Balanced ionic equations (20 marks)

metal + very dilute nitric acid → metal nitrate + hydrogen
metal + dilute nitric acid → metal nitrate + nitrogen monoxide + water
metal + concentrated nitric acid → metal nitrate + nitrogen dioxide + water
metal + concentrated sulphuric acid → metal sulphate + sulphur dioxide + water
請先記住以上四條 equations
(1)
zinc + very dilute nitric acid → zinc nitrate + hydrogen
(i) 轉做 chemical equation
Zn + 2HNO3 → Zn(NO3)2 + H2
(ii) 轉做 ionic equation, soluble compounds 拆做 ions, atoms/molecules 不變
Zn + 2H+ + 2NO3- → Zn2+ + 2NO3- + H2
(iii) 約簡/cancel overlappings
Zn + 2H+ → Zn2+ + H2
(2)
copper + conc. nitric acid → copper nitrate + nitrogen dioxide + water
(i) 轉做 chemical equation
Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O
(ii) 轉做 ionic equation, soluble compounds 拆做 ions, atoms/molecules 不變
Cu + 4H+ + 4NO3- → Cu2+ + 2NO3- + 2NO2 + 2H2O
(iii) 約簡/cancel overlappings
Cu + 4H+ + 2NO3- → Cu2+ + 2NO2 + 2H2O
(3)
copper + dilute nitric acid → copper nitrate + nitrogen monoxide + water
(i) 轉做 chemical equation
3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O
(ii) 轉做 ionic equation, soluble compounds 拆做 ions, atoms/molecules 不變
3Cu + 8H+ + 8NO3- → 3Cu2+ + 6NO3- + 2NO + 4H2O
(iii) 約簡/cancel overlappings
3Cu + 8H+ + 2NO3- → 3Cu2+ + 2NO + 4H2O
(4)
potassium permanganate + sodium sulphite
由於佢係 redox reaction
only show the reaction between permanganate ion and sulphite ion is okay
MnO4- + SO32- → Mn2+ + SO42-
change of ON of Mn = 2-7 = -5
change of ON of S = 6-4 = +2
(i) multiply the equation in order to balance the electrons needed
2MnO4- + 5SO32- → 2Mn2+ + 5SO42-
(ii) find the total charge of both sides and add positive charge
left = -2-10 = -12
right = +4-10 = -6
that means left + 6 = right
2MnO4- + 5SO32- + 6H+ → 2Mn2+ + 5SO42-
(iii) balance the hydrogen ion added
2MnO4- + 5SO32- + 6H+ → 2Mn2+ + 5SO42- + 3H2O
(5)
magnesium + conc. sulphuric acid → magnesium sulphate + sulphur dioxide + water
(i) 轉做 chemical equation
Mg + 2H2SO4 → MgSO4 + SO2 + 2H2O
(ii) 轉做 ionic equation, soluble compounds 拆做 ions, atoms/molecules 不變
Mg + 4H+ + 2SO42- → Mg2+ + SO42- + SO2 + 2H2O
(iii) 約簡/cancel overlappings
Mg + 4H+ + SO42- → Mg2+ + SO2 + 2H2O

definition: ionic equations only display the chemical species which participate into the reaction while ignore those who does not so that is so called "spectator" species.

instruction: in orther words, all spectator ions has to be eliminated from both sides of whole equation to form ionic equation.

1. Zinc and very dilute nitric acid
in this case, very dilute nitric acid performs its acidic properties so that H+ attack zinc metal while NO3- ion does not participate into the reaction and is a spectator ion. hence, NO3- is not shown in the ionic equation.
whole equation: Zn(s) + 2HNO3(aq) --> Zn(NO3)2(aq) + H2(g)
expanded: Zn(s) + 2H+(aq) + 2NO3-(aq) --> Zn2+(aq) + 2NO3-(aq) + H2(g)
NO3- is eliminated from both sides,
ionic equation: Zn(s) + 2H+(aq) --> Zn2+(aq) + H2(g)

