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Question
The EMF of the cell, Cd(s)|CdCl2(aq,0.1 M)||AgCl(s)|Ag(s) in which the cell reaction is
Cd(s)+2AgCl(s)→2Ag(s)+Cd2+(aq)+2Cl–(aq) is 0.6915 V at 0°C and 0.6753 V at 25°C. The ΔS and ΔH of the reaction at 25°C is:
Take, F=96485 C mol−1
The EMF of the cell, Cd(s)|CdCl2(aq,0.1 M)||AgCl(s)|Ag(s) in which the cell reaction is
Cd(s)+2AgCl(s)→2Ag(s)+Cd2+(aq)+2Cl–(aq) is 0.6915 V at 0°C and 0.6753 V at 25°C. The ΔS and ΔH of the reaction at 25°C is:
Take, F=96485 C mol−1
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Solution
The correct option is A ΔS=−123.8 J K−1 and ΔH=−167.72 kJ
The cell reaction requires 2 faradays of electricity for its completion since n=2
We know,
ΔG=−nFE
ΔG=−2 mol×96485 C mol−1×0.6753 V
ΔG=−130.33 kJ
EMF of cell is given by
E=−ΔH2F+T(∂E∂T)P
In this case, EMF decreases with increase in temperature i.e., (∂E∂T)P is negative.
Thus,
(∂E∂T)P=−(0.6915−0.675325)=−0.00065 V K−1
Substituting values, we get,
0.6753 V=−ΔH2 mol×96485 C mol−1+298 K×(−0.00065 V K−1)
ΔH=−167717J=−167.72 kJ
The entropy change is related to enthalpy and free energy change by the well known thermodynamic expression, ΔG=ΔH−TΔS
−ΔS=ΔG−ΔHT=−130.33−(−167.72)298=0.1238 kJ K−1=123.8 J K−1
ΔS=−123.8 J K−1
The cell reaction requires 2 faradays of electricity for its completion since n=2
We know,
ΔG=−nFE
ΔG=−2 mol×96485 C mol−1×0.6753 V
ΔG=−130.33 kJ
EMF of cell is given by
E=−ΔH2F+T(∂E∂T)P
In this case, EMF decreases with increase in temperature i.e., (∂E∂T)P is negative.
Thus,
(∂E∂T)P=−(0.6915−0.675325)=−0.00065 V K−1
Substituting values, we get,
0.6753 V=−ΔH2 mol×96485 C mol−1+298 K×(−0.00065 V K−1)
ΔH=−167717J=−167.72 kJ
The entropy change is related to enthalpy and free energy change by the well known thermodynamic expression, ΔG=ΔH−TΔS
−ΔS=ΔG−ΔHT=−130.33−(−167.72)298=0.1238 kJ K−1=123.8 J K−1
ΔS=−123.8 J K−1
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