Part B Solution (Mg(s) ---> Mg2+ + 2e-) X 3 (Al(OH)4- + 3e- ---> Al(s) + 4OH-) X
ID: 683361 • Letter: P
Question
Part BExplanation / Answer
(Mg(s) ---> Mg2+ + 2e-) X 3 (Al(OH)4- + 3e- ---> Al(s) + 4OH-) X 2 --------------------------------------... 3Mg(s) + 2(Al(OH4)) + 6e- ---> 3Mg2+ + 6e- + 2Al(s) + 8OH- We have 6 e- in this reaction, so n=6 And Mg is being oxidized and Al is being reduced. E std= (-2.31) - (-2.356)= .046V It is at nonideal conditions (concentrations are not at 1 Molar) soyou have to find Q: Q= ([Mg2+]^3)([OH-)^8))/[Al(OH)4]^2 = ((.020)^3 (.042)^8)/(.30)^2 = 8.61* 10^-16 So plug everything into equation: Ecell= (.046)- (.0592/6)logQ =0.195V