Question

Question #dea0b

Answer 1

I'll assume the reaction for this cell is,

`2Cu^(2+) + 2H_2O(l) rightleftharpoons 4H^(+)(aq) + O_2(g) + 2Cu(s)`

However, this reaction will be spontaneous (`+1.57V`, per reference tables), so I don't know why an electric current is passed through, especially since the author didn't mention its magnitude. This looks like more of a difficult stoichiometry problem coupled with gas laws to me.

`0.404g * (mol)/(63.6g) approx 6.35*10^-3mol`

of copper are produced. Hence,

`6.35*10^-3mol* (O_2)/(2Cu) approx 3.18*10^-3mol`

of oxygen gas are reduced from water. We'll derive the answer from the ideal gas equation. Hence,

`1atm * V = 3.18*10^-3mol * (0.08206L*atm)/(mol*K) * 273K`
`therefore V approx 71.2mL`

of oxygen gas are produced in this specific case.