ElectrochemistryMedium
Question
The standard EMF of a Daniel cell at 298 K is E1. When the concentration of ZnSO4 is 1.0 M and that of CuSO4 is 0.01 M, the EMF becomes E2 at 298 K. The correct relationship between E1 and E2 is
Options
A.E1 = E2
B.E2 = 0
C.E1 > E2
D.E1 < E2
Solution
$E_{2} = E_{1} - \frac{0.059}{2}.\log\frac{1.0}{0.01} = E_{1} - 0.059$
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