AgBr(s) --> Ag⁺ + Br^- ; E^o = -0.72V.
Value of logK_sp ?

The solution is as follows:

For an equilibrium reaction:

      AgBr (aq) $\leftrightharpoons$ Ag⁺(aq) + Br^-(aq)

     K_sp is the equilibrium constant.

   Formula:

       $$\begin{gathered} ∆G = -RT \log K_{sp} \\ -nF{E}_{cell} = -RT \log K_{sp} \\ Thus, \\ K_{sp } = 1.65\times {10}^{-28} \end{gathered}$$