1
$\begingroup$

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:

$$\ce{2 Sn^2+(aq) + 4 Cl-(aq) -> Sn(s) + SnCl4(aq)}$$

and

$$\ce{Sn^2+ (aq) + 2 Cl- (aq) -> Sn(s) + Cl2}$$

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?

$\endgroup$
3
$\begingroup$

Let's try to look at those electrons. C - cathode, A - anode.

$$ \begin{align} &\mathrm{C(-):} &\ce{Sn^2+ + 2e- &→ Sn^0} \\ &\mathrm{A(+):} &\ce{2Cl^- &→ Cl2^0 + 2e-} \end{align} $$

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

$$\ce{Sn + 2 Cl2 → SnCl4}$$

But I should mention that this reaction is for $\pu{115 °C}$, so don't worry, it won't go for your conditions.

Another way is

$$\ce{SnCl2 + Cl2 → SnCl4}$$

It requires a lot of work, bubbling $\ce{Cl2}$ through aqueous solution and distillation in the end, so you also shouldn't worry.

You can basically think that all you $\ce{Sn}$ that becomes a part of new structure goes into solid $\ce{Sn}$. It's acceptable since synthesis of $\ce{SnCl4}$ is not an easy thing.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.