$$\ce{C2H6O(l) + Cr2O7^{2-}(aq) -> C2H4O2(aq) + Cr^3+(aq)}$$

Basically, I have written the half equations, but I have having trouble with converting ethanol to acetic acid ($\ce{C2H6O -> C2H4O2}$).

I do not know how to balance this. I tried adding $\ce{H2O}$ to either side but I didn't get an equal number of oxygens or hydrogens on either side.

  • $\begingroup$ Duplicate. Does this help? chemistry.stackexchange.com/questions/140462/… $\endgroup$ – user55119 Nov 23 '20 at 14:10
  • $\begingroup$ @user55119 yes it is duplicate. $\endgroup$ – Nilay Ghosh Nov 23 '20 at 14:17
  • $\begingroup$ More usual is using structural formulas of ethanol and acetic acid, as $\ce{C2H5OH + CH3COOH}$ $\endgroup$ – Poutnik Nov 23 '20 at 14:40

Of course you must add $\ce{H2O}$ on the left hand side, in order to compensate the excess of $\ce{O}$ atoms at right. Then add $\ce{4 H+}$ on the right hand side ! And $4$ electrons to compensate the charges. And that's over. $\ce{C2H6O + H2O -> C2H4O2 + 4 H+ + 4 e-}$. Where is the problem ?

  • $\begingroup$ How come you added 4H+ to the RHS, shouldn't it be 2H+ because we only have 2 hydrogen in oxygen $\endgroup$ – Chongzzy Nov 23 '20 at 13:29
  • $\begingroup$ Oh wait i think i see, because we need to balance the H+ ?? $\endgroup$ – Chongzzy Nov 23 '20 at 13:33

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