Question

A 1.0 M solution of Cd2+ is added ti excess iron and the system is allowed to reach equillibrium. What is the concentration of Cd2+?


Cd2+(aq)+Fe(s) gives Cd +Fe2+;E°cell =0.037

Answer 1

Given : A 1M solution of Cd²⁺ is added with excess iron and the system is allowed to reach equilibrium.

To find : The concentration of Cd²⁺ ions.

solution : chemical reaction is....

Cd²⁺(aq) + Fe(s) ⇔Cd(s) + Fe²⁺ (aq)

at t = 0, 1 excess 0 0

at eql 1 - x x

so, equilibrium constant, K = [Fe²⁺]/[Cd²⁺] = x/(1 - x)

from Nernst equation,

E = E° - 0.0591/n logK

⇒0 = 0.037 - 0.0591/2 logK [ n = 2 as you can see in reaction ]

⇒0.037 × 2 = 0.0591 logK

⇒ 0.074 = 0.0591 logK

⇒logK = 1.252

⇒K = 17.86

⇒x/(1 - x) =17.86

⇒x = 17.86 - 17.86x

⇒18.86x = 17.86

⇒x ≈ 947

so, 1 - x = 1 - 0.947 = 0.053

Therefore the concentration of Cd²⁺ is 0.053 M