I just tried conducting an experiment during lunch today. Now, I calculated this reaction 4 days ago dealing with electrochemistry,
$$\ce{SnCl2 + CuCl2 -> Cu(s) + SnCl4}$$
Theoretically, I got these 1/2 cell reactions from a couple different sources (one of which are posted below):
$$\ce{Sn^2+ -> Sn^4+ + 2e-} \quad E° = -0.15~\mathrm{V}$$
$$\ce{Cu_{(s)} -> Cu^2+ + 2e-} \quad E° = -0.34~\mathrm{V}$$
Using basic electrochemistry, we know that Copper prefers to be copper metal. Due to its overbearing power over the tin(II) ion, the copper(II) ion forces the tin(II) ion to become tin(IV).
I did not check the reduction or oxidation potential of the chloride ion (since the chloride ion is the most stable form of chlorine due to its (relative to the other compounds in this experiment) high ionization potential. I concluded that chloride would be a spectator ion.) Finally resulting in this net ionic reaction.
$$\ce{Sn^2+ + Cu^2+ -> Cu_{(s)} + Sn^4+} \quad E^\circ_{\rm cell} = 0.19~\mathrm{V}$$
I mixed the compounds in $150~\mathrm{ml}$ of deionised water. Later combining them only to end up with a milky-green colored solution. I saw some brown-red precipitate at the bottom of the beaker. however, only a small amount of what I think is copper precipitated. Tomorrow I will place the beaker with the solution onto a hot plate in hopes of speeding up the reaction.
I was wondering if a smaller difference in electrical potential between the two half cells results in a slower reaction. The other thing I was wondering concerns any other ideas or misconceptions I may have faced while conducting this experiment.
