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Question 6 0.5 pts Use tabulated half-cell potentials (found in appendix IV-D) to calculate Eºcell and...
Use tabulated half-cell potentials (found in appendix IV-D) to calculate Eºcell and then calculate AGºrxn for the f...
Use tabulated half-cell potentials (found in appendix IV-D) to calculate Eºcell and then calculate AGºrxn for the following reaction at 25 °C. 2Fe3+ (aq) + 3Sn(s) → 2Fe(s) + 3Sn2+ (aq) Hint: the standard potential of Fe3+ + 3e + Fe(s) is -0.036 and the standard potential of Sn2+ + 2e → Sn(s) is -0.140. AG° = -5 kJ/mol AG° = -10 kJ/mol AG° = -20 kJ/mol AG° = -30 kJ/mol
Calculate the equilibrium constant for each of the reactions at 25∘C. Standard Electrode Potentials at 25...
Calculate the equilibrium constant for each of the reactions at 25∘C. Standard Electrode Potentials at 25 ∘C Reduction Half-Reaction E∘(V) Fe3+(aq)+3e− →Fe(s) -0.036 Sn2+(aq)+2e− →Sn(s) -0.14 Ni2+(aq)+2e− →Ni(s) -0.23 O2(g)+2H2O(l)+4e− →4OH−(aq) 0.40 Br2(l)+2e− →2Br− 1.09 I2(s)+2e− →2I− 0.54 A) 2Fe3+(aq)+3Sn(s)→2Fe(s)+3Sn2+(aq) (answers are not 4.1x10^5, 3.3x10^3, 2.7x10^10, or 2.6x10^10) B) O2(g)+2H2O(l)+2Ni(s)→4OH−(aq)+2Ni2+(aq) C) Br2(l)+2I−(aq)→2Br−(aq)+I2(s) (answer is not 1.7x10^18)
2. Using the information provided, calculate the standard cell potential, Eºcell, for the reaction below: Half-reaction...
2. Using the information provided, calculate the standard cell potential, Eºcell, for the reaction below: Half-reaction Cr3+ (aq) + 3e- → Cr(s) Fe2+ (aq) + 2e- → Fe(s) Fe3+ (aq) + e- + Fe2+ (s) Sn4+ (aq) + 2e- + Sn2+ (aq) E° (V) -0.74 -0.440 +0.771 +0.154 3Sn(aq) +2Cr(s) →2Cr" (aq) +35n²+ (aq)
A) Use tabulated electrode potentials to calculate ΔG∘ for the reaction. 2K(s)+2H2O(l)→H2(g)+2OH−(aq)+2K+(aq) B) (Refer to the...
A) Use tabulated electrode potentials to calculate ΔG∘ for the reaction. 2K(s)+2H2O(l)→H2(g)+2OH−(aq)+2K+(aq) B) (Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V) An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:Ni(s)VO2+(aq,0.024M)+2H+(aq,1.4M)+e−→→Ni2+(aq,1.8M)+2e−VO2+(aq,1.8M)+H2O(l) Calculate the cell potential under these nonstandard concentrations. C) 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...
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)...
use tabulated standard electrode potential to calculate the standard cell potential for the reaction occurring in an el...
use tabulated standard electrode potential to calculate the
standard cell potential for the reaction occurring in an
electrochemical cell at 25 C. (The equation is balanced.)
3Ni^2+(aq)+2Cr(s)--->3Ni(s)2Cr^3+(aq)
Express your answer to two significant figures and include the
appropriate units.
em 26 E (V) -0.45 -0.50 -0.73 -0.76 -1.18 Standard reduction half-cell potentials at 25°C Half-reaction E° (V) Half-reaction Aul+ (aq) + 3e +Au(s) 1.50 Fe2+ (aq) + 2eFe(s) Ag+ (aq) +e-Ag(s) 0.80 Cr3+ (aq) + Cr²+ (aq) Fe+(aq) + 3e...
Standard reduction half-cell potentials at 25°C E (V) E (V) 1.50 -0.45 0.80 -0.50 0.77 -0.73...
Standard reduction half-cell potentials at 25°C E (V) E (V) 1.50 -0.45 0.80 -0.50 0.77 -0.73 0.52 -0.76 0.34 -1.18 Half-reaction Aut (aq) + 3e +Au(s) Ag+ (aq) + +Ag(s) Fe3+ (aq) +34 Fo+ (aq) Cut(aq) + Cu(s) Cu²+ (aq) + 2e +Cu(s) 2H+ (aq) - 2e +H2 (6) Fe3+ (aq) + 3e Fe(s) Pb2+ (aq) + 2e →Pb(s) Sn-(aq) + 2e +Sn(s) Ni2+ (aq) + 2e →Ni(s) Co2(aq) +2e + Co(s) ca? (aq) + 2e +Cd(s) 0.00 Half-reaction Fe(aq)...
