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Q.
$E^{\circ}=\frac{R T}{n F} \ln K_{e q}$, this equation is called
Chhattisgarh PMTChhattisgarh PMT 2005
Solution:
To determine the electrode potential (E) produced, Nemst gave the following equal:
$E=E^{\circ}-\frac{R T}{n F} \log _{n} K$
$E=E^{\circ}$
$E^{\circ}=\frac{R T}{n F} \log _{n} K$
or $E^{\circ}=\frac{2.303 R T}{n F} \log _{10}\left[M^{n+}\right]$
$E =$ electrode potential developed
$E^{\circ}=$ standard reduction potential
$R =$ gas constant $(8.3\, J / K / mol)$
$T =$ temperature,
$F =$ Faraday constant
$n =$ number of electron taking pans in reduction and oxidation