Q25 / 160 · TS EAMCET 2019 (06 May Shift 1) · Chemistry

If ' $A$ ' is the reactant and ' $P$ ' is the product, which one of the following is the correct form of Nernst equation?

  • A
    $\frac{[A]}{[P]}=\exp \left(\frac{R T}{n F}\left(E-E^{\circ}\right)\right)$
  • Correct answer
    B
    $\frac{[A]}{[P]}=\exp \left(\frac{n F}{R T}\left(E-E^{\circ}\right)\right)$
  • C
    $\frac{[A]}{[P]}=\exp \left(-\frac{n F}{R T}\left(E-E^{\circ}\right)\right)$
  • D
    $E=E^{\circ}-\frac{R T}{n F} \ln \frac{[A]}{[P]}$

Solution

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.

Hence, the correct answer is option: 'B'.

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