Half Reaction F,+ 2e →2F Ag* + e → Ag Co3 + e + CO2- H2O2 + 2H+ + 2e → 2H,0 Ce4+ + e → Ce+ PbO, + 4H+ + SO42- + 2e → PbSO, + 2H,0 Mno, + 4H+ + 3e → MnO2 + 2H,0 2e + 2H+ + 10, → 103 + H2O Mn0, +8H+ + 5e → Mn2+ + 4H,0 Aul+ + 3e → Au Cl2 + 2e...
1. What is E° for the reaction Cu+ + Ag+ → Cu2+ + Ag ? 2....
1. What is E° for the reaction Cu+ + Ag+ → Cu2+ + Ag ? 2. What is the standard cell potential at 298K for the reaction: 2IO3- + 2H+ + 5H2O2→ I2 + 5O2 + 6H2O 3. Will Cu react with 1M HCl?
5. Consider the following set of half reactions Ag (aq)Ag (s) Cu2(a)2 e- Cu (s) Ai...
5. Consider the following set of half reactions Ag (aq)Ag (s) Cu2(a)2 e- Cu (s) Ai (aq)3eAl (s) Eo0.7996 V E0.342 V Eo-1.6632 V i. Which pair of half reactions would make the battery with the largest E cell? i. Identify the cathode and the anode for this battery, and calculate the value of E cell. i. Write the overall redox reaction that would take place in the battery above.
Use the tabulated electrode potentials to calculate K for the oxidation of nickel by H+: Ni(s)+2H+(aq)→Ni2+(aq)+H2(g)...
Use the tabulated electrode potentials to calculate K for the oxidation of nickel by H+: Ni(s)+2H+(aq)→Ni2+(aq)+H2(g) Express your answer using two significant figures. Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V) Half-reaction E∘ (V) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) −0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) −0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) −0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) −0.76 Cu2+(aq)+2e−→Cu(s) 0.34 Mn2+(aq)+2e−→Mn(s) −1.18 2H+(aq)+2e−→H2(g) 0.00 Al3+(aq)+3e−→Al(s) −1.66 Fe3+(aq)+3e−→Fe(s) −0.036 Mg2+(aq)+2e−→Mg(s) −2.37 Pb2+(aq)+2e−→Pb(s) −0.13 Na+(aq)+e−→Na(s) −2.71 Sn2+(aq)+2e−→Sn(s) −0.14 Ca2+(aq)+2e−→Ca(s) −2.76 Ni2+(aq)+2e−→Ni(s) −0.23 Ba2+(aq)+2e−→Ba(s) −2.90 Co2+(aq)+2e−→Co(s) −0.28 K+(aq)+e−→K(s) −2.92 Cd2+(aq)+2e−→Cd(s)...
Calculate the E°cell for the following reaction: Cu2+ (aq) + Ni (s) → Cu (s)...
Calculate the E°cell for the following reaction: Cu2+ (aq) + Ni (s) → Cu (s) + Ni2+ (aq) A) (-0.59 ± 0.01) V B) (-0.09 ± 0.01) V C) (0.59 ± 0.01) V D) (0.09 ± 0.01) V --------------------------------------------------------------------------------------------------------- What is the proper line notation for the following reaction? Cu (s) + 2Ag+ (aq) → Cu2+ (aq) + 2Ag (s) A) Cu2+ | Cu || Ag | Ag+ B) Ag+ | Ag || Cu | Cu2+ C) Cu |...
please show work and how they got this answer 12. Consider the cell reaction Cu(s) +...
please show work and how they got this answer
12. Consider the cell reaction Cu(s) + 2Agt → Cu2+ + 2Ag(s). If [Ag+] = 1.0M and Ecell = 0.62 V, then what is the concentration of Cu2+7 Eºcell = +0.46 (0.80 -0.34), Q = x/1.02, x = [Cu2+] = 3.9x10-6
Half-Reaction Fe2+(aq) + 2e Fe(s) Hg2+ (aq) + 2e Hg() Ag+ (aq) + e + Ag(s) Cu2+ (aq) + 2e + Cu(s) Zn2+ (aq) + 2e → Zn(s) E (V) -0.44 0.86 0.80 0.34 - 0.76 Using the table, calculate Eºcell for the following electrochemical cell under standard conditions voltmeter a) 1.24 V Fe. salt bridge Ag b) -1.24 V c) 2.04 V d) - 2.04 V Ag a b С
Fill in the Blanks 0.34 Half Reaction E. (V) Half Reaction Erd® (V) F + 2e →2F 0+ 2H + 2e →H,02 0.68 Ag* + e → Ag Cu* + e → Cu 0.52 Co? + e Co? O2 + 2H,0 + 4e → 40H 0.40 H,O, + 2H + 2e → 2H,0 Cu2+ + 2e → Cu Cet+e → Ces Cu2+ + e → Cut 0.16 PbO, + 4H+ +50,2 +2e → PbSO, + 2H,0 2H* + 2e →...
The % difference = Standard E° (using table) = (show calculation) Ee=Fe+2/ Fe= -0.44V Eo=Cu2+/ Cu= +0.34 V Eo cell=Eocathode-Eganode = Ee Cu+2/cu- Eo Fe2+/ Fe = +0.34+ 0.44 V = +0.78 V Note that the list of reduction reactions includes the reduction of ironfil) to elemental iron, and also the reduction of iron(III) to iron(II). Is your data more consistent with the formation of Fe? or Felt? Explain your answer.
2.87 Ered® (V) 0.68 0.52 0.40 0.34 0.16 Half Reaction F,+ 2e →2F Ag* + e → Ag Co3 + e + CO2- H2O2 + 2H+ + 2e → 2H,0 Ce4+ + e → Ce+ PbO, + 4H+ + SO42- + 2e → PbSO, + 2H,0 Mno, + 4H+ + 3e → MnO2 + 2H,0 2e + 2H+ + 10, → 103 + H2O Mn0, +8H+ + 5e → Mn2+ + 4H,0 Aul+ + 3e → Au Cl2 + 2e...
5. Consider the following set of half reactions Ag (aq)Ag (s) Cu2(a)2 e- Cu (s) Ai (aq)3eAl (s) Eo0.7996 V E0.342 V Eo-1.6632 V i. Which pair of half reactions would make the battery with the largest E cell? i. Identify the cathode and the anode for this battery, and calculate the value of E cell. i. Write the overall redox reaction that would take place in the battery above.
please show work and how they got this answer
12. Consider the cell reaction Cu(s) + 2Agt → Cu2+ + 2Ag(s). If [Ag+] = 1.0M and Ecell = 0.62 V, then what is the concentration of Cu2+7 Eºcell = +0.46 (0.80 -0.34), Q = x/1.02, x = [Cu2+] = 3.9x10-6
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