---

tags: chemistry
title: Mn-Redox

---
# Mn 氧還

## Question

$$
\ce{$a$MnO4^2- + $b$H2O <=> $c$MnO4- + $d$MnO2 + $e$OH-}
$$

:::info
:memo: Note:
Latex mhchem(using tag `\ce`) package is a wonderful chemistry representation tools. See more in [MathJax-mhchem](https://mhchem.github.io/MathJax-mhchem/)
:::

1. Charge Conservation
Only $\ce{Mn}$ changes oxidation number, so we just need to deal with it.
$$
\ce{Mn}: 6a = 7c + 4d
$$

2. Mass Conservation
$$
\begin{align*}
\ce{Mn}&: a = c + d\\
\ce{O} &: 4a + b = 4c + 2d + e\\
\ce{H} &: 2b = e
\end{align*}
$$

That is, we just need to solve the following equation:
$$
\left\{
\begin{array}{rl}
6a &=& 7c + 4d \\
a &=& c + d\\
4a + b &=& 4c + 2d + e\\
2b &=& e
\end{array}
\right. , \\
$$
$$
\Rightarrow a:b:c:d:e = 3:2:2:1:4
$$
The answer is:
$$
\ce{3MnO4^2- + 2H2O <=> 2MnO4- + MnO2 + 4OH-}
$$