# How do you represent the oxidation of Cu^+ by MnO_4^(-)?

Nov 30, 2015

$C {u}^{+} \rightarrow C {u}^{2 +} + {e}^{-}$ OXIDATION
$M n {O}_{4}^{-} + 8 {H}^{+} + 5 {e}^{-} \rightarrow M {n}^{2 +} + 4 {H}_{2} O$ REDUCTION

#### Explanation:

We can write the overall redox equation by adding $5$ $\times$ the oxidation equation to $1$ of the reduction reaction to give:

$5 C {u}^{+} + M n {O}_{4}^{-} + 8 {H}^{+} \rightarrow M {n}^{2 +} + 5 C {u}^{2 +} + 4 {H}_{2} O \left(l\right)$

This reaction is stoichiometrically balanced with respect to mass and charge, and it is self indicating. Why? Because permanganate ion, $M n {O}_{4}^{-}$, has a strong purple colour, whereas $M {n}^{2 +}$ is almost colourless. The colour of the permanganate ion will persist at the endpoint.

Concentration of cuprous ion: $\frac{{n}_{C {u}^{+}}}{2.49 \times {10}^{-} 2 \cdot L}$

Moles of permanganate ion $=$ $0.128 \cdot m o l \cdot {\cancel{L}}^{- 1} \times 3.19 \times {10}^{- 2} \cancel{L}$ $=$ $4.08 \times {10}^{- 3} m o l$. Given the stoichiometry there were $5 \times 4.08 \times {10}^{- 3} m o l$ $=$ $2.042 \times {10}^{- 2}$ $m o l$ of cuprous ion.

SO the concentration of the original solution was $\frac{2.042 \times {10}^{- 2} \cdot m o l}{2.49 \times {10}^{-} 2 \cdot L}$ $=$ $0.820$ $m o l \cdot {L}^{- 1}$ in $C {u}^{+}$ ion.