CHEM titration

a)
The equation is:
2MnO4-(aq) + 6H+(aq) + 5H2O2(aq) → 2Mn2+(aq) + 8H2O(l) + 5O2(g)

The two half equations of the redox reaction are:
MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l) ...... (1)
H2O2(aq) → 2H+(aq) + O2(g) + 2e- ...... (2)

The overall equation can be obtained by adding 2 times of (1) and 5 times of (2).


b)
The colour of the solution changes from colourless to pale pink.

On the addition of acidified KMnO4 solution, the redox reaction gives colourless products. At the equivalent point, all the H2O2 is completely reacted. A slightly excess of KMnO4 would not cause any reaction. The presence of a small amount of purple MnO4- cause the appearance of pale pink colour.
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呢題有無限多種解法, 我舉三種
(a) 5 H2O2 + 2 MnO4- + 6 H+ ---> 2 (Mn 2+) + 5 O2+ 8 H2O
或4H2O2 + 4MnO4- + 12H+ ---> 4Mn 2+ + 7O2 + 10 H2O
或
3 H2O2 + 2 MnO4- + 6H+ ---> 2 (Mn 2+) + 4 O2 + 6 H2O

H2O2 + MnO4 + H+ ---> Mn+2 + O2 + H2O
[1] 先睇有冇element係兩邊都只有一個ion/molecule擁有, 咁呢個case係「Mn」, 假設MnO4- 同 Mn2+都係1, 條equation暫時不變
[2] 再睇有冇element係咪是但一邊都只出現在同一個element, 呢個case係「右邊的H」, 呢刻左邊有三個, 右邊得兩個, 咁「H2O」起碼要「3/2」個先夠balance, 條equation變做：

H2O2 + MnO4 + H+ ---> Mn+2 + O2 + (3/2)H2O

[3] 再睇下charge有冇唔balance, 右邊冇得減, 唯有左邊加charge, 即係要set「H+」為3, 條equation 變做：

H2O2 + MnO4 + 3H+ ---> Mn+2 + O2 + (3/2)H2O

[4] 之後重複運用[1],[2],[3], 個H又郁左, 我地再加大右邊的「H2O」：

H2O2 + MnO4- + 3H+ ---> Mn+2 + O2 + (5/2)H2O

[5] 呢個時候我地睇淨個「O」, 左邊有六個, 右邊得4.5個, 我地郁「O2」的影響最細：

H2O2 + MnO4- + 3H+ ---> Mn+2 + (7/4)O2 + (5/2)H2O

[6] 每邊乘返4：
4H2O2 + 4MnO4- + 12H+ ----> 4Mn+2 + 7O2 + 10H2O

(b) from purple to pink/red