# Does electrolysis of aqueous tin (II) chloride produce Cl2 or SnCl4? Or both?

When applying a current to an aqueous solution of tin (II) chloride, tin crystals grow from the cathode. However, I'm not sure about the anode. It seems like there are two possible reactions:

$$\ce{2Sn^{2+}_{(aq)} + 4Cl^{-}_{(aq)} -> Sn_{(s)} + SnCl_4_{(aq)}}$$ and

$$\ce{Sn^{2+}_{(aq)} + 2Cl^{-}_{(aq)} -> Sn_{(s)} + Cl_2}$$

Online sources that I've found—including an educational lab and YouTube videos of the experiment being performed—give inconsistent answers. I'm thinking that both products are produced; if that is the case, how can I predict exactly how much of each will be produced? Can other factors—e.g. current, electrode material, pH—change the results?

Apologies if this is too elementary of a question.

Let's try to look at those electrons. C - cathode, A - anode. $$\ce{C(-): Sn^2+ +2e=Sn^0 A(+): 2Cl^--2e=Cl2^0}$$
$$\ce{Sn + 2Cl2=SnCl_4}$$. But I should mention that this reaction is for $$115 °C$$, so don't worry, it won't go for your conditions.
Another way is $$\ce{SnCl_2 + Cl2=SnCl_4}$$. It requires a lot of work, bubbling $$\ce{Cl_2}$$ through aq solution and distillation in the end, so you also shouldn't worry.
You can basically think that all you Sn that becomes a part of new structure goes into solid Sn. It's acceptable since synthesis of $$\ce{SnCl_4}$$ is not an easy thing.
• Arrows can be made using $\ce{->, <=>}$ $$\ce{->, <=>}$$ using  centre aligns equations whereas  is used for inline formatting. Other than that 1 2 3 should get you started. – Avnish Kabaj Dec 6 '18 at 19:01