1. Tardigrade
  2. Question
  3. Chemistry
  4. Statement 1: On cooling a freezing mixture, colour of the mixture turns to pink from deep blue for a reaction. Co(H2O)2+6(aq)+4Cl-(aq)leftharpoons CoCl2-4(aq)+6H2O(l) Statement 2: Reaction is endothermic so on cooling, the reaction moves to backward direction

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Q. Statement 1: On cooling a freezing mixture, colour of the mixture turns to pink from deep blue for a reaction.
$Co\left(H_{2}O\right)^{2+}_{6}{\left(aq\right)}+4Cl^{-}{\left(aq\right)}\rightleftharpoons CoCl^{2-}_{4}{\left(aq\right)}+6H_{2}O_{\left(l\right)}$
Statement 2: Reaction is endothermic so on cooling, the reaction moves to backward direction

Equilibrium

Solution:

Both statements are true and statement 2 is the correct explanation of statement 1. $Co\left(H_{2}O\right)^{2+}_{6}$ + (Pink) while $CoCl^{2-}_{4}$ (blue). So, on cooling because of Le-Chatelier’ s principle the reaction tries to overcome the effect of temperature.