Jawabanmu

2013-12-13T10:27:38+07:00
Cu( NO_{3)} elektroda platina
katoda= 2 H_{2}O+2 e^{-} -> H_{2}+ 2OH^{-}
anoda       =2H_{2}O->4 H^{+}+ O_{2}+4 e^{-}
w=e.f
12,7= \frac{63,5}{2} .f->f=2/5f
dimana 1f=1e
jadi volum  O_{2}= \frac{1}{4} .\frac{2}{5}  .22.4=2,24liter
1 5 1
2013-12-13T10:57:36+07:00
 elektroda platina
katoda
anoda       =
w=e.f

dimana 1f=1e
jadi volum 


semoga berguna :D