What are the products for the redox reaction HCl + SnCl₂ + KMnO₄? In the redox reaction, where $\ce{HCl}$ is the excess reactant do these reactions produce following half reactions?
\begin{aligned}
(1)&amp;&amp;\ce{SnCl2 &amp;-&gt; Sn^{4+} + 2e-}\\
(2)&amp;&amp;\ce{8H+ + KMnO4 + 5e- &amp;-&gt; Mn^{2+} + 4H2O}
\end{aligned}

Where does the $\ce{HCl}$ come into play here, or do you consider it at the end when you have a final equation?

This is a redox reaction where $\ce{KMnO4}$ was added to $\ce{HCl}$ and $\ce{SnCl2}$ (hydrochloric acid was simply used as an excess reagent)
When using $\ce{MnO4-}$ as an oxidant, pH is crucial.

In acidic medium: $\ce{Mn(VII) -&gt; Mn(II)}$

$$\ce{8 H3O+ + MnO4^{-} + 5 e- -&gt; Mn^{2+} + 12 H2O}$$

Under neutral conditions: $\ce{Mn(VII) -&gt; Mn(IV)}$

$$\ce{2 H2O + MnO4^{−} + 3 e− -&gt; MnO2 + 4 OH^{−}}$$

In strongly alkaline medium: $\ce{Mn(VII) -&gt; Mn(VI)}$

$$\ce{MnO4^{−} + e− -&gt; MnO4^{2−}}$$

As far as an excess of $\ce{HCl}$ is concerned, one might take into account that $\ce{Sn(II)}$ and $\ce{Sn(IV)}$ possibly exist as their anionic chloro complexes 
$\ce{[Sn(Cl)4]^{2-}}$ and $\ce{[Sn(Cl)6]^{2-}}$, respectively.