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Q. The nature of curve of $E_{c e l l}^{o}$ against log $K_{c}$ is

NTA AbhyasNTA Abhyas 2020Electrochemistry

Solution:

$E=E^\circ -\frac{0.059}{n}log_{10}Q$

At equilibrium, $E=0,Q=K_{c}$

$E^\circ =\frac{0.059}{n}log_{10}K_{c}$

This equation is of the form y = mx, i.e., straight line, passing through origin.