Consider the following reaction
$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{+ 2} + 4 H_{2} O, E^{o} = 1.51 V .$

The quantity of electricity required in Faraday to reduced five moles of $M n O_{4}^{-}$ is
(Integer answer)

25.0

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25.00

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Solution

$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{+ 2} + 4 H_{2} O; E^{o} = 1.51 V$

$∵$ 1 mole of $M n O_{4}^{-}$ requires 5 moles of electrons or 5 F electricity.

$∴$ 5 moles of $M n O_{4}^{-}$ requires 25 F electricity.

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Similar questions

$M n {O_{4}}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O E^{o} = 1.51 V$
The quantity of electricity required in Faraday to reduce five moles of $M n O_{4}^{-}$ is ____.

M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O; E^{O} = 1.51 V; Δ G_{1}^{0} = - 5 \times 1.51 \times F

M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O; E^{O} = 1.23 V; Δ G_{2}^{0} = - 5 \times 1.23 \times F

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

Q. The following two redox reactions are given:
i)

M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O; E^{o} = X_{1} V

ii)

M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O; E^{o} = X_{2} V

Find

E^{o}

for the following reaction.

M n O_{4}^{-} + 4 H^{+} + 3 e^{-} \to M n O_{2} + 2 H_{2} O

Q. Calculate the quantity of electricity ( in Faraday) that would be required to reduce 12.3 g of nitrobenzene to aniline, if current efficiency is 50% (write answer in nearest integer).

Q. In the following redox reaction:

5 {F e}^{2 +} + M n O_{4}^{-} + 8 H^{+} ⇌ {M n}^{2 +} + 5 {F e}^{3 +} + 4 H_{2} O

Given:

{F e}^{3 +} + e^{-} ⟶ {F e}^{2 +}, {E_{1}}^{o}

M n O_{4}^{-} + 8 H^{+} + 5 e^{-} ⟶ {M n}^{2 +} + 4 H_{2} O, {E_{2}}^{o}

Potential at the equivalence point is :

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Consider the following reaction MnO−4+8H++5e−→Mn+2+4H2O,Eo=1.51 V. The quantity of electricity required in Faraday to reduced five moles of MnO−4 is (Integer answer)

MnO−4+8H++5e−→Mn+2+4H2O ;Eo=1.51 V ∵ 1 mole of MnO−4 requires 5 moles of electrons or 5 F electricity. ∴ 5 moles of MnO−4 requires 25 F electricity.

Consider the following reaction
$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{+ 2} + 4 H_{2} O, E^{o} = 1.51 V .$

The quantity of electricity required in Faraday to reduced five moles of $M n O_{4}^{-}$ is
(Integer answer)

$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{+ 2} + 4 H_{2} O; E^{o} = 1.51 V$

$∵$ 1 mole of $M n O_{4}^{-}$ requires 5 moles of electrons or 5 F electricity.

$∴$ 5 moles of $M n O_{4}^{-}$ requires 25 F electricity.