# When the redox equation Cr^(3+)(aq) + 3Mn(s) -> Mn^(2+)(aq) + Cr(s) is completely balanced, what will the coefficient of Cr^(3+)(aq) be?

Jun 10, 2017

$2. \ldots \ldots \ldots .$

#### Explanation:

Manganese is oxidized:

$\stackrel{0}{M} n \left(s\right) \rightarrow M {n}^{2 +} + 2 {e}^{-}$ $\left(i\right)$

Chromium is reduced:

$C {r}^{3 +} + 3 {e}^{-} \rightarrow \stackrel{0}{C} r \left(s\right)$ $\left(i i\right)$

And so we take $3 \times \left(i\right) + 2 \times \left(i i\right)$ to eliminate the electrons:

$3 \stackrel{0}{M} n \left(s\right) + 2 C {r}^{3 +} \rightarrow 3 M {n}^{2 +} + 2 \stackrel{0}{C} r$

Charge is balanced, and mass is balanced, and so this is a reasonable representation of reality. Whether the reaction actually works, I don't know, and cannot be bothered to look up redox tables.