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

Hint: 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.