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atom360

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Half-equations

« on: July 09, 2010, 02:03:38 am » HelloI am having trouble with the following questionsQuestion: Ethanol is the alchohol component of wines, beer and spirits. When a bottle of wine is opened, the ethanol is exposed to oxygen in the atmosphere. If the wine is left open for a long period of time, the wine iwll acquire a strong taste of vinegar. This is because the ethanol becomes oxidised into acetic acid.(a) Write a balanced half-equation for the oxidation of ehtanol to acetic acid.CH3CH2OH --> CH3COOHCH3CH2OH + H2O--> CH3COOHCH3CH2OH + H2O --> CH3COOH + 4H+ + 4 e-However the answer isCH3CH2OH --> CH3COOH + 2H+ + 2e-How come it is 2H+ added not 4H+ as the number of hydrogens in each side of equaiton are not balanced? there are 8 hydrogens on the left and 6 on the right. Can someone please explain this for me.(b)Write a balanced half-equation for the reduction of oxygen gas to hydroxide ions.I need help in this one. I got my answer slightly wrong(c) write a combined redox reaction (I can do this once I understand a and b)Help is greatly appreciated and respectedThank you Logged

AWK

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Re: Half-equations

« Reply #1 on: July 09, 2010, 03:10:00 am » O2 + _H2O +..__..= _OH- Logged AWK

eleven6eleven

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Re: Half-equations

« Reply #2 on: August 03, 2010, 11:54:11 am » For part a, it should beCH3CH2OH + 1/2 O2 --> CH3COOH + 2H+ + 2e-because this involves oxygen and not just water. If it involved only water then it would be oxidized regardless of whether the bottle was opened or not with an appropriate oxidizing agent (besides oxygen).For part b, it should beO2 + 2H+ + 4e- --> 2OH-So part c should beCH3CH2OH + O2 --> CH3COOH + H2OHope this helped! Let me know if there are any mistakes. Logged

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