Question

a student uses the following equation to calculate the precise mass of Zn needed to reduce the [Cu(H2O)6]^2+ ions in his solution. His calculated recovery of Cu was were low. What error might account for this low recovery

Answer 1

[Cu(H2O)6]^2+(aq) + Zn(s) --> Cu(s) + Zn^2+(aq)+6H2O(l)

Answer 2

"At the same time, some of the zinc metal, which is present in excess, reduces hydronium ions to H2.

Zn (s) + 2 H3O+ (aq) --> Zn2+ (aq) + H2 (g) + 2 H2O (l)"

http://web.lemoyne.edu/~giunta/chm151L/copper.html

Answer 3

Hi @aaronq! I actually stubbled upon the same website, but why would that cause the recover to be low? because the zinc is still present??

Answer 4

hey, so there would be a low recovery because the copper ions wouldn't get reduced to elemental copper (thus they would still be soluble). Instead of the Zn donating it's electrons to reduce copper, they were taken by \(H_3O^+\) to be reduced to water (i.e. they electrons were used up in another process), and since the person running the experiment added stoichiometric amounts (exact amounts) there was a deficiency in the electrons needed to reduce all the \(Cu^{2+}\) present.

Answer 5

aagh. @aaronq idk how you do it! definitely appreciate alllll your help

Answer 6

haha no problem, really