Question

Using the potential given below, calculate the activity of cl^- in 1M KCl solution.

Answer 1

Answer:

Explanation:

E° (calomel electrode) = 0.268 v

E( calomel electrode, 1M KCl) = 0.280 v

For calomel electrode

$E = E^0 _( Hg^+/Hg ) - \frac{.059}{2} \times log( a_{cl^-} )^2$

$.280 = .268 - \frac{.059}{2} \times log( a_{cl^-} )^2$

$.012= - \frac{.059}{2} \times log( a_{cl^-} )^2$

$log( a_{cl^-})^2= - .4067$

$a_{cl^-_}$  = 0.626 .