# Question #dea0b

##### 1 Answer
Dec 3, 2017

I'll assume the reaction for this cell is,

$2 C {u}^{2 +} + 2 {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s 4 {H}^{+} \left(a q\right) + {O}_{2} \left(g\right) + 2 C u \left(s\right)$

However, this reaction will be spontaneous ($+ 1.57 V$, 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.

$0.404 g \cdot \frac{m o l}{63.6 g} \approx 6.35 \cdot {10}^{-} 3 m o l$

of copper are produced. Hence,

$6.35 \cdot {10}^{-} 3 m o l \cdot \frac{{O}_{2}}{2 C u} \approx 3.18 \cdot {10}^{-} 3 m o l$

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

$1 a t m \cdot V = 3.18 \cdot {10}^{-} 3 m o l \cdot \frac{0.08206 L \cdot a t m}{m o l \cdot K} \cdot 273 K$
$\therefore V \approx 71.2 m L$

of oxygen gas are produced in this specific case.