ElectrochemistryEasy
Question
The emf of Daniel cell at 298 K is E1Zn|ZnSO4(0.01M)||CuSO4(1.0M)|Cu When the concentration of ZnSo4 is 1.0 M and that of CuSO4 is 0.01 M the emf changed to E2
What in the relation between E1 and E2 ?
What in the relation between E1 and E2 ?
Options
A.E1 = E2
B.E2 = 0 ≠ E2
C.E1 > E2
D.E1< E2
Solution
Using the relatio
Ecell = Eocell -
= Eocell -
Substituting the given values in two cases.
E1 = E0 -
= E0 -
log 10-2
= E0 + × or (E0 + 0.0591)V
E2 = E0 -
= E0 -
log 102
= E0 -
or (E0 - 0.0591)V
Thus, E1 > E2
Ecell = Eocell -
= Eocell -
Substituting the given values in two cases.
E1 = E0 -
= E0 -
log 10-2= E0 +
× or (E0 + 0.0591)VE2 = E0 -
= E0 -
log 102 = E0 -
or (E0 - 0.0591)VThus, E1 > E2
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