In the Fehling's solution test for an aldehyde, the red precipitates is due to the formation of

In the Fehling's solution test for an aldehyde, the red precipitates is due to the formation of
  1. cupric nitrate
  2. copper
  3. cupric oxide
  4. cuprous oxide

Solution

The reaction between copper $\mathrm{(II)}$ ions and aldehyde in Fehling's solution is represented as;
$\mathrm{RCHO}+2 \mathrm{Cu}^{2+}+5 \mathrm{OH}^{-} ightarrow \mathrm{RCOO}^{-}+\mathrm{Cu}_{2} \mathrm{O}+3 \mathrm{H}_{2} \mathrm{O}$
When tartrate is added, the reaction can be written as:
$\mathrm{RCHO}+2 \mathrm{Cu}\left(\mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}ight)_{2}{ }^{2-}+5 \mathrm{OH}^{-} ightarrow \mathrm{RCOO}^{-}+\mathrm{Cu}_{2} \mathrm{O}+4 \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}^{2-}+3 \mathrm{H}_{2} \mathrm{O}$
When the redox reaction is completed, the copper $\mathrm{(II)}$ ions are reduced to Copper $\mathrm{(I)}$ oxide, which forms a red precipitate and is insoluble in water. A positive test result is indicated by the presence of this red precipitate.

Asked in: JEE-TOPICTESTS-CHEMISTRY

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