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. $\ce{C(-): Sn^2+ +2e=Sn^0 A(+): 2Cl^--2e=Cl2^0}$

So, your second reaction is right. But the first is wrong: Sn is reduced, but there are no oxidation. You should use

$\ce{Sn + 2Cl2=SnCl_4}$. 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.

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    $\begingroup$ 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. $\endgroup$ – Avnish Kabaj Dec 6 '18 at 19:01

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