Print calculator Periodic Table astion 22 of 22 General Chemistry 4th Edition Un
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Question
Print calculator Periodic Table astion 22 of 22 General Chemistry 4th Edition University Science Books presented by Sapling Leaming The cell diagram for the lead-acid cell that is used in automobile and truck batteries is where the comma between Pbods and Pbso4(s denotes a heterogeneous mixture of the two solids and the right-hand lead electrode is nonreactive. (a) write the balanced equation for the net cell reaction (b) Look up standard potentials for the oxidation and the (c) Calculate the value of AG rom. reduction reactions, then calculate the value of Eoel. kJ mol (d) Calculate the value of Eeel at 25 Cif IH2SO4l 10.0 M e) How many lead acid cells are in a 12 V car attery? Round to the nearest integer o Previous check Answer 0 Next JiExt submil AssignmentExplanation / Answer
Solution:
In each cell in the truck battery has the cathode which is a thick, porous plate of metallic lead. Anode is a plate consisting mostly of porous lead dioxide paste, supported on a thin metal grid. In between the plates we have concentrated sulfuric acid.
As the battery discharges, the following half-cell reaction takes place at the anode(-ve plate)
Pb (s) + HSO4 (aq) PbSO4(s)+H+(aq) +2e
[Pb0 Pb2+ loss of 2 electrons (oxidation)] electrode potential = 0.1262V
The other half of the discharge reaction taking place at the reduction (+ ve plate) can be written as
PbO2(s) + 3H+ + HSO4 + 2e PbSO4(s)+ 2H2O
[Pb4+ + 2e Pb2+ gain of 2 electrons (reduction)] electrode potential = 1.455V
Adding the two half-cell reactions together, we get the full-cell discharge reaction as
Pb (s) + PbO2(s) + 2H2SO4 PbSO4(s)+ 2H2O
b) E0 value for oxidation
PbO2(s) + 3H+ + HSO4 + 2e PbSO4(s)+ 2H2O
E0 value for reduction
Ecell = oxidation potential + reduction potential
Ecell = 0.1262V + 1.455V = 1.5812V
c) The cell potential is the energy per unit charge which is available to do work when charge is transferred through an external circuit. This maximum work is equal to the change in Gibbs free energy, G, in the reaction. These relationships can be expressed as
Maximum work = G = -nFE°cell
Where n is the number of electrons taking part in the reaction ( here it is 2) and F is the Faraday constant(96,485 coul/mole).
G = -nFE°cell = -2 x 96,485 coul/mole x 1.5812 joule/coul = -305124.164J or 305 kJ
[1V=1joule/coul]
d) E = E0 +[2 x2.303 RT/ n F]x {log a H2SO4}
log a H2SO4 concentration of H2SO4
=1.5812V[2x 2.303 x 8.314 J/molx298/ 2x96,485 coul/mole ] x log 10
=1.5812V[11411.70/ 2x96,485 coul/mole ] x log 10
=1.5812V[11411.70/ 192970] x log 10
=1.5812V[0.059] x log 10
=1.5812V[0.059] x 1=0.093V
The other half of the discharge reaction taking place at the reduction (+ ve plate) can be written as
PbO2(s) + 3H+ + HSO4 + 2e PbSO4(s)+ 2H2O
[Pb4+ + 2e Pb2+ gain of 2 electrons (reduction)] electrode potential = 1.455V
Adding the two half-cell reactions together, we get the full-cell discharge reaction as
Pb (s) + PbO2(s) + 2H2SO4 PbSO4(s)+ 2H2O
b) E0 value for oxidation
PbO2(s) + 3H+ + HSO4 + 2e PbSO4(s)+ 2H2O
E0 value for reduction
Ecell = oxidation potential + reduction potential
Ecell = 0.1262V + 1.455V = 1.5812V
c) The cell potential is the energy per unit charge which is available to do work when charge is transferred through an external circuit. This maximum work is equal to the change in Gibbs free energy, G, in the reaction. These relationships can be expressed as
Maximum work = G = -nFE°cell
Where n is the number of electrons taking part in the reaction ( here it is 2) and F is the Faraday constant(96,485 coul/mole).
G = -nFE°cell = -2 x 96,485 coul/mole x 1.5812 joule/coul = -305124.164J or 305 kJ
[1V=1joule/coul]
Solution:
In each cell in the truck battery has the cathode which is a thick, porous plate of metallic lead. Anode is a plate consisting mostly of porous lead dioxide paste, supported on a thin metal grid. In between the plates we have concentrated sulfuric acid.
As the battery discharges, the following half-cell reaction takes place at the anode(-ve plate)
Pb (s) + HSO4 (aq) PbSO4(s)+H+(aq) +2e
[Pb0 Pb2+ loss of 2 electrons (oxidation)] electrode potential = 0.1262V
The other half of the discharge reaction taking place at the reduction (+ ve plate) can be written as
PbO2(s) + 3H+ + HSO4 + 2e PbSO4(s)+ 2H2O
[Pb4+ + 2e Pb2+ gain of 2 electrons (reduction)] electrode potential = 1.455V
Adding the two half-cell reactions together, we get the full-cell discharge reaction as
Pb (s) + PbO2(s) + 2H2SO4 PbSO4(s)+ 2H2O
b) E0 value for oxidation
PbO2(s) + 3H+ + HSO4 + 2e PbSO4(s)+ 2H2O
E0 value for reduction
Ecell = oxidation potential + reduction potential
Ecell = 0.1262V + 1.455V = 1.5812V
c) The cell potential is the energy per unit charge which is available to do work when charge is transferred through an external circuit. This maximum work is equal to the change in Gibbs free energy, G, in the reaction. These relationships can be expressed as
Maximum work = G = -nFE°cell
Where n is the number of electrons taking part in the reaction ( here it is 2) and F is the Faraday constant(96,485 coul/mole).
G = -nFE°cell = -2 x 96,485 coul/mole x 1.5812 joule/coul = -305124.164J or 305 kJ
[1V=1joule/coul]
Solution:
In each cell in the truck battery has the cathode which is a thick, porous plate of metallic lead. Anode is a plate consisting mostly of porous lead dioxide paste, supported on a thin metal grid. In between the plates we have concentrated sulfuric acid.
As the battery discharges, the following half-cell reaction takes place at the anode(-ve plate)
Pb (s) + HSO4 (aq) PbSO4(s)+H+(aq) +2e
[Pb0 Pb2+ loss of 2 electrons (oxidation)] electrode potential = 0.1262V
The other half of the discharge reaction taking place at the reduction (+ ve plate) can be written as
PbO2(s) + 3H+ + HSO4 + 2e PbSO4(s)+ 2H2O
[Pb4+ + 2e Pb2+ gain of 2 electrons (reduction)] electrode potential = 1.455V
Adding the two half-cell reactions together, we get the full-cell discharge reaction as
Pb (s) + PbO2(s) + 2H2SO4 PbSO4(s)+ 2H2O
b) E0 value for oxidation
PbO2(s) + 3H+ + HSO4 + 2e PbSO4(s)+ 2H2O
E0 value for reduction
Ecell = oxidation potential + reduction potential
Ecell = 0.1262V + 1.455V = 1.5812V
c) The cell potential is the energy per unit charge which is available to do work when charge is transferred through an external circuit. This maximum work is equal to the change in Gibbs free energy, G, in the reaction. These relationships can be expressed as
Maximum work = G = -nFE°cell
Where n is the number of electrons taking part in the reaction ( here it is 2) and F is the Faraday constant(96,485 coul/mole).
G = -nFE°cell = -2 x 96,485 coul/mole x 1.5812 joule/coul = -305124.164J or 305 kJ
[1V=1joule/coul]
Solution:
In each cell in the truck battery has the cathode which is a thick, porous plate of metallic lead. Anode is a plate consisting mostly of porous lead dioxide paste, supported on a thin metal grid. In between the plates we have concentrated sulfuric acid.
