equation of the reaction of ethanol with acidified potassium dichromate?
can you please tell me the two half equations ( oxidation and reduction)
K2Cr2O7 & 14 H+ & 6 electrons –> 2 Cr+3 & 2 K+ & 7 H2O
CH3CH2OH & H2O –> CH3CO2H & 4H+ & 4 electrons
to baalance electrons, double the reduction & triple the oxidation, which gives:
2K2Cr2O7 & 16H+ & 3CH3CH2OH –> CH3CO2H & 4Cr+3 & 4K+ & 11 H2O
it had 2 different possibilities:
1.if you use an excess of alcohol & distill off the aldehyde as fast as it is made, ethanal can be your product
2.if you reflux the reaction with an excess of K2CrO7,… You will produce ethanoic acid, & that’s what I think they want:
CH3CH2OH (with an oxidizer added)–> CH3CO2H
did I think this up , not hardly, your reference:
What is the coefficient of H2O?
The “Breathalyzer” utilizes the oxidation-reduction reaction between potassium dichromate and ethanol. When the following redox equation is properly balanced what is the coefficient of H2O
K2Cr2O7 + C2H5OH + H2SO4 –> Cr2(SO4)3 + K2SO4 + HC2H3O2
please explain how you got your answer,
The equation is already balanced (I think), there’s no negative or positive charge on either side. So what I did was I add all the hydrogen on one side and oxygen and combine them to make H2O. And I made 4, so d) was my answer, I doubt that I’m correct.
It is d] 11
You’ll find this clearly explained and the full equation on the excellent Chemguide site.
it’s well down the page.
Please help me answer this question for my homework.. ?
In the class, we did an experiment using potassium dichromate, ethanol and sulphuric acid to make chrome alum crystals.
1. A student suggests that sulphuric acid is a catalyst in the reaction that you have performed. Do you agree with this suggestion? Explain your answer.
Since you wind up with
and the sulfate can only come from H2SO4
the sulfuric acid does NOT remain unchanged – it is a reactant – not a catalyst
Need help with titration and equations.?
Okay so I’ve got this chemistry project and it was all sorted out but then the chemicals had to be changed so it messed up all of my process and now I’m stuck on the equations.
Ethanol reacts with dichromate ions according to the equation:
2Cr2O72- + 3C2H5OH + 16H+ —> 4Cr3+ + 3CH3CO2H + 11H2O
The equation for the reaction of iodine with the un-reacted dichromate is:
6I- + Cr2O72- + 14H+ —> 3I2 + 2Cr3+ + 7H2O
The iodine produced is then titrated with standard sodium thiosulfate solution. It reacts according to the equation below:
I2 + 2S2O32- —-> 2I- + S4O62-
Since starch turns black/blue in the presence of iodine, it is an indicator for this final reaction.
Okay so that’s what it was originally. I was using 0.0800 Mol/L potassium dichromate and 0.0200 Mol/L sodium thiosulfate. but now I’ve had to change the Sodium thiosulfate to 0.500 Mol/L.
Does this mean that those equations will have to change?
Certainly not. All the equations will remain exactly the same. All that will change is the details of your calculations that depend on the concentration of Na2S2O3.
Chemistry…i need help with percentage yields?
ill give yu a bag of cookies if yu can answer this ^_^
ethanol can also be oxidised to produce ethanoic acid by heating it with an oxidising agent such as acidified potassium dichromate(VI) solution:
CH3CH2OH —reflux—Cr2O7^2-/H+(aq)—> CH3OOH
calculate the percentage yield of this reaction if the starting amount of ethanol was 23.0g and 20.0g of ethanoic acid were produced.
20/23 x 100 = 86.956
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