# How do you do redox titration problems?

Jul 11, 2017

Well you require a stoichiometrically balanced equation, which represents the redox transfer....... determination of oxalic acid by potassium permanganate is a good example.

#### Explanation:

$\text{Permanganate ion,}$ $M n {O}_{4}^{-}$ is a commonly used reagent for redox titrations........Why? Well because as $M n \left(V I I +\right)$ it is strongly coloured RED; and $M n \left(I I +\right)$, its reduction product is almost colourless.....and we must supply the reduction half equation....

Deep purple permanganate ion, $M n \left(V I I +\right)$, is reduced to colourless $M {n}^{2 +}$:

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

Charge and mass are balanced as required......

Meanwhile oxalic acid is oxidized to carbon dioxide.......$C \left(+ I I I\right) \rightarrow C \left(+ I V\right)$.........

${C}_{2} {O}_{4}^{2 -} \rightarrow 2 C {O}_{2} \left(g\right) \uparrow + 2 {e}^{-}$ $\left(i i\right)$

We takes $2 \times \left(i\right) + 5 \times \left(i i\right)$ to eliminate the electrons.......

$2 M n {O}_{4}^{-} + 6 {H}^{+} + 5 H O \left(O =\right) C - C \left(= O\right) O H \left(a q\right) \rightarrow 2 M {n}^{2 +} + 10 C {O}_{2} \left(g\right) \uparrow + 8 {H}_{2} O \left(l\right)$

Vizualization of the endpoint is made easy by the disappearance of the colour of permanganate.... The stoichiometry follows the given equation.