# How do we represent the oxidation of Cr^(3+) ion to CrO_4^(2-) by hydrogen peroxide, using the method of half-equations?

Apr 8, 2017

By the method of half-equations............

$2 C {r}^{3 +} + 2 {H}_{2} O + 3 {H}_{2} {O}_{2} \rightarrow 2 C r {O}_{4}^{2 -} + 10 {H}^{+}$

#### Explanation:

$\text{Chromic chloride}$, $C r C {l}_{3}$, is oxidized from $C r \left(+ I I I\right)$ to $C r \left(+ V I\right)$ in chromate:

$C {r}^{3 +} + 4 {H}_{2} O \rightarrow \stackrel{V I +}{C r} {O}_{4}^{2 -} + 8 {H}^{+} + 3 {e}^{-}$ $\left(i\right)$

Is this balanced with respect to mass and charge? It is your problem not mine.

Hydrogen peroxide, ${H}_{2} {O}_{2}$, is reduced from ${O}^{- I}$ to ${O}^{- I I}$.

${H}_{2} {O}_{2} + 2 {H}^{+} + 2 {e}^{-} \rightarrow 2 {H}_{2} O$ $\left(i i\right)$

Again is the thing balanced?

To give the overall redox equation, we cross-multiply the individual redox equations: $2 \times \left(i\right) + 3 \times \left(i i\right)$:

$2 C {r}^{3 +} + \cancel{8} 2 {H}_{2} O + 3 {H}_{2} {O}_{2} + \cancel{6 {H}^{+}} + \cancel{6 {e}^{-}} \rightarrow \cancel{6 {H}_{2} O} + 2 C r {O}_{4}^{2 -} + \cancel{16} 10 {H}^{+} + \cancel{6 {e}^{-}}$

And then we SUBTRACT the superfluous reagents from each side:

$2 C {r}^{3 +} + 2 {H}_{2} O + 3 {H}_{2} {O}_{2} \rightarrow 2 C r {O}_{4}^{2 -} + 10 {H}^{+}$

But you wanted the reaction done under basic conditions. And so we add $10 \times H {O}^{-}$ to BOTH SIDES OF THE REACTION:

$2 C {r}^{3 +} + 3 {H}_{2} {O}_{2} + 10 H {O}^{-} \rightarrow 2 C r {O}_{4}^{2 -} + 8 {H}_{2} O$

Is this balanced with respect to mass and charge?