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Integration Using Substitution

PbO 2 oxidise...

Question

${P b O}_{2}$ oxidises ${M n O}_{2}$ (black) to :

{M n O}_{4}^{2 -}

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{M n O}_{4}^{-}

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{M n}^{2 +}

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{M n}_{2} O_{7}

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Solution

The correct option is B ${M n O}_{4}^{-}$
${P b O}_{2}$ oxidises ${M n O}_{2}$ (black) to ${M n O}_{4}^{-}$ .
$2 M n O_{2} + 6 H N O_{3} + 3 P b O_{2} ⇌ = 2 H_{2} O + 2 H M n O_{4} + 3 P b (N O_{3})_{2}$

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Similar questions

{M n O_{4}}^{-}

M n^{2 +}

M n O_{2}

x

x

K M n O_{4}

M n O_{4}^{2 -}, M n O_{2}, {M n}_{2} O_{3}

{M n}^{2 +}

K M n O_{4}

M n O_{4}^{2 -}, M n O_{2}, {M n}_{2} O_{3}

{M n}^{2 +}

E_{M n O_{4}^{-} | M n O_{2}}^{\circ}

M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O

E^{\circ} = 1.51 V

M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O

E^{\circ} = 1.23 V

M n O_{4}^{-} / M n O_{2}

M n O_{4}^{-} / M n^{2 +} = 1.51 V

M n O_{2} / M n^{2 +} = 1.23 V

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