2. Copper and conc nitric acid
concentrated HNO3 performs its oxidizing properties so that NO3- is the effective oxidizing agent to participate in the reaction and is NOT a spectator ion.
hence, NO3- cannot be ignored in the ionic equation.
H+, while, participate in to form water.
Cu is oxidised to Cu2+
both Cu, H+ cannot be elminated from the whole equation.
whole equation: Cu(s) + 4HNO3(aq) --> Cu(NO3)2(aq) + 2H2O(l) + 2NO2(g)
however, from the whole equation, some NO3-(aq) remains unreacted and become a part of Cu(NO3)2. the NO3- in Cu(NO3)2 is eliminated.
expanded: Cu(s) + 4H+(aq) + 4NO3-(aq) --> Cu2+(aq) + 2NO3-(aq) + 2H2O(l)
+ 2NO2(g)
ionic equation: Cu(s) + 4H+(aq) + 2NO3-(aq) --> Cu2+(aq) + 2H2O(l) + 2NO2(g)

3. Copper and dilute nitric acid
dilute nitric acid possess oxidising power as well in which NO3- is reduced to NO
again, Cu is oxidised Cu2+ while H+ is to form H2O.
all reactants participate into the reaction.
similar to (2), some NO3- remains unreacted.
whole: 3Cu(s) + 8HNO3(aq) --> 3Cu(NO3)2(aq) + 4H2O(l) + 2NO(g)
expanded: 3Cu(s) + 8H+(aq) + 8NO3-(aq) --> 3Cu2+(aq) 6NO3-(aq) + 4H2O(l)
+ 2NO(g)
ionic: 3Cu(s) + 8H+(aq) + 2NO3-(aq) --> 3Cu2+(aq) + 4H2O(l) + 2NO(g)

4. Potassium permanganate solution added to sodium sulphite solution
MnO4- is reduced to Mn2+;
SO32- is oxidised to SO42-;
H+ forms H2O
Na+ and K+ do not participate into the reaction and can be eliminated from the whole equation
whole: 5Na2SO3(aq) + 2KMnO4(aq) + 6H+(aq) --> 5Na2SO4(aq) + 2Mn2+(aq)
+ 2K+(aq) + 3H2O(l)
expanded: 10Na+(aq) + 5SO32-(aq) + 2K+(aq) + 2MnO4-(aq) + 6H+(aq)
--> 10Na+(aq) + 5SO42-(aq) + 2Mn2+(aq) + 2K+(aq) + 3H2O(l)
ionic: 5SO32-(aq) + 2MnO4-(aq) + 6H+(aq) --> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l)

5. Magnesium added to conc sulphuric acid
conc H2SO4 also performs its oxidising property.
Mg is oxidised to Mg2+ while SO42- is reduced to SO2 and H+ is to form H2O
whole/ionic: Mg(s) + 2H2SO4(l) --> Mg2+(aq) + SO42-(aq) + SO2(g) + 2H2O(l)
SO42- is not eliminated because concentrated acid has its percentage by mass 90% or above. a container of conc H2SO4 consists of mainly H2SO4 molecules and only dissociate in a little extent. conc. sulphuric acid is written as H2SO4(l) instead of
2H+(aq) + SO42-(aq)
however, after reaction, water is formed to solvate SO42-(aq).
therefore, SO42- can not be elminated.

1:2Zn + 2HNO3 > Zn(NO3)2 + H2
2Zn + 2H+ > 2Zn2+ H2 (ionic)
2: Cu + 4HNO3 > Cu(NO3)2 + 2NO2 + 2H2O
3: 3Cu + 8HNO3 > 3Cu(NO3)2 + 2NO + 4H2O
4: 唔識
5: Mg + 2H2SO4 > MgSO4 + 2H2O + SO2

我記得我化學o既老師話
如果o係狀態,電荷,氧化數..有改變o既時候,就要放成舊野落equation度
(只係唔要冇改變過o既野)
例如: CaCO3 + 2HCl > CaCl2 + H2O + CO2
咁ionic equation就係 CaCO3 + 2H+ > Ca2+ + H2O + CO2
(因為Cl- 從來都係冇改變過,一係都係Cl-,而且係溶水
CaCO3因為係固態,但係反應後係液態,所以成個都要放落地,儘管Ca一直係2+)

Redox方面就係唔係好識(我未學)
我睇書話
首先分開2邊,reduction同oxidation
之後分別balance金屬離子o既數量,O o既數量(加H2O),跟住balance H o既數量(加H+),仲有balance個charge
要跟住依個步驟去做,然後聯立(好似做數學果d咁)
約埋一樣o既野就得嫁啦