Consider the following reaction MnO4-+8H++5e- arrow Mn+ 2+4H2O,E° =1.51V . The quantity of electricity required in Faraday, to reduce five moles of MnO4- is.

Question

Consider the following reaction MnO4-+8H++5e- arrow Mn+ 2+4H2O,E° =1.51V . The quantity of electricity required in Faraday, to reduce five moles of MnO4- is. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

MnO4-+8H++5e- arrow Mn+ 2+4H2O 1 mole of MnO4- require 5 Faraday charge 5 moles of MnO4- will require 25 Faraday charge.

Q. Consider the following reaction $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{+ 2} + 4 H_{2} O, E^{\circ} = 1.51 V$ .
The quantity of electricity required in Faraday, to reduce five moles of $M n O_{4}^{-}$ is______.

$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{+ 2} + 4 H_{2} O$
$1$ mole of $M n O_{4}^{-}$ require $5$ Faraday charge
$5$ moles of $M n O_{4}^{-}$ will require $25$ Faraday charge.

Q. Consider the following reaction MnO4−​+8H++5e−→Mn+2+4H2​O,E∘=1.51V . The quantity of electricity required in Faraday, to reduce five moles of MnO4−​ is______.

MnO4−​+8H++5e−→Mn+2+4H2​O 1 mole of MnO4−​ require 5 Faraday charge 5 moles of MnