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The equilibrium constant, K, for a redox reaction is related to the standard potential, Eº, by...
Cell Potential and Equilibrium Standard reduction potentials The equilibrium constant, K, for a redox reaction is...
Cell Potential and Equilibrium Standard reduction potentials The equilibrium constant, K, for a redox reaction is related to the standard cell potential, Ecel, by the equation Reduction half-reaction (V) Ag+ (aq) + e-→Ag(s) Cu2+ (aq) + 2e-→Cu(s) 0.34 Sn (a) 4e-Sn(s 0.15 2H' (aq) + 2e-→H2 (g) Ni2+ (aq) + 2e-→Ni(s)-0.26 Fe2+ (aq) + 2e-→Fe(s)-0.45 Zn2+ (aq) + 2e-→Zn(s)-0.76 Al3+ (aq) +3e-→Al(s) -1.66 Mg2+ (aq) + 2e-→Mg(s) -2.37 0.80 n FEcell where n is the number of moles of electrons...
Review I Constants I Periodic Table The equilibrium constant, K, for a redox reaction is related...
Review I Constants I Periodic Table The equilibrium constant, K, for a redox reaction is related to the standard potential, E°, by the equation Standard reduction potentials nFE RT In K E° (V) Reduction half-reaction Agt(aq)eAg(s) Cu2+(aq)2eCu(s) 0.80 where n is the number of moles of electrons transferred, F (the Faraday constant) is equal to 96,500 C/(mol e), R (the gas constant) is equal to 8.314 J/(mol - K) , and T is the Kelvin temperature. 0.34 Sn4t (aq)4eSn(s) 0.15...
The equilibrium constant, K. for a redox reaction is related to the standard potential, E, by...
The equilibrium constant, K. for a redox reaction is related to the standard potential, E, by the equation Fe(s) + Ni+ (aq) +Fe?+ (aq) + NI(s) FE In K = Express your answer numerically. View Available Hints) where n is the number of moles of electrons transferred, F (the Faraday constant) is equal to 96,500 C/(mole). R (the gas constant) is equal to 8.314 J/(mol-K). and T is the Kelvin temperature. ΟΙ ΑΣΦ h ? KK- Submit Previous Answers *...
The equilibrium constant, K, for a redox reaction is related to the standard potential, E∘, by...
The equilibrium constant, K, for a redox reaction is related to the standard potential, E∘, by the equation lnK=nFE∘RT where n is the number of moles of electrons transferred, F (the Faraday constant) is equal to 96,500 C/(mol e−) , R (the gas constant) is equal to 8.314 J/(mol⋅K) , and T is the Kelvin temperature. Calculate the standard cell potential (E∘) for the reaction X(s)+Y+(aq)→X+(aq)+Y(s) if K = 5.51×10−3.
Free-energy change, AGº, is related to cell potential, Eº, by the equation AG° = -nFE° where...
Free-energy change, AGº, is related to cell potential, Eº, by the equation AG° = -nFE° where n is the number of moles of electrons transferred and F = 96,500 C/(mol e ) is the Faraday constant. When Eº is measured in volts, AGⓇ must be in joules since 1 J =1C.V. Part A Calculate the standard free-energy change at 25°C for the following reaction: Mg(s) + Fe2+ (aq)Mg2+ (aq) + Fe(s) Express your answer to three significant figures and include...
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...
- cellpotentials and standard reduction potential table. -Determining the Nernst equation and finding the Faraday constant....
- cellpotentials and standard reduction potential table.
-Determining the Nernst equation and finding the Faraday
constant.
If you can explain how to solve for each part please.
25°C Standard Reduction Potentials in Aqueous solution a 2.87 Reduction half-reaction 2F (a 1.77 2H2O 1.692 2e 2H (a Au (s 1.085 PbSO4 (s) 2H20. Au (ag) 2e 1.51 4H20 Mn 1.50 5e 8H (a Mno4 (a Au(S 1.36 3e 2Cl (aq 1.33 2e 2Cr3 (ag) 7H20 C12 6e- 1.229 14H 2H2O 1.08...
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)...
