Question

Write the value of 2.303 RT F in the nernst equation

Answer 1

Answer:

The value of  $\frac{2.303\ R\ T}{F}$ in the Nernst equation is $0.0592$·

Explanation:

We know that the Nernst Equation,

  $E_{(cell)}\ =\ E^0_{(cell)} \ -\ \frac{2.303\ R\ T}{n\ F}$   log  $\frac{[Products]}{[Reactants]}$

where,

$E_{(cell)}$ $=$ The cell potential

$E^0_{(cell)}$ $=$ The standard cell potential

R $=$  The universal gas constant

T $=$  Temperature in kelvin

n $=$  The number of electrons

F $=$  Faraday's constant

Here, we want to find out the value of $\frac{2.303\ R\ T}{F}$ in the Nernst equation·

For that,

The value of R $=$ $8.314\ J\ K^{-1}\ mol^{-1}$

The temperature we take as the standard value  $=$ $298$ K

The value of F  $=$ $96500$ C

On substituting these values in the expression we get,

⇒  $\frac{2.303\ R\ T}{F}$ $=$ $\frac{2.303\ *\ 8.314\ *\ 298}{96500}$

⇒  $\frac{2.303\ R\ T}{F}$ $=$ $0.0592$

Therefore,

The value of  $\frac{2.303\ R\ T}{F}$ in the Nernst equation is $0.0592$·