Question

In dilute aqueous $H_2\text{S}{\text{O}}_4$ , the complex diaquodioxalatoferrate(II) is oxidized by $Mn{O}_4^{-}$ . For this reaction, the ratio of the rate of change of $\left[H^{+}\right]$ to the rate of change of $\left[Mn{O}_4^{-}\right]$ is

Answer 1

Answer: 8

The balanced chemical equation for the oxidation of the complex is as given below
$\text{Mn}{\text{O}}_4^{-}+5[\text{Fe}\left(H_{2}O\right){\left(C_2{O}_4\left({)}_2\right]\right)}^{2-}+\text{8}{\left(\text{H}\right)}^{+}\to 5{\left(\left[\text{Fe}\left(H_{2}O\right){\left(C_2{O}_4\right)}_2\right]\right)}^{-}+\text{M}{\left(\text{n}\right)}^{2+}+\text{4}{\left(\text{H}\right)}_{2}O$
Hence $-\frac{d}{dt}\left[Mn{O}_4^{-}\right]=-\frac{1}{8}\frac{d}{dt}\left[H^{+}\right]$
$\frac{\frac{d}{dt}\left[H^{+}\right]}{\frac{d}{dt}\left[Mn{O}_4^{-}\right]}=8$
Rate of change of $\left[H^{+}\right]$ is 8 times the rate of change of $\left[Mn{O}_4^{-}\right]$ .