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If ' $A$ ' is the reactant and ' $P$ ' is the product, which one of the following is the correct form of Nernst equation?
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Verified Answer
The correct answer is:
$\frac{[A]}{[P]}=\exp \left(\frac{n F}{R T}\left(E-E^{\circ}\right)\right)$
According to Nernst equation :
$\begin{aligned} & E=E^{\circ}-\frac{R T}{n F} \ln \frac{[\text { Product }]}{[\text { Reactant }]} \\ & \text { or, } \quad E=E^{\circ}-\frac{R T}{n F} \ln \frac{[P]}{[A]} \\ & \text { or, } \quad E-E^{\circ}=-\frac{R T}{n F} \ln \left[\frac{P}{A}\right] \\ & \text { or, } \quad\left[\frac{A}{P}\right]=e^{\left[\frac{n F}{R T}\left(E-E^{\circ}\right)\right]} \\ & \text { or } \quad \frac{[A]}{[P]}=\exp ^{\left[\frac{n F}{R T}\left(E-E^{\circ}\right)\right]} \\ & \end{aligned}$
Hence, option (b) is the correct answer.
$\begin{aligned} & E=E^{\circ}-\frac{R T}{n F} \ln \frac{[\text { Product }]}{[\text { Reactant }]} \\ & \text { or, } \quad E=E^{\circ}-\frac{R T}{n F} \ln \frac{[P]}{[A]} \\ & \text { or, } \quad E-E^{\circ}=-\frac{R T}{n F} \ln \left[\frac{P}{A}\right] \\ & \text { or, } \quad\left[\frac{A}{P}\right]=e^{\left[\frac{n F}{R T}\left(E-E^{\circ}\right)\right]} \\ & \text { or } \quad \frac{[A]}{[P]}=\exp ^{\left[\frac{n F}{R T}\left(E-E^{\circ}\right)\right]} \\ & \end{aligned}$
Hence, option (b) is the correct answer.
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