Question

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

Answer 1

Hint: The copper ion is complexed with tartrate or citrate ion to keep it from encouraging as \[Cu{\left( {OH} \right)_2}\] . Formaldehyde is a particularly incredible reducing agent in that the complexed copper \[\left( {II} \right)\] ions are reduced to metallic copper. The formaldehyde is oxidized to formic acid, which exists as a formate particle in the basic arrangement.

Complete step by step answer:
They might be utilizing Fehling's test or Benedict's test for the presence of an aldehyde.
The two tests utilize an answer of \[C{u^{2 + }}\] in essential arrangement. The copper ion is complexed with tartrate or citrate ion to keep it from precipitating as \[Cu{\left( {OH} \right)_2}\] .
Formaldehyde is a particularly incredible reducing agent in that the complexed copper(\[II\]) ions are decreased to metallic copper.
The blue solution shapes a copper mirror inside the test tube.
\[C{u^{ + + }}\left( {complexed} \right)\; + \;2{e^ - }\;\;\] → \[Cu\]
The formaldehyde is oxidized to formic acid, which shows as a formate particle in the fundamental solution.
\[{H_2}CO\; + {\text{ }}3O{H^ - }\;\;\;\] → \[HCO{O^ - }\;\; + 2{H_2}O\; + 2{e^ - }\]
Consolidating the two half-reaction, we get
\[{H_2}CO + C{u^{2 + }} + 3O{H^ - } → HCO{O^ - } + Cu + 2{H_2}O\]

Note:
The reaction between copper(II) ion and aldehyde in Fehling's solution is spoken to as;
\[RCHO{\text{ }} + {\text{ }}2{\text{ }}C{u^{2 + }} + {\text{ }}5{\text{ }}OH\] → \[RCO{O^ - } + {\text{ }}C{u_2}O{\text{ }} + {\text{ }}3{\text{ }}{H_2}O\]
At the point when tartrate is added, the reaction can be composed as:
\[RCHO{\text{ }} + {\text{ }}2{\text{ }}Cu{\left( {{C_4}{H_4}{O_6}} \right)_2}^{2 - } + {\text{ }}5{\text{ }}O{H^ - }\;\] → \[RCO{O^ - } + {\text{ }}C{u_2}O{\text{ }} + {\text{ }}4{\text{ }}{C_4}{H_4}{O_6}^{2 - } + {\text{ }}3{\text{ }}{H_2}O\]
At the point when the redox reaction is finished, the copper \[II\] ions are reduced to Copper\[\;I\] oxide, which forms a red ppt. and is insoluble in water. A positive test outcome is shown by the presence of this red a ppt. The sodium salt of the acid is abandoned in solution.