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This question is solved by using the simple concept of calculation of equilibrium constant using the redox reaction.
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30) Use the tabulated half-cell potentials below to calculate the equilibrium constant (K) for the following...
Use the tabulated half-cell potentials below to calculate AG ( in Kilojoules for the following balanced...
Use the tabulated half-cell potentials below to calculate AG ( in Kilojoules for the following balanced redox reaction. F= 96,500 J/. mole e Pb2+(aq) + Cu(s) - Pb(s) + Cu2+(aq) Cu2+ (aq) + 2e -> Cu(s) E". Cu = -0.34 Volt Pb2+ (aq) + 2e-> Pb(s) E,Pb = -0.13 volt a. -41 kJ b. -81 kJ C. +46 kJ d. +91 kl e..91 kb 24 Calculate AS®rxn for the following reaction. The S' for each species is shown below the...
25) Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced...
25) Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced redox reaction at 25°C. 2 Al(s)+3 Mg2+(aq) A) 1.1 x 1072 B) 8.9 x 10-73 C) 1.1 x 10-72 D) 1.0 x 1024 E) 4.6 x 1031 2 Al3+(aq) +3 Mg(s)
Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced redox reacti...
Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced redox reaction at 25°C. 3 I2(s) + 2 Fe(s) → 2 Fe3+(aq) + 6 I(aq) A.8.9 × 10-18 B.1.1 × 1017 C.1.7 × 1029 D.2.4 × 1058 E.3.5 × 10-59
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)...
se the tabulated half-cell potentials to calculate ??-for the following ,edo-m-mm in (7 points ) Tabulated...
se the tabulated half-cell potentials to calculate ??-for the following ,edo-m-mm in (7 points ) Tabulated half-cell potentials (Reduction) Zar" (aq) + 2 e-? Zn(s) E.--0763 V Cl:(g) +2e2 CT(a)+1.358 F-96,485 C/mol. electron, 13-1 C.V and 1kJ -1000 C). (7 points) 4) Calculate K for the oxidation of copper by H' (at 25 Cu(s) + 2 r(aq) ? Cu2+(aq) + H2(g) Tabulated half-cell potentials (Reduction) Cu2 (aq) + 2 e-Al(s) E-0.34 v 2H+(aq) + 2 e--+ H2 (g) E#20.00 V...
Write the half reactions and overall reaction for each cell with calculated overall potentials as shown...
Write the half reactions and overall reaction for each cell with calculated overall potentials as shown in Table 5-1. (Note: for the iron solutions the Nernst equation must be used) Pb(s) | Pb(NO3)2 (0.1M) || Cu(NO3)2 (0.1M) Cu(s) Zn(s) | Zn(NO3)2(0.1M) || Cu(NO3)2 (0.1M) Cu(s) Cds) | Ca(NO3)2 (0.1M) || Cu(NO3)2 (0.1M) | Cu(s) Cu() Cu(NO3)2(0.1M) Il Fe (0.1M/Fe? (0.1M graphite Pb(s) Pb(NO3)2(0.1M) Il Fe3(aq) (0.1M)/ Fe2(aq) (0.1MI graphite(s) Zns | Zn(NO3)2 (0.1M) || Pb(NO3)2 (0.1M) | Pb(s) Cdis Ca(NO3)2...
Using the standard reduction potentials given below, choose the reaction than can only be achieved through...
Using the standard reduction potentials given below, choose the reaction than can only be achieved through electrolysis. Cu2+(aq) + 2e → Cu(s) E° = 0.34 V Pb2+(aq) + 2e + Pb(s) E° = -0.13 V Fe2+(aq) + 2e Fe(s) E° = -0.44 V Zn2+(aq) + 2e + Zn(s) E° = -0.77 V Zn2+(aq) + Pb(s) → Zn(s) + Pb2+(aq) o Fe2+(aq) + Zn(s) → Fe(s) + Zn2+(aq) Pb2+(aq) + Fe(s) → Pb(s) + Fe2+(aq) Cu2+(aq) + Fe(s) → Cu(s) +...
