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Balancing Redox Equations

In dilute aqu...

Question

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

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Solution

$8 H^{+} + M n O_{4}^{-} + [F e (H_{2} O)_{2} (O X)_{2}]^{2 -} \to M n^{2 +} + F e^{3 +} + 4 C O_{2} + 6 H_{2} O$
$\frac{rate of change of [H^{+}]}{rate of change of [M n O_{4}^{-}]} = 8$

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Similar questions

M n O_{4}^{-}

H_{2} S O_{4}

M n O_{4}^{-}

M n O_{4}^{-}

H_{2} O_{2}

(a)

H_{2} O_{2} + 2 H^{+} + 3 I^{-} ⟶ {I_{3}}^{-} + 2 H_{2} O

- \frac{d [I^{-}]}{d t} =

- \frac{d [H^{+}]}{d t} =

(b)

16 H^{+} + 2 {M n O_{4}}^{-} + 10 I^{-} ⟶ 2 {M n}^{2 +} + 8 H_{2} O + 5 I_{2}

- \frac{d [M n {O_{4}}^{-}]}{d t} =

(c)

4 N H_{3} + 5 O_{2} ⟶ 4 N O_{2} + 6 H_{2} O

- \frac{d [N H_{3}]}{d t} =

M n O_{4}^{⊖} + C_{2} O_{4}^{2 -} + H^{\oplus} \to M n^{+ 2} + C O_{2} + H_{2} O

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