Asked by Ben
Wirte the half equation for the reduction of acetic acid.
I Know that a colourless odourless gas is formed in the process but I can't seem to find an appropriate product to accompany it. Pleasae help.
I Know that a colourless odourless gas is formed in the process but I can't seem to find an appropriate product to accompany it. Pleasae help.
Answers
Answered by
bobpursley
I think I would look for carbon dioxide.
Answered by
Ben
So would the half equation be
CH3COOH -> CO2 + CH4
and if so, where is the oxidation/reduction - electron transfer?
CH3COOH -> CO2 + CH4
and if so, where is the oxidation/reduction - electron transfer?
Answered by
DrBob222
I don't know about the products; however, for what you have written the redox is like this.
In CH3COOH:
O is -2 each for total -4.
H is +1 each for total +4.
Therefore each C = 0
In CO2, O is -2 and C is +4. Carbon has gone from 0 to +4 which is a loss of electrons which makes it oxidized.
In CH4, C is -4; C has gone from 0 to -4 which is a gain of electrons which makes it reduced.
In CH3COOH:
O is -2 each for total -4.
H is +1 each for total +4.
Therefore each C = 0
In CO2, O is -2 and C is +4. Carbon has gone from 0 to +4 which is a loss of electrons which makes it oxidized.
In CH4, C is -4; C has gone from 0 to -4 which is a gain of electrons which makes it reduced.
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