As the battery discharges, the following half-cell reaction takes place at the anode(-ve plate)
Pb (s) + HSO4 (aq) PbSO4(s)+H+(aq) +2e
[Pb0 Pb2+ loss of 2 electrons (oxidation)] electrode potential = 0.1262V
The other half of the discharge reaction taking place at the reduction (+ ve plate) can be written as
PbO2(s) + 3H+ + HSO4 + 2e PbSO4(s)+ 2H2O
[Pb4+ + 2e Pb2+ gain of 2 electrons (reduction)] electrode potential = 1.455V
Adding the two half-cell reactions together, we get the full-cell discharge reaction as
Pb (s) + PbO2(s) + 2H2SO4 PbSO4(s)+ 2H2O
b) E0 value for oxidation
PbO2(s) + 3H+ + HSO4 + 2e PbSO4(s)+ 2H2O
E0 value for reduction
Ecell = oxidation potential + reduction potential
Ecell = 0.1262V + 1.455V = 1.5812V
c) The cell potential is the energy per unit charge which is available to do work when charge is transferred through an external circuit. This maximum work is equal to the change in Gibbs free energy, G, in the reaction. These relationships can be expressed as
Maximum work = G = -nFE°cell
Where n is the number of electrons taking part in the reaction ( here it is 2) and F is the Faraday constant(96,485 coul/mole).
G = -nFE°cell = -2 x 96,485 coul/mole x 1.5812 joule/coul = -305124.164J or 305 kJ
[1V=1joule/coul]
Solution:
In each cell in the truck battery has the cathode which is a thick, porous plate of metallic lead. Anode is a plate consisting mostly of porous lead dioxide paste, supported on a thin metal grid. In between the plates we have concentrated sulfuric acid.
As the battery discharges, the following half-cell reaction takes place at the anode(-ve plate)
Pb (s) + HSO4 (aq) PbSO4(s)+H+(aq) +2e
[Pb0 Pb2+ loss of 2 electrons (oxidation)] electrode potential = 0.1262V
The other half of the discharge reaction taking place at the reduction (+ ve plate) can be written as
PbO2(s) + 3H+ + HSO4 + 2e PbSO4(s)+ 2H2O
[Pb4+ + 2e Pb2+ gain of 2 electrons (reduction)] electrode potential = 1.455V
Adding the two half-cell reactions together, we get the full-cell discharge reaction as
Pb (s) + PbO2(s) + 2H2SO4 PbSO4(s)+ 2H2O
b) E0 value for oxidation
PbO2(s) + 3H+ + HSO4 + 2e PbSO4(s)+ 2H2O
E0 value for reduction
Ecell = oxidation potential + reduction potential
Ecell = 0.1262V + 1.455V = 1.5812V
c) The cell potential is the energy per unit charge which is available to do work when charge is transferred through an external circuit. This maximum work is equal to the change in Gibbs free energy, G, in the reaction. These relationships can be expressed as
Maximum work = G = -nFE°cell
Where n is the number of electrons taking part in the reaction ( here it is 2) and F is the Faraday constant(96,485 coul/mole).
G = -nFE°cell = -2 x 96,485 coul/mole x 1.5812 joule/coul = -305124.164J or 305 kJ
[1V=1joule/coul]
Solution:
In each cell in the truck battery has the cathode which is a thick, porous plate of metallic lead. Anode is a plate consisting mostly of porous lead dioxide paste, supported on a thin metal grid. In between the plates we have concentrated sulfuric acid.
As the battery discharges, the following half-cell reaction takes place at the anode(-ve plate)
Pb (s) + HSO4 (aq) PbSO4(s)+H+(aq) +2e
[Pb0 Pb2+ loss of 2 electrons (oxidation)] electrode potential = 0.1262V
The other half of the discharge reaction taking place at the reduction (+ ve plate) can be written as
PbO2(s) + 3H+ + HSO4 + 2e PbSO4(s)+ 2H2O
[Pb4+ + 2e Pb2+ gain of 2 electrons (reduction)] electrode potential = 1.455V
Adding the two half-cell reactions together, we get the full-cell discharge reaction as
Pb (s) + PbO2(s) + 2H2SO4 PbSO4(s)+ 2H2O
b) E0 value for oxidation
PbO2(s) + 3H+ + HSO4 + 2e PbSO4(s)+ 2H2O
E0 value for reduction
Ecell = oxidation potential + reduction potential
Ecell = 0.1262V + 1.455V = 1.5812V
c) The cell potential is the energy per unit charge which is available to do work when charge is transferred through an external circuit. This maximum work is equal to the change in Gibbs free energy, G, in the reaction. These relationships can be expressed as
Maximum work = G = -nFE°cell
Where n is the number of electrons taking part in the reaction ( here it is 2) and F is the Faraday constant(96,485 coul/mole).
G = -nFE°cell = -2 x 96,485 coul/mole x 1.5812 joule/coul = -305124.164J or 305 kJ
[1V=1joule/coul]
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