Question

In dilute aqueous ${H_2}{O_4}$, the complex diaquadioxalatochromate $\left( {II} \right)$ is oxidized by $MnO_4^ -$. What will be the ratio of rate of change of $\left[ {{H^ + }} \right]$ to the rate of change of $\left[ {MnO_4^ - } \right]$ for this reaction?

Answer 1

First of all, make the chemical equation which is taking place in the question. Then, balance that chemical equation.
From the chemical equation, find the rate of change permanganate $\left[ {MnO_4^ - } \right]$ and rate of change of $\left[ {{H^ + }} \right]$. Then, find their ratio by dividing rate of change of $\left[ {{H^ + }} \right]$ by rate of change of permanganate.

Complete step by step answer:
The permanganate is the chemical compound whose molecular formula is $\left[ {MnO_4^ - } \right]$. This chemical compound contains the manganate $(VII)$ ion. The permanganate is the strong oxidizing agent because manganese has an oxidation state of $+ 7$. the colour of permanganate is purple and becomes stable when they are neutral or in slightly alkaline media. The permanganate ion has a tetrahedral geometry.
Now, according to the question, the permanganate oxidized the complex diaquodioxalatoferrate in dilute aqueous ${H_2}{O_4}$. This can be represented in the form of chemical equation as –
$MnO_4^ - + {\left[ {Fe\left( {{H_2}O} \right){{\left( {{C_2}{O_4}} \right)}_2}} \right]^{2 - }} + {H^ + } \to {\left[ {Fe\left( {{H_2}O} \right){{\left( {{C_2}{O_4}} \right)}_2}} \right]^ - } + M{n^{2 + }} + {H_2}O$
Now, we have to balance the above given chemical equation which can be done as –
$MnO_4^ - + 5{\left[ {Fe\left( {{H_2}O} \right){{\left( {{C_2}{O_4}} \right)}_2}} \right]^{2 - }} + 8{H^ + } \to 5{\left[ {Fe\left( {{H_2}O} \right){{\left( {{C_2}{O_4}} \right)}_2}} \right]^ - } + M{n^{2 + }} + 4{H_2}O$
Now, this chemical equation is the balanced chemical equation which is given in the question.
So, now we have to find the rate of change of $\left[ {{H^ + }} \right]$ and $\left[ {MnO_4^ - } \right]$.
Hence, the rate of change of $\left[ {{H^ + }} \right]$ can be calculated as $\Rightarrow - \dfrac{1}{8}\dfrac{d}{{dt}}\left[ {{H^ + }} \right]$ , and
the rate of change of $\left[ {MnO_4^ - } \right]$ can be calculated as $\Rightarrow - \dfrac{d}{{dt}}\left[ {MnO_4^ - } \right]$
To calculate the ratio of rate of change of $\left[ {{H^ + }} \right]$ and $\left[ {MnO_4^ - } \right]$ we have to divide the rate of change of $\left[ {{H^ + }} \right]$ by rate of change of $\left[ {MnO_4^ - } \right]$. This can be done as –
$\Rightarrow \dfrac{{\dfrac{1}{8}\dfrac{d}{{dt}}\left[ {{H^ + }} \right]}}{{\dfrac{d}{{dt}}\left[ {MnO_4^ - } \right]}} \\\ \Rightarrow \dfrac{{\dfrac{d}{{dt}}\left[ {{H^ + }} \right]}}{{\dfrac{d}{{dt}}\left[ {MnO_4^ - } \right]}} = 8 \\\$

Hence, the ratio of rate of change of $\left[ {{H^ + }} \right]$ and $\left[ {MnO_4^ - } \right]$ is $8$.

Note:

Permanganate is the strong oxidizer. So, it is used in redox reactions which needs the qualitative analysis. It is also strong enough to oxidize water.
Permanganate is manufactured by the oxidation of manganese compounds.