Use the provided reduction potentials to calculate the
equilibrium constant (K) for the following balanced redox
reaction at 25 °C:
2Al(s) + 3Mg2+(aq) →
2Al3+(aq) + 3Mg(s)
E°(Al3+/Al) = -1.66 V and
E°(Mg2+/Mg) = -2.37 V
| 1.0 × 1024 |
| 9.7 × 10-73 |
| 1.1 × 1072 |
| 4.6 × 1031 |
| 8.9 × 10-70 |
Use the provided reduction potentials to calculate the equilibrium constant (K) for the following balanced redox...
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)
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
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...
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
30) Use the tabulated half-cell potentials below to calculate the equilibrium constant (K) for the following balanced redox reaction at 25°C. Pb2+(aq) + Cu(s) → Pb(s) + Cu2+(aq) Pb2+(aq) + 2e → Pb(s) Cu2+ (aq) +2e → Cu(s) E° = -0.13 V E = 0.34 V C) 7.9 x 1015 A) 7.9 x 10-8 D) 1.3 x 10-16 B) 8.9 x 107 E) 1.1 x 10-8
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 standard reduction potential for Al3+ to Al is ?°red = −1.662 ? and the standard reduction potential for Co2+ to Co is ?°red = −0.277. Given this information calculate the maximum amount of work, in kJ/mol rxn for the balanced redox reaction from 2Al(s)+3Co2+(aq)----> 2Al3+(aq)+3Co(s) (T = 25.0 °C). Is this a spontaneous reaction? why or why not
6. Calculate the equilibrium constant (K) for the following redox reaction at 25 ?C. Ecell = -0.71 V 2 Al(s) + 3 Mg2+(aq) --> 2 Al3+(aq) + 3 Mg(s
The equilibrium constant, K, for a redox reaction is related to the standard potential, Eº, by the equation In K = 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. Standard reduction potentials Reduction half-reaction E° (V) Ag+ (aq) + e +Ag(s) 0.80 Cu²+ (aq) + 2e + Cu(s) 0.34...
Calculate the equilibrium constant for this reaction at 25º C, Al (s) + Fe3+ (aq) —> Al3+ (aq) + Fe (s) given the following standard reduction potentials for the two half-reactions. Fe3+ (aq) + 3 e- —> Fe (s) Eº = 0.77 V Al3+ (aq) + 3 e- —> Al (s) Eº = -1.66 V