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好心人請救救我一個萬試萬靈0既 balance redox equation 0既方法

好心人請救救我一個萬試萬靈0既 balance redox equation 0既方法

我睇書 /notes 講過 half-equation 或 oxidation number method,但一做到0的怪 redox 就兩個方法都唔 work!!!!
好似:

H2SO4 (l) + HI (g) --> H2S (g) + I2 (s)
BrO3- (aq) + Br- (aq) --> Br2 (aq)  in acidic medium

以上兩個方法我都做唔到呢兩條囉!!!
可唔可以教0下我呢??

仲有, oxidation number method 係咪唔係理咩 medium 都得0架??                                                                                                                                                                                                                                                                               

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H2SO4 (l) + HI (g) --> H2S (g) + I2 (s)
Method 1:
first check the oxidation no. of different species to know which ones are oxidizing and reducing agents.
then you know sulphuric acid is oxidizing agent while hydrogen iodide is reducing agent.
Reduction:
Step1: H2SO4 + e- > H2S + 4H2O
Stpe2 (balance both the charge and hydrogen): 8H+ + H2SO4 + 8e- > H2S + 4H2O
Oxidation:
2HI > I2 + 2e- + 2H+
Then you get the answer: H2SO4(l) + 8 HI(g) > H2S(g) + 4 I2(s) + 4H2O(l)

Method 2:
The oxidation no. of S decreases from +6 to -2 and that of I increases from -1 to 0.
Then
Step 1: H2SO4 + 8HI > H2S + 4I2
Step 2: H2SO4 + 8HI > H2S + 4I2 + 4H2O


Similiarly two methods can be used to balance BrO3- (aq) + Br- (aq) --> Br2 (aq)  in acidic medium
The oxidation no. of Br in B3O3- decreases from +5 to 0 and that of Br in Br- increases from -1 to 0.
Step 1: BrO3- + 5Br- > 3Br2
Step 2 (balance oxygen) :
BrO3- + 5 Br- > 3Br2 + 3H2O
Step 3( balance charge and hydrogen by adding H+ to the left which suits the condition in the question):
6H+(aq) + BrO3-(aq) + 5Br-(aq) > 3Br2(aq) + 3H2O(l)

Try to add sth that can serve multi-purposes such as balancing both the charge and some particles.
Of course you should pay attention to the condition provided in the question, which usually also serves as a hint.

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麻煩你可唔可以解多少少......
H2SO4 0個條 Method 2 Step1 點整出0黎??
原本 HI --> I2 係 lose 1粒 e- ,要做到佢 lose 8 粒,點解唔係兩邊 x 8 ??
左邊 x8 但又右邊就係為0左 balance 佢??

我又有幾條用以上0既方法 bal 唔到:

Cr2O72- (aq) + H2O2 (aq) --> Cr3+ (aq) + O2 (g)       in acidic medium

MnO4- (aq) + SO2 (g) --> MnO2 (s) + SO42- (aq)       in alkaline medium

                                                                                                                                                                                                                                                                               

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Method 2:
The oxidation no. of S decreases from +6 to -2 and that of I increases from -1 to 0.
Then
Step 1: H2SO4 + 8HI > H2S + 4I2
Step 2: H2SO4 + 8HI > H2S + 4I2 + 4H2O
______________________________________________________
It is reminded that it is just the oxidation number that decreases but not the no. of electron.
To balance the equation, i figure out it this way:
the total oxidation number changed is 8*-1 - 4*0=-8
multiplying both sides by two will not balance the equation since the I is not balanced, i.e. 8HI...8I2<-strange!

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