Question

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

Answer 1

Answer: The standard potential is -0.141 V

Step-by-step explanation:

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

`\Delta G=-RTlnK`

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:

`\Delta G=40853J`

Also
`\Delta G=-nFE^0`

where n = no of electrons gained or lost = 3

F = Faradays constant = 96500 C

`E^0` = standard potential = ?

`40853=3* 96500* E^0`

`E^0=-0.141V`

Thus the standard potential is -0.141 V