# 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)&&\ce{SnCl2 &-> Sn^{4+} + 2e-}\\ (2)&&\ce{8H+ + KMnO4 + 5e- &-> 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)

• Welcome to Chemistry.SE. I edited your post to make the chemical equations pretty using MathJax. More information here:chemistry.stackexchange.com/help/notation – Ben Norris Feb 25 '14 at 3:07
• @BenNorris Dear Ben, I believe it is $H_2O$ not $H_{20}$. :) – Jun-Goo Kwak Feb 25 '14 at 3:08
• @Jun-GooKwak - fixed – Ben Norris Feb 25 '14 at 3:08

$$\ce{KMnO4 + SnCl2 + HCl -> KCl + MnCl2 + SnCl4 + H2O}$$

Oxidation states are for reactants: \begin{aligned}\ce{ K &= +1\\ Mn &= +7\\ O &= -2 \\ Sn &= +2\\ Cl &= -1\\ H &= +1\\ }\end{aligned}

Oxidation states for products: \begin{aligned}\ce{ K &= +1\\ Cl &= -1\\ Mn &= +2\\ Sn &= +4\\ H &= +1\\ O &= -2\\ }\end{aligned}

This is a redox reaction carried out in acidic solution. $\ce{SnCl2}$ and $\ce{KMnO4}$ are titrated with hydrochloric acid. What is being oxidized and what is being reduced? Potassium is being reduced, magnesium and tin are oxidized.

Let's write out our half reactions:

\begin{aligned}\ce{ HCl + SnCl2 &-> SnCl\\ HCl + KMnO4 &-> MnCl2 + KCl + H2O }\end{aligned}

If we look at the oxidation states of our reactants, and products, $\ce{H}$ has +1, $\ce{Cl}$ -1, and for the products $\ce{Sn}$ +2, $\ce{Cl}$ -1 for the first reaction. The first reaction gains two electrons and thus is reduced.

For the 2nd reaction, $\ce{H}$ again has +1 and $\ce{Cl}$ -1. Potassium has +1 and the permanganate ion has -2. Specifically, $\ce{Mn}$ has a theoretical oxidation state of +7 and oxygen -2. In the products side, magnesium has an oxidation state of +2 and $\ce{Cl}$ -1, $\ce{H}$ is +1, $\ce{O}$ is 2-, and $\ce{K}$ becomes +1. Chlorine as an element has an oxidation state of 0.

For redox reactions, we have to make sure our reactants and products are first balanced for our elements other than H and O in acidic solution. It is clear that the chlorines are not balanced, therefore:

\begin{aligned}\ce{ 2HCl + SnCl2 &-> SnCl4 \\ 3HCl + KMnO4 &-> MnCl2 + KCl + H2O }\end{aligned}

We balance oxygen atoms with adding molecules of water:

\begin{aligned}\ce{ 2HCl + SnCl2 &-> SnCl4\\ 3HCl + KMnO4 &-> MnCl2 + KCl + 4H2O }\end{aligned}

Now we add H+ to balance H:

\begin{aligned}\ce{ 2HCl + SnCl2 &-> SnCl4 + 2H+\\ 3HCl + KMnO4 + 5H+ &-> MnCl2 + KCl + 4H2O }\end{aligned}

We now add electrons to balance charge:

\begin{aligned}\ce{ 2HCl + SnCl2 + 2e- &-> SnCl4 + 2H+\\ 3HCl + KMnO4 + 5H+ &-> MnCl2 + KCl + 4H2O + 5e- }\end{aligned}

We multiply our half-reactions by integers so that the electrons are the same number:

\begin{aligned}\ce{ 10HCl + 5SnCl2 + 10e- &-> 5SnCl4 + 10H+\\ 6 HCl + 2KMnO4 + 10H+ &-> 2MnCl2 + 2KCl + 8H2O + 10e- }\end{aligned}

Now we add 'em up and cancel:

\begin{aligned}\ce{ 16HCl + 5SnCl2 + 2KMnO4 &-> 5SnCl4 + 2MnCl2 + 2KCl + 8H2O }\end{aligned}

• I think Sn loses 2 electrons, while Mn gains 5 electrons – Jonie Feb 25 '14 at 2:06
• Ah... this question is long. I almost got it. I think I wrote one of the half-reactions wrong, let me look it up. – Jun-Goo Kwak Feb 25 '14 at 2:13
• The final equation should be 16H+ + 2KMnO4 + 5SnCl2 ==> 5SnCl4 + 2MnCl2 + 8H2O + 2KCl – Jonie Feb 25 '14 at 2:15
• @Jonie Sn loses 2. Mn gains 4. – Jun-Goo Kwak Feb 25 '14 at 2:16
• no Mn gains 5, it goes from +7 to +2 – Jonie Feb 25 '14 at 2:17

When using $\ce{MnO4-}$ as an oxidant, pH is crucial.

In acidic medium: $\ce{Mn(VII) -> Mn(II)}$

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

Under neutral conditions: $\ce{Mn(VII) -> Mn(IV)}$

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

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

$$\ce{MnO4^{−} + e− -> MnO4^{2−}}$$

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.