chemical cell

電子由x->y
x做oxidation ,anode
y做reduction, cathode

A) Carbon,sulphate 唔會放electron, 一係Zn2+ +2e- ->Zn (reduction) 所以基本上不太可能Zn2+再electron
B) Zn -> Zn2+ + 2e-
Mg2+ +2e- ->Mg
Not possible, Mg is stronger reducing agent, electron flow inverse.
C) not possible. No electron flow
D) Zn ,x,strong reducing agent-> oxidation
Ag2+, y, strong oxidizing agent-> reduction
So, Zn -> Zn2+ +2e- release electron and Ag2+ + 2e-->Ag

2) 呢題我估要考慮H2O2個structure, O有一對lone pair electron, attract H+ attack, or 果d咩自由基,freeradical,
H2O2-> 2HO. , 咁係acidic medium H2O2 就係strong oxidizing agent.
至於點先知個medium,, 有時d題目有比alkaline, 咁你balance 公式就要用OH- 代替.個人認為考試最盡係alkaline 做redox, 因為一般都係acidic.
至於H2O2 係specical case, 應該要知一知.

2014-05-18 01:10:34 補充：
因為一般公式表都係用H+ balance.

有錯請糾正

2014-05-18 01:19:42 補充：
,,第2條唔洗咁複雜,
題目有寫變做水,
H2O2 ->H2O
做reduction, gain electron
H2O2 + __+ Ae- ->H2O, ,所以只有一個可能性就係加酸先可以balance條式

2014-05-21 17:46:20 補充：
depends on oxidation number,
increase O.N->oxidation,
drecase O.N.-> reduction,
H2O2 -> H2O (hint by question)
find the O.N in H2O2
H2O : O=-2 ,H=1 , H2O2: H=1, O=-1
change in o.n of O= -2-(-1)=-1 ->reduction
(most cases, O=-2, H=1)
reduction ->gain electron
oxidation-> loss electron

1.
The answer is : D

A. is incorrect.
Carbon is inert electrode. At electrode X, in the absence of zinc metal, theonly possible reaction is the reduction of zinc ion.
Zn²⁺(aq) + 2e⁻ → Zn(s)
However, silver is a weaker reducing agent than zinc. Hence, silver would notundergo oxidation in this case.
As a result, there is no electron flow.

B. is incorrect.
Magnesium is a stronger reducing agent than zinc. Hence, electrons flow from magnesium (Y) tozinc (X).

C. is incorrect.
Carbon is inert electrode. Both zinc ion and copper(II) ion are oxidizing agents.In the absence of reducing agent, there is no electron flow.

D. is correct.
zinc is stronger reducing agent than silver.
At electrode X, zinc metal is oxidized to zinc ion.
Zn(s) → Zn²⁺ (aq) + 2e⁻
At electrode Y, silver ion is reduced to silver.
Ag⁺ (aq)+ e⁻ → Ag(s)
Hence, electrons flow from electrode X to electrode Y.


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2.
The oxidation number of O in H2O­2 is -1, while that in H2Ois -2.
Due to the decrease in oxidation number when H2O2 isdecomposed to H2O, the reaction is reduction (gain electrons).

In an acidic medium, H2O2 can reduced to H2O.
H2O2(aq) + 2H⁺(aq)+ 2e⁻ → 2H2O(l)

In an alkaline medium, H2O2 can only be reduced to OH⁻.
H2O2(aq) + 2e⁻ → 2OH⁻(aq)

Hence, the decomposition of H2O2 to H2O mustbe done in an acidic medium.