Draw a mechanism for the reaction of water with formic acid. In the box to the left, draw any necessary curved arrows. Show the products of the reaction in the box to the right. Include any nonzero formal charges and all lone pairs of electrons. Finally, indicate which side of the reaction is favored at equilibrium.

See explanation

Explanation:

When water reacts with formic acid, The following equilibrium is set up;

HCOOH(aq) + H20(l) ⇄ HCOO-(aq) + H30+(aq)

This is because, the water abstracts a proton from formic acid to form its conjugate base, formate ion.

At equilibrium,  the forward is favored.

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