Answer:
Explanation:
E° (calomel electrode) = 0.268 v
E( calomel electrode, 1M KCl) = 0.280 v
For calomel electrode
[tex]E = E^0 _( Hg^+/Hg ) - \frac{.059}{2} \times log( a_{cl^-} )^2[/tex]
[tex].280 = .268 - \frac{.059}{2} \times log( a_{cl^-} )^2[/tex]
[tex].012= - \frac{.059}{2} \times log( a_{cl^-} )^2[/tex]
[tex]log( a_{cl^-})^2= - .4067[/tex]
[tex]a_{cl^-_}[/tex] = 0.626 .
