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)

(  JEE MAIN 2021)

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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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Q. 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)

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

Find 10 rational numbers between $\frac{- 2}{5}$ and $\frac{1}{2}$ .

Which of the following rational numbers does not lie between $\frac{14}{20}$ and $\frac{37}{40}$ .

A.  $\frac{3}{4}$ 

B.  $\frac{7}{8}$ 

C.  $\frac{9}{10}$ 

D.  $\frac{3}{5}$

Represent $\frac{19}{8}$ on the number line.

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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) (  JEE MAIN 2021)

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.

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

$∵$ 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.