about92年ga chem past paper
有個circuit,咁anode支electrode係iron nail which 浸左係iron(ll) sulphate solution到
而cathode支electrode係metalM (unknow ga)浸左係solution of ions of M 到
when the circuit is closed, the green iron (ll) sulphate solution becomes darker gradually?
write a half equation for the reaction that occurs in the half-cell of iron nail and iron(ll) sulphate solution.
- 莫愁Lv 61 十年前最愛解答
I finally find the Marking scheme I kept and the answers are as follows.
我想問sectionB 的第2bi題,點解係出NO而5係NO2 GA?
This question is so "dog好鬼狗". I think it 15 mins. Actually, the question said that a dropwise of conc. HNO3 into saturated Fe(II)SO4 soln.本來它是濃的HNO3,但是現在加在
saturated Fe(II)SO4 soln裏,由於只是Fe(II)SO4 是 saturated 在這溶液中,但這溶液是有水的,所以濃的HNO3立即被dilute了當加入去時,所以變成以浠的HNO3反應,所以出NO.
在+ve pole會有OH- 及 (SO4)2- ions ;-ve pole 會有H+ 及Fe2+ ions.
Fe iron nail undergoes oxidation Fe------->Fe2+ + 2e- (式一)
Fe2+ ions in Iron(II) sulphate soln undergoes reduction Fe2+ + 2e- ------->Fe
and OH- undergoes oxidation : 4OH- -> O2 + 2H2O + 4e
The removal of OH- causing a increase in conc. of (SO4)2-
The increase in the conc. of Fe2+(式一) and (SO4)2- makes solution darker.
FeSO4 is pale green colour due to increase in conc. darker colour.
(SO4)2- can't be removed by electrolysis.
+ve = positive
-ve = negative.