close
標題:
chemistry electrochemical cell
發問:
1.)Use the half-reaction shown below to set up the concentration cell as illustrated: click the link and see the image:https://access1.lon-capa.uiuc.edu/enc/37/82464006cf978e2483607eaeaecc875eea327d24f5ec42af528162dd40331e8cdd3e51951cb9a0fee8ec80af3711034b.gifLi+ (aq) + e- → Li (s) ξo... 顯示更多 1.)Use the half-reaction shown below to set up the concentration cell as illustrated: click the link and see the image: https://access1.lon-capa.uiuc.edu/enc/37/82464006cf978e2483607eaeaecc875eea327d24f5ec42af528162dd40331e8cdd3e51951cb9a0fee8ec80af3711034b.gif Li+ (aq) + e- → Li (s) ξo = -3.05 V There are Li (s) electrodes in both beakers. There is also a salt bridge joining the two beakers and a voltmeter to measure the potential of this concentration cell, at 25 Celsius. The beaker on the left contains: [Li+] = 0.2 M The beaker on the right contains: [Li+] = 0.4 M The initial potential of this concentration cell will be: a.)3.050 V b.)0.036 V c.)0.000 V d.)0.009 V e.)0.018 V Electrons will flow: a.)from the left beaker to the right beaker b.)from the right beaker to the left beaker c.)there will be no flow of electrons 2.)Consider a cell constructed by using the following half-reactions. Fe3+ + e- → Fe 2+ (aq) ξo = 0.77 V Fe2+ (aq) + 2 e- → Fe (s) ξo = -0.44 V By changing the initial concentration of Fe3+ from 1.0 M to 2.0 M, but leaving all of the other species at standard concentrations, you would change the cell potential. Under these, non-standard conditions, the cell potential, ξ, at 25o C would be: a.)1.25 V b.)1.23 V c.)1.22 V d.)1.30 V e.)1.19 V 3.)If you built a standard galvanic cell using the following half-reactions, what is the maximum work that could be done by this standard cell? (ΔGo= wmax). Pb2+ (aq) + 2 e- → Pb (s) ξo = - 0.13 V PbSO4 (s) + 2 e- → Pb (s) + SO42- (aq) ξo = - 0.35 V a.)-46 kJ b.)-93 kJ c.)-42 kJ d.)-21 kJ e.)-17 kJ Hope fun and enjoy the multiple choice. thank you
最佳解答:
(1) By the Nernst equation, potential will be: 2.303 x RT/(zF) x log (0.4/0.2) = 2.303 x 8.314 x 298/(1 x 96500) x log 2 = 0.018 V Since the right beaker has a higher [Li+], its potential is more negative and hence it will be the negative terminal, i.e. electrons flow from right to left. (2) In standard cell, the emf is 0.77 + 0.44 = 1.21 V When [Fe3+] changes from 1 M to 2 M, extra potential will be: 2.303 x RT/(zF) x log 2 = 2.303 x 8.314 x 298/(1 x 96500) x log 2 = 0.018 V So new potential = 1.21 + 0.018 = 1.23 V (3) Cell emf of the standard cell = 0.22 V No. of moles of electrons involved for each mole of cell reaction = 2 So max. work done = 0.22 x 96500 x 2 = 42 kJ
chemistry electrochemical cell
發問:
1.)Use the half-reaction shown below to set up the concentration cell as illustrated: click the link and see the image:https://access1.lon-capa.uiuc.edu/enc/37/82464006cf978e2483607eaeaecc875eea327d24f5ec42af528162dd40331e8cdd3e51951cb9a0fee8ec80af3711034b.gifLi+ (aq) + e- → Li (s) ξo... 顯示更多 1.)Use the half-reaction shown below to set up the concentration cell as illustrated: click the link and see the image: https://access1.lon-capa.uiuc.edu/enc/37/82464006cf978e2483607eaeaecc875eea327d24f5ec42af528162dd40331e8cdd3e51951cb9a0fee8ec80af3711034b.gif Li+ (aq) + e- → Li (s) ξo = -3.05 V There are Li (s) electrodes in both beakers. There is also a salt bridge joining the two beakers and a voltmeter to measure the potential of this concentration cell, at 25 Celsius. The beaker on the left contains: [Li+] = 0.2 M The beaker on the right contains: [Li+] = 0.4 M The initial potential of this concentration cell will be: a.)3.050 V b.)0.036 V c.)0.000 V d.)0.009 V e.)0.018 V Electrons will flow: a.)from the left beaker to the right beaker b.)from the right beaker to the left beaker c.)there will be no flow of electrons 2.)Consider a cell constructed by using the following half-reactions. Fe3+ + e- → Fe 2+ (aq) ξo = 0.77 V Fe2+ (aq) + 2 e- → Fe (s) ξo = -0.44 V By changing the initial concentration of Fe3+ from 1.0 M to 2.0 M, but leaving all of the other species at standard concentrations, you would change the cell potential. Under these, non-standard conditions, the cell potential, ξ, at 25o C would be: a.)1.25 V b.)1.23 V c.)1.22 V d.)1.30 V e.)1.19 V 3.)If you built a standard galvanic cell using the following half-reactions, what is the maximum work that could be done by this standard cell? (ΔGo= wmax). Pb2+ (aq) + 2 e- → Pb (s) ξo = - 0.13 V PbSO4 (s) + 2 e- → Pb (s) + SO42- (aq) ξo = - 0.35 V a.)-46 kJ b.)-93 kJ c.)-42 kJ d.)-21 kJ e.)-17 kJ Hope fun and enjoy the multiple choice. thank you
最佳解答:
(1) By the Nernst equation, potential will be: 2.303 x RT/(zF) x log (0.4/0.2) = 2.303 x 8.314 x 298/(1 x 96500) x log 2 = 0.018 V Since the right beaker has a higher [Li+], its potential is more negative and hence it will be the negative terminal, i.e. electrons flow from right to left. (2) In standard cell, the emf is 0.77 + 0.44 = 1.21 V When [Fe3+] changes from 1 M to 2 M, extra potential will be: 2.303 x RT/(zF) x log 2 = 2.303 x 8.314 x 298/(1 x 96500) x log 2 = 0.018 V So new potential = 1.21 + 0.018 = 1.23 V (3) Cell emf of the standard cell = 0.22 V No. of moles of electrons involved for each mole of cell reaction = 2 So max. work done = 0.22 x 96500 x 2 = 42 kJ
此文章來自奇摩知識+如有不便請留言告知
其他解答:A9A3995907B431A4
文章標籤
全站熱搜
留言列表