1. Use the tabulated half-cell potentials to calculate AG for the following balanced redox reaction. SHow...
1. Use the tabulated half-cell potentials to calculate AG for the following balanced redox reaction. SHow your work 2 Fe3+(aq) + 6 1(aq) 3 12(s)+ 2 Fe(s)
need help for half cell potentials pls calculate step by step (NOTE - Remember that the...
need help for half cell potentials pls calculate step by step
(NOTE - Remember that the positive electrode is attached to the red wire and the negative electrode is attached to the black wire.) Electrode Systems Used Anode (oxidation) Negative Cathode (reduction) Positive Measured Potential (V) Positive (Ecu) Copper and silver Cu() - Cu2+ + 2e" Ag+ +1e Ag) 0.432 V Zinc + Silver Zn cs + 2n**+ Zé dat + leº nAg (s) 1.484 V Copper & Zinc Zn(s)...
8. (2 pts) Use the tabulated half-cell potentials to calculate AGⓇ for the following balanced redox...
8. (2 pts) Use the tabulated half-cell potentials to calculate AGⓇ for the following balanced redox reaction. 3 Mg2+(aq) + 2 Al(s) 3 Mgs) + 2 A13+(aq) E° Mg2(aq) + 2e → Mg(s) R -2.37 V AP (aq) + 3e + Al(s) O -1.66 V Cathode Anode a. - 2.3 x 10² kJ b. +4.1 x 102 kJ c. +1.4 x 10² kJ d. - 7.8 x 10 kJ
Use tabulated half-cell potentials (found in appendix IV-D) to calculate Eºcell and then calculate AGºrxn for the following reaction at 25 °C. 2Fe3+ (aq) + 3Sn(s) → 2Fe(s) + 3Sn2+ (aq) Hint: the standard potential of Fe3+ + 3e + Fe(s) is -0.036 and the standard potential of Sn2+ + 2e → Sn(s) is -0.140. AG° = -5 kJ/mol AG° = -10 kJ/mol AG° = -20 kJ/mol AG° = -30 kJ/mol
2. Using the information provided, calculate the standard cell potential, Eºcell, for the reaction below: Half-reaction Cr3+ (aq) + 3e- → Cr(s) Fe2+ (aq) + 2e- → Fe(s) Fe3+ (aq) + e- + Fe2+ (s) Sn4+ (aq) + 2e- + Sn2+ (aq) E° (V) -0.74 -0.440 +0.771 +0.154 3Sn(aq) +2Cr(s) →2Cr" (aq) +35n²+ (aq)
use tabulated standard electrode potential to calculate the
standard cell potential for the reaction occurring in an
electrochemical cell at 25 C. (The equation is balanced.)
3Ni^2+(aq)+2Cr(s)--->3Ni(s)2Cr^3+(aq)
Express your answer to two significant figures and include the
appropriate units.
em 26 E (V) -0.45 -0.50 -0.73 -0.76 -1.18 Standard reduction half-cell potentials at 25°C Half-reaction E° (V) Half-reaction Aul+ (aq) + 3e +Au(s) 1.50 Fe2+ (aq) + 2eFe(s) Ag+ (aq) +e-Ag(s) 0.80 Cr3+ (aq) + Cr²+ (aq) Fe+(aq) + 3e...
Standard reduction half-cell potentials at 25°C E (V) E (V) 1.50 -0.45 0.80 -0.50 0.77 -0.73 0.52 -0.76 0.34 -1.18 Half-reaction Aut (aq) + 3e +Au(s) Ag+ (aq) + +Ag(s) Fe3+ (aq) +34 Fo+ (aq) Cut(aq) + Cu(s) Cu²+ (aq) + 2e +Cu(s) 2H+ (aq) - 2e +H2 (6) Fe3+ (aq) + 3e Fe(s) Pb2+ (aq) + 2e →Pb(s) Sn-(aq) + 2e +Sn(s) Ni2+ (aq) + 2e →Ni(s) Co2(aq) +2e + Co(s) ca? (aq) + 2e +Cd(s) 0.00 Half-reaction Fe(aq)...
1. Use the tabulated half-cell potentials to calculate AG for the following balanced redox reaction. SHow your work 2 Fe3+(aq) + 6 1(aq) 3 12(s)+ 2 Fe(s)
need help for half cell potentials pls calculate step by step
(NOTE - Remember that the positive electrode is attached to the red wire and the negative electrode is attached to the black wire.) Electrode Systems Used Anode (oxidation) Negative Cathode (reduction) Positive Measured Potential (V) Positive (Ecu) Copper and silver Cu() - Cu2+ + 2e" Ag+ +1e Ag) 0.432 V Zinc + Silver Zn cs + 2n**+ Zé dat + leº nAg (s) 1.484 V Copper & Zinc Zn(s)...
8. (2 pts) Use the tabulated half-cell potentials to calculate AGⓇ for the following balanced redox reaction. 3 Mg2+(aq) + 2 Al(s) 3 Mgs) + 2 A13+(aq) E° Mg2(aq) + 2e → Mg(s) R -2.37 V AP (aq) + 3e + Al(s) O -1.66 V Cathode Anode a. - 2.3 x 10² kJ b. +4.1 x 102 kJ c. +1.4 x 10² kJ d. - 7.8 x 10 kJ
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