A soluion of $10\mathrm{.}0$ mL of $0\mathrm{.}0500$ mol$/$L AgNO$3$ is titrated with $0\mathrm{.}0250$ mol$/$L NaBr

in the cell:

VKE$||$onbekend$e$ oplossing$|$Ag$($s$)$

Calculate the potential of the cell after adding $0\mathrm{.}1$ mL$,10\mathrm{.}0$ mL$,20\mathrm{.}0$ mL and $30\mathrm{.}0ml$ of

titrant. E$\backslash$deg $($Ag$+,$Ag$=0\mathrm{.}799$ V; E$($VKE$=0\mathrm{.}244$ V; Ksp$($ABr$)=5\mathrm{.}0*{10}^{-13}$