It has to be predicted whether the potential for the voltaic cell will be higher, lower or same as the standard potential and the verification of the prediction also has to be done. 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
It has to be predicted whether the potential for the voltaic cell will be higher, lower or same as the standard potential and the verification of the prediction also has to be done. 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 voltaic cell's maximum cell potential is directly related to the free energy difference between the reactants and products in the cell.
Science that deals with the amount of energy transferred from one equilibrium state to another equilibrium state.
Chapter 19, Problem 84GQ
Interpretation Introduction
Interpretation:
It has to be predicted whether the potential for the voltaic cell will be higher, lower or same as the standard potential and the verification of the prediction also has to be done.
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.
There is an instrument in Johnson 334 that measures total-reflectance x-ray fluorescence (TXRF) to do elemental analysis (i.e., determine what elements are present in a sample). A researcher is preparing a to measure calcium content in a series of well water samples by TXRF with an internal standard of vanadium (atomic symbol: V). She has prepared a series of standard solutions to ensure a linear instrument response over the expected Ca concentration range of 40-80 ppm. The concentrations of Ca and V (ppm) and the instrument response (peak area, arbitrary units) are shown below. Also included is a sample spectrum. Equation 1 describes the response factor, K, relating the analyte signal (SA) and the standard signal (SIS) to their respective concentrations (CA and CIS).
Ca, ppm
V, ppm
SCa, arb. units
SV, arb. units
20.0
10.0
14375.11
14261.02
40.0
10.0
36182.15
17997.10
60.0
10.0
39275.74
12988.01
80.0
10.0
57530.75
14268.54
100.0…
A mixture of 0.568 M H₂O, 0.438 M Cl₂O, and 0.710 M HClO are enclosed in a vessel at 25 °C.
H₂O(g) + C₁₂O(g) = 2 HOCl(g)
K = 0.0900 at 25°C
с
Calculate the equilibrium concentrations of each gas at 25 °C.
[H₂O]=
[C₁₂O]=
[HOCI]=
M
Σ
M
What units (if any) does the response factor (K) have? Does the response factor (K) depend upon how the concentration is expressed (e.g. molarity, ppm, ppb, etc.)?
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