Selective Reduction The standard reduction potential for the half-reaction: Sn4+ + 2e - Sn2+ is +0.15...
Selective Reduction The standard reduction potential for the half-reaction: Sn4+ + 2e - Sn2+ is +0.15 V. Consider data from the table of standard reduction potentials for common half-reactions, in your text. For a galvanic cell under standard conditions, which of the following anodic half reactions would produce, at the cathode a spontaneous reduction of Sn4+ to Sn2+ but not Sn2+ to Sn. no yes yes yes yes yes Fe — Fe2+ + 2e- Sn2+ Sn4+ + 2e- Sn Sn2+...
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...
Cell Potential and Equilibrium Standard reduction potentials The equilibrium constant, K, for a redox reaction is related to the standard cell potential, Ecel, by the equation Reduction half-reaction (V) Ag+ (aq) + e-→Ag(s) Cu2+ (aq) + 2e-→Cu(s) 0.34 Sn (a) 4e-Sn(s 0.15 2H' (aq) + 2e-→H2 (g) Ni2+ (aq) + 2e-→Ni(s)-0.26 Fe2+ (aq) + 2e-→Fe(s)-0.45 Zn2+ (aq) + 2e-→Zn(s)-0.76 Al3+ (aq) +3e-→Al(s) -1.66 Mg2+ (aq) + 2e-→Mg(s) -2.37 0.80 n FEcell where n is the number of moles of electrons...
Review I Constants I Periodic Table The equilibrium constant, K, for a redox reaction is related to the standard potential, E°, by the equation Standard reduction potentials nFE RT In K E° (V) Reduction half-reaction Agt(aq)eAg(s) Cu2+(aq)2eCu(s) 0.80 where n is the number of moles of electrons transferred, F (the Faraday constant) is equal to 96,500 C/(mol e), R (the gas constant) is equal to 8.314 J/(mol - K) , and T is the Kelvin temperature. 0.34 Sn4t (aq)4eSn(s) 0.15...
The equilibrium constant, K. for a redox reaction is related to the standard potential, E, by the equation Fe(s) + Ni+ (aq) +Fe?+ (aq) + NI(s) FE In K = Express your answer numerically. View Available Hints) where n is the number of moles of electrons transferred, F (the Faraday constant) is equal to 96,500 C/(mole). R (the gas constant) is equal to 8.314 J/(mol-K). and T is the Kelvin temperature. ΟΙ ΑΣΦ h ? KK- Submit Previous Answers *...
Free-energy change, AGº, is related to cell potential, Eº, by the equation AG° = -nFE° where n is the number of moles of electrons transferred and F = 96,500 C/(mol e ) is the Faraday constant. When Eº is measured in volts, AGⓇ must be in joules since 1 J =1C.V. Part A Calculate the standard free-energy change at 25°C for the following reaction: Mg(s) + Fe2+ (aq)Mg2+ (aq) + Fe(s) Express your answer to three significant figures and include...
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...
- cellpotentials and standard reduction potential table.
-Determining the Nernst equation and finding the Faraday
constant.
If you can explain how to solve for each part please.
25°C Standard Reduction Potentials in Aqueous solution a 2.87 Reduction half-reaction 2F (a 1.77 2H2O 1.692 2e 2H (a Au (s 1.085 PbSO4 (s) 2H20. Au (ag) 2e 1.51 4H20 Mn 1.50 5e 8H (a Mno4 (a Au(S 1.36 3e 2Cl (aq 1.33 2e 2Cr3 (ag) 7H20 C12 6e- 1.229 14H 2H2O 1.08...
Selective Reduction The standard reduction potential for the half-reaction: Sn4+ + 2e - Sn2+ is +0.15 V. Consider data from the table of standard reduction potentials for common half-reactions, in your text. For a galvanic cell under standard conditions, which of the following anodic half reactions would produce, at the cathode a spontaneous reduction of Sn4+ to Sn2+ but not Sn2+ to Sn. no yes yes yes yes yes Fe — Fe2+ + 2e- Sn2+ Sn4+ + 2e- Sn Sn2+...
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