Why didn't copper act as a reducing agent

In lab, we took $\ce{CuCl\_2(aq)}$ and poured it into a cup. Then we took some aluminum foil and put it inside and started to stir. I could see the aluminum getting eaten by the copper because it was a better reducing agent. However, when I put silver instead of aluminum, nothing happened. I know it's because the standard cell potential of silver is greater than that of copper, but why didn't the copper act as a reducing agent?




Copper is already oxidized as $\ce{Cu^2+(aq)}$ and silver reduced as $\ce{Ag(s)}$. Copper would act as a reducing agent in $\ce{Cu(s)}$ and $\ce{AgNO3(aq)}$ scenario.


Using $\ce{CuCl2}$ instead of $\ce{CuSO4}$ is a mistake, if $\ce{Ag}$ is involved, as forming insoluble $\ce{AgCl(s)}$ would complicate things. Eventually, $\ce{Cl-(aq)}$ may keep $\ce{Ag/Ag+}$ potential below or near $\ce{Cu/Cu^2+}$ potential:


$$E = E\_{\ce{Ag/Ag+}}^\circ + 0.059 \log{[\ce{Ag+}]} = E\_{\ce{Ag/Ag+}}^\circ + 0.059 \log{\frac{K\_{\mathrm{s},\ce{AgCl}}}{[\ce{Cl-}]}}$$
(simplified replacing activities by concentrations)


But, forming $\ce{AgCl(s)}$ layer would freeze it kinetically.


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The answer lies in the redox potential chart:


Reactions:


$\ce{2 Alumimum metal + 3 Cu++ -> 2 Al+++ + 3 Copper metal}$
  
$\ce{2 Silver metal + Cu++ -> 2 Ag+ + Copper metal no reaction}$


Drawing the half reactions:


$\ce{Al+++ -> Aluminum metal - 1.66 V}$
  
$\ce{Cu++ -> Copper metal + 0.34 V}$
  
$\ce{Ag+ -> Silver metal + 0.80 V}$


We can see why Copper ions react with Aluminum metal but not Silver metal.
Silver metal has a higher voltage potential to **keep** its electrons than both Copper and Aluminum metals. Aluminum metal (and Sodium) are strong electron *donors*.


In this case Copper metal sits right between Silver and Aluminum metal. Copper metal will **give** electrons to Silver ions, and Copper ions will **take** electrons from Aluminum metal. Silver metal and Copper ions will not react, nor will Silver metal and Aluminum ions.


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