Question

In dilute aqueous $H_{2}\text{S}\text{O}_{4}$ , the complex diaquodioxalatoferrate(II) is oxidized by $MnO_{4}^{-}$ . For this reaction, the ratio of the rate of change of $\left[H^{+}\right]$ to the rate of change of $\left[M n O_{4}^{-}\right]$ is

Answer 1

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