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The potential of the Daniel cell, Zn |(Z...

The potential of the Daniel cell, `Zn |(ZnSO_(4))||(CuSO_(4)|Cu` was reported by Buckbee, Surdzial and Metz as `E^(@) = 1.1028 - 0.641 xx 10^(-3)T` + `0.72 xx 10^(-5)T^(2)`, where T is the temperature in degree celcius, Calcualte `DeltaS^(@0` for the cell reaction at `25^(@)C`.

A

`-45.32 EU`

B

`-34.53 EU`

C

`-25.43 EU`

D

`54.23 EU`

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The correct Answer is:
D
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The potential of the Daniell cell, Zn|(ZnSO_4),((1M))||(CuSO_4),((1M))|Cu was reported by Buckbee, Surdzial, and Metz at E^@ = 1 . 1028 - 0. 41 xx 10^(-3) T + 0. 72 xx 10^(-5) T^2 where T is the Celsius temperature. Calculate Delta S^@ for the cell reaction at 235^@ C ,.

EMF of the cell Zn ZnSO_(4)(a =0.2)||ZnSO_(4)(a_(2))|Zn is -0.0088V at 25^(@)C . Calculate the value of a_(2) .

The temperature dependence of the emf of a standard electrochemical cell is given by E=1.02-4.0xx10^(-5) (T-20)-9.0xx10^(-7) (T-20)^(2) where, T is in .^(@)C and E is in volts. The temperature coefficient of the emf at 30^(@)C is :

Calculate EMF of following cell at 298 K Zn|ZnSO_(4) (0.1 M) ||CuSO_(4) (1.0 M)|Cu (s) if E_(cell) = 2.0 V

The emf of a Daniell cell at 298 K is E_(1) Zn|ZnSO_(4)(0.01 M)||CuSO_(4)(1.0M)|Cu When the concentration of ZNSO_(4) is 1.0 M and that of CuSO_(4) is 0.01 M , the emf changed to E_(2) . What is the relationship between E_(1) and E(2) ?

GRB PUBLICATION-ELECTROCHEMISTY-Thermodynamics in Electrochemistry
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