The celll potential for the given cell has to be determined in which [H + (aq)] is1 .0×10 -7 M and has to be identify whether the reaction is to be more or less favourable at the lower pH. Concept introduction: According to the first law of thermodynamics , the change in internal energy of a system is equal ti the heat added to the sysytem minus the work done by the system. The equation is as follows. ΔU = Q - W ΔU = Change in internal energy Q = Heat added to the system W=Work done by the system In voltaic cell, the maximum cell potential is directly related to the free energy difference between the reactants and products in the cell. ΔG 0 = -nFE 0 n = Number of moles transferred per mole of reactant and products F = Faradayconstant=96485C/mol E 0 = Volts = Work(J)/Charge(C) The relation between standard cell potential and equilibrium constant is as follows. lnK = nE 0 0 .0257 at 298K The relation between solubility product K sp and equilibrium constant is as follows. K sp = e +lnK
The celll potential for the given cell has to be determined in which [H + (aq)] is1 .0×10 -7 M and has to be identify whether the reaction is to be more or less favourable at the lower pH. Concept introduction: According to the first law of thermodynamics , the change in internal energy of a system is equal ti the heat added to the sysytem minus the work done by the system. The equation is as follows. ΔU = Q - W ΔU = Change in internal energy Q = Heat added to the system W=Work done by the system In voltaic cell, the maximum cell potential is directly related to the free energy difference between the reactants and products in the cell. ΔG 0 = -nFE 0 n = Number of moles transferred per mole of reactant and products F = Faradayconstant=96485C/mol E 0 = Volts = Work(J)/Charge(C) The relation between standard cell potential and equilibrium constant is as follows. lnK = nE 0 0 .0257 at 298K The relation between solubility product K sp and equilibrium constant is as follows. K sp = e +lnK
Solution Summary: The author explains that the cell potential for a given cell is determined based on the first law of thermodynamics, wherein the change in internal energy is equal ti the heat added to the system minus the work done
Science that deals with the amount of energy transferred from one equilibrium state to another equilibrium state.
Chapter 19, Problem 85GQ
Interpretation Introduction
Interpretation:
The celll potential for the given cell has to be determined in which [H+(aq)] is1.0×10-7 M and has to be identify whether the reaction is to be more or less favourable at the lower pH.
Concept introduction:
According to the first law of thermodynamics, the change in internal energy of a system is equal ti the heat added to the sysytem minus the work done by the system.
The equation is as follows.
ΔU = Q - WΔU = Change in internal energyQ = Heat added to the systemW=Work done by the system
In voltaic cell, the maximum cell potential is directly related to the free energy difference between the reactants and products in the cell.
ΔG0= -nFE0n = Number of moles transferred per mole of reactant and productsF = Faradayconstant=96485C/mol E0= Volts = Work(J)/Charge(C)
The relation between standard cell potential and equilibrium constant is as follows.
lnK = nE00.0257 at 298K
The relation between solubility product Ksp and equilibrium constant is as follows.
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell