Electrochemistry: Ksp calculations

Use the following reduction potentials to calculate the Ksp of La(OH)3(s)


La(OH)3(s)+3eβˆ’β†’La(s)+3 OH(aq)βˆ’Eo=βˆ’2.90VLa(aq)3++3eβˆ’β†’La(s)Eo=βˆ’2.379 V\begin{array}{llllll}&La(OH)_{3(s)}&+&3e-&\rightarrow& La_{(s)}&+&3\ OH^-_{(aq)}& &&E^o=-2.90 V \\ &La^{3+}_{(aq)}&+&3e^-&\rightarrow& La_{(s)}&&&&& E^o=-2.379\ V \end{array}


More Galvanic Cells Questions:
Electrochemistry: Reduction potentials in a Galvanic Cell
Use the standard reduction potentials below to answer the following question: which metal could be used to create a sacrificial anode for an aluminum pipe?
Reduction halfβˆ’reactionEΒ°(V)πΆπ‘Ž2+(π‘Žπ‘ž)+2π‘’βˆ’β‡ŒπΆπ‘Ž(𝑠)βˆ’2.87π‘π‘Ž+(π‘Žπ‘ž)+π‘’βˆ’β‡Œπ‘π‘Ž(𝑠)βˆ’2.71𝐴𝑙3+(π‘Žπ‘ž)+3π‘’βˆ’β‡Œπ΄π‘™(𝑠)βˆ’1.66𝑀𝑛2+(π‘Žπ‘ž)+2π‘’βˆ’β‡Œπ‘€π‘›(𝑠)βˆ’1.18𝐢𝑒2+(π‘Žπ‘ž)+2π‘’βˆ’β‡ŒπΆπ‘’(𝑠)+0.15𝐴𝑔+(π‘Žπ‘ž)+π‘’βˆ’β‡Œπ΄π‘”(𝑠)+0.80\begin{array}{c:c}\rm Reduction\ half-reaction& EΒ° (V)\\ \hline πΆπ‘Ž^{2+} (π‘Žπ‘ž)+2 𝑒^βˆ’β‡ŒπΆπ‘Ž (𝑠)& -2.87\\ \hline π‘π‘Ž^+ (π‘Žπ‘ž)+𝑒^βˆ’β‡Œπ‘π‘Ž (𝑠)&-2.71\\ \hline 𝐴𝑙^{3+} (π‘Žπ‘ž)+3 𝑒^βˆ’β‡Œπ΄π‘™(𝑠) &-1.66\\ \hline 𝑀𝑛^{2+} (π‘Žπ‘ž)+2 𝑒^βˆ’β‡Œπ‘€π‘› (𝑠)&-1.18\\ \hline 𝐢𝑒^{2+} (π‘Žπ‘ž)+2 𝑒^βˆ’ β‡ŒπΆπ‘’ (𝑠) &+0.15\\ \hline 𝐴𝑔^+ (π‘Žπ‘ž)+𝑒^βˆ’ β‡Œπ΄π‘” (𝑠)&+0.80\end{array}