How does the reaction between formaldehyde and copper hydroxide take place?

1 Answer
Apr 30, 2017

Answer:

Here's what I think.

Explanation:

They may be using Fehling's test or Benedict's test for the presence of an aldehyde.

Both tests use a solution of #"Cu"^"2+"# in basic solution. The copper ion is complexed with tartrate or citrate ions to prevent it from precipitating as #"Cu(OH)"_2#.

Formaldehyde is such a powerful reducing agent that the complexed copper(II) ions are reduced to metallic copper.

The blue solution forms a copper mirror inside the test tube.

#underbrace("Cu"_text(complexed)^"2+")_color(red)("blue solution") + "2"e^"-" → underbrace("Cu")_color(red)("Cu mirror")#

The formaldehyde is oxidized to formic acid, which exists as formate ion in the basic solution.

#underbrace("H"_2"CO")_color(red)("formaldehyde") + "3OH"^"-" → underbrace("HCOO"^"-")_color(red)("formate ion") + 2"H"_2"O" + 2e^"-"#

Combining the two half-reactions, we get

#underbrace("H"_2"CO")_color(red)("formaldehyde") + underbrace("Cu"_text(complexed)^"2+")_color(red)("blue solution") + "3OH"^"-" → underbrace("HCOO"^"-")_color(red)("formate ion") + underbrace("Cu")_color(red)("Cu mirror") + 2"H"_2"O"#