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...
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 the tabulated half-cell potentials to calculate ΔG° for the following balanced redox reaction. (F =...
Use the tabulated half-cell potentials to calculate ΔG° for the following balanced redox reaction. (F = 96,485 C/mol e) 2Li( s) + Cl 2 ( g) → 2 Cl - ( aq) + 2Li +( aq) Eº Li+(aq) + e- → Li(s) -3.04V Cl2 (g) + 2e- → 2 Cl- (aq) +1.36 V a. -425 kJ b. -849 kJ c. -8.49 x 10 5 kJ d. +324 kJ
Use the tabulated half-cell potentials below to calculate AG ( in Kilojoules for the following balanced redox reaction. F= 96,500 J/. mole e Pb2+(aq) + Cu(s) - Pb(s) + Cu2+(aq) Cu2+ (aq) + 2e -> Cu(s) E". Cu = -0.34 Volt Pb2+ (aq) + 2e-> Pb(s) E,Pb = -0.13 volt a. -41 kJ b. -81 kJ C. +46 kJ d. +91 kl e..91 kb 24 Calculate AS®rxn for the following reaction. The S' for each species is shown below the...
25) Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced redox reaction at 25°C. 2 Al(s)+3 Mg2+(aq) A) 1.1 x 1072 B) 8.9 x 10-73 C) 1.1 x 10-72 D) 1.0 x 1024 E) 4.6 x 1031 2 Al3+(aq) +3 Mg(s)
se the tabulated half-cell potentials to calculate ??-for the following ,edo-m-mm in (7 points ) Tabulated half-cell potentials (Reduction) Zar" (aq) + 2 e-? Zn(s) E.--0763 V Cl:(g) +2e2 CT(a)+1.358 F-96,485 C/mol. electron, 13-1 C.V and 1kJ -1000 C). (7 points) 4) Calculate K for the oxidation of copper by H' (at 25 Cu(s) + 2 r(aq) ? Cu2+(aq) + H2(g) Tabulated half-cell potentials (Reduction) Cu2 (aq) + 2 e-Al(s) E-0.34 v 2H+(aq) + 2 e--+ H2 (g) E#20.00 V...
Write the half reactions and overall reaction for each cell with calculated overall potentials as shown in Table 5-1. (Note: for the iron solutions the Nernst equation must be used) Pb(s) | Pb(NO3)2 (0.1M) || Cu(NO3)2 (0.1M) Cu(s) Zn(s) | Zn(NO3)2(0.1M) || Cu(NO3)2 (0.1M) Cu(s) Cds) | Ca(NO3)2 (0.1M) || Cu(NO3)2 (0.1M) | Cu(s) Cu() Cu(NO3)2(0.1M) Il Fe (0.1M/Fe? (0.1M graphite Pb(s) Pb(NO3)2(0.1M) Il Fe3(aq) (0.1M)/ Fe2(aq) (0.1MI graphite(s) Zns | Zn(NO3)2 (0.1M) || Pb(NO3)2 (0.1M) | Pb(s) Cdis Ca(NO3)2...
Using the standard reduction potentials given below, choose the reaction than can only be achieved through electrolysis. Cu2+(aq) + 2e → Cu(s) E° = 0.34 V Pb2+(aq) + 2e + Pb(s) E° = -0.13 V Fe2+(aq) + 2e Fe(s) E° = -0.44 V Zn2+(aq) + 2e + Zn(s) E° = -0.77 V Zn2+(aq) + Pb(s) → Zn(s) + Pb2+(aq) o Fe2+(aq) + Zn(s) → Fe(s) + Zn2+(aq) Pb2+(aq) + Fe(s) → Pb(s) + Fe2+(aq) Cu2+(aq) + Fe(s) → Cu(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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