# How did the temperature in the solution with a lower concentration of CuCl2 increased in a faster rate?

My lab assignment is about the reaction between $$\ce{CuCl2}$$ and Aluminium and we tracked the temperature of the solution every 15 seconds and we needed to make our own experiment to check how different things affect the temperature of the solution every 15 seconds. So I chose the concentration of $$\ce{CuCl2}$$ and thus we had four cups, one with $$\ce{0.1 M}$$, one with $$\ce{0.2 M}$$, one with $$\ce{0.5 M}$$ and one with $$\ce{0.8 M}$$. Obviously the temperature in the cups with a lower concentration of $$\ce{CuCl2}$$ barely changed and throughout most of the experiment stayed at the exact same state.

Now, the cup with the $$\ce{0.5 M}$$ concentration of $$\ce{CuCl2}$$ changed temperature in a faster rate than the cup with the $$\ce{0.8 M}$$ concentration. Not only that, at the end of the experiment, the temperature in the cup with $$\ce{0.8 M}$$ was 28° C meanwhile the temperature of the cup with the $$\ce{0.5 M}$$ concentration of $$\ce{CuCl2}$$ was 29° C .

Can somebody explain to me how that happened?

The reaction $$\ce{Cu^2+ + 4 Cl- <=> CuCl4^2-}$$ may lead to slowing down reaction with aluminium, decreasing the equilibrium concentration of $$\ce{Cu^2+}$$ in $$\pu{0.8 M}$$ $$\ce{CuCl2}$$. That all would lead to lower peak temperature due bigger heat dissipation before reaching maximal temperature.
• decreasing of chemical activity of water for the electrode surface reaction $$\ce{Al(s) + 6 H2O -> [Al(H2O)6]^3+(aq) + 3 e-}$$. These electrons, charging the electrode, then reduce copper: $$\ce{[Cu(H2O)n]^2+(aq) + 2 e- -> Cu(s) + n H2O}$$