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Equation for the reaction of ethanol with dichromate(VI) ions under acidic condition:
If this is a question in inorganic chemistry, an ionic equation can be written as usual by combining two half-equations.
The half-equation of the reduction of dichromate(VI) ions is:
Cr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l) ... (*)
Ethanol is oxidized to ethanal (CH3CHO) and then ethanoic acid (CH3COOH).
(a) If the product is ethanal :
(a) The half-equation of the oxidation of ethanol to ethanal is:
(a) CH3CH2OH(aq) → CH3CHO(aq) + 2H+(aq) + 2e- .... (#)
(a) The overall equation is obtained by (*) + 3 x (#)
(a) Cr2O72-(aq) + 8H+(aq) + 3CH3CH2OH(aq) → 2Cr3+(aq) + 7H2O(l) + 3CH3CHO(aq)
(b) If the product is ethanic acid :
(a) The half-equation of the oxidation of ethanol to ethanal is:
(a) CH3CH2OH(aq) + H2O(l) → CH3COOH(aq) + 4H+(aq) + 4e- ... (##)
(a) The overall equation is obtained by 2 x (*) + 3 x (##)
(a) 2Cr2O72-(aq) + 16H+(aq) + 3CH3CH2OH(aq) → 4Cr3+(aq) + 11H2O(l) + 3CH3COOH(aq)
However, if this is a question in organic chemistry, the equation would be written as follows:
CH3CH2OH Cr2O72-/H+ - CH3CHO -- Cr2O72-/H+
CH3CH2OH ----------→ CH3CHO --------------→ CH3COOH
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Equation for the reaction of H2O2 with MnO4- ions under acidic condition.
The half-equation of the reduction of the manganate(VII) ions is:
MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l) ... (1)
The half-equation of the oxidation of hydrogen peroxide is:
H2O2(aq) → 2H+(aq) + O2(g) + 2e- ... (2)
The overall equation is obtained by 2 x (1) + 5 x (2):
2MnO4-(aq) + 6H+(aq) + 5H2O2(aq) → 2Mn2+(aq) + 8H2O(l) + 5O2(g)
Consider the half-equation: H2O2(aq) → 2H+(aq) + O2(g) + 2e-
H2O2 is oxidized because the oxidation number increases from -1 (in H2O2) to 0 (in O2).
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2008-07-10 15:24:11 補充：
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