Calculate the standard potential, ∘, for this reaction from its equilibrium constant at 298 K. X(s)+Y3+(aq)↽−−⇀X3+(aq)+Y(s)=6.90×1 0−8 X ( s ) + Y 3 + ( aq ) ↽ − − ⇀ X 3 + ( aq ) + Y ( s ) K = 6.90 × 10 − 8

The standard potential is  -0.141 V

Explanation:

To calculate the Gibbs free energy for given value of equilibrium constant we use the relation:

where,

= standard Gibbs free energy = ?

R = Gas constant = 8.314 J/Kmol

T = temperature = 298 K

K = equilibrium constant =

Putting values in above equation, we get:

Also

where n = no of electrons gained or lost = 3

F = Faradays constant = 96500 C

= standard potential = ?

Thus the standard potential is -0.141 V

answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.

it is a good conductor

explanation: