Question

The emf of the cell is found to be 0.20v at 298k cd^2+ /cd //ni^2(2.0)/ni what is molar concentration of cd ^2+ ion ecd =-0.40 and eni = -0.25

Answer 1

The molar concentration of cadmium^2+ is 98

Explanation:

E ° = -025-(04)

= 0.15

Ecell= E₀ - $\frac{0.059}{2} log\frac{[Cd 2+]}{[Ni2+]}$

0.2=0.15 - $\frac{0.059}{2} log\frac{[Cd2+]}{[Ni2+]}$

            =    $\frac{0.05 X 2}{0.059} log \frac{[Cd2+]}{[Ni2+]}$

      1.69 = $log \frac{[Cd2+]}{[Ni2+]}$

      10¹°⁶⁹ = $\frac{[Cd2+]}{2}$

                =  97.9 ≅ 98.