How does permanganate ion oxidize oxalate dianion to give carbon dioxide, and Mn^(2+)? Given 0.1467*g of "sodium oxalate", what was the concentration with respect to "permanganate" if 28.85*mL were required for equivalence?

Aug 9, 2017

Well, we needs a redox equation............

$5 {C}_{2} {O}_{4}^{2 -} + 2 M n {O}_{4}^{-} + 16 {H}^{+} \rightarrow 10 C {O}_{2} + 2 M {n}^{2 +} + 8 {H}_{2} O \left(l\right)$

I gets approx. $1.5 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$ with respect to permanganate.

Explanation:

Permanganate ion is reduced to COLOURLESS $M {n}^{2 +}$ ion:

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

The abrupt colour change acts as the indicator for this reaction........

And oxalate ion is oxidized to give 2 equiv of carbon dioxide.......take that climate!

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

Are these individual redox equations balanced with respect to mass and charge? Do not trust my arithmetic........

And we adds the redox equation so that electrons, conceptual particles, do not appear in the final equation. And thus $2 \times \left(i\right) + 5 \times \left(i i\right)$ gives..........

$5 {C}_{2} {O}_{4}^{2 -} + 2 M n {O}_{4}^{-} + 16 {H}^{+} + \cancel{10 {e}^{-}} \rightarrow 10 C {O}_{2} + \cancel{10 {e}^{-}} + 2 M {n}^{2 +} + 8 {H}_{2} O \left(l\right)$

$5 {C}_{2} {O}_{4}^{2 -} + 2 M n {O}_{4}^{-} + 16 {H}^{+} \rightarrow 2 M {n}^{2 +} + 10 C {O}_{2} + 8 {H}_{2} O \left(l\right)$

The which is balanced with respect to mass and charge, and thus a reasonable representation of reality.........

$\text{Moles of sodium oxalate} = \frac{0.1467 \cdot g}{134.00 \cdot g \cdot m o {l}^{-} 1} = 1.095 \times {10}^{-} 3 \cdot m o l$

And thus there were $\frac{2}{5} \times 1.095 \times {10}^{-} 3 \cdot m o l$ of permanganate ion in the given solution......

And so we can calculate $\left[K M n {O}_{4} \left(a q\right)\right]$

$= \text{Moles of permanganate (mol)"/"Volume of solution (L)}$

=(2/5xx1.095xx10^-3*mol)/(28.85*mLxx10^-3*L*mL^-1)=??*mol*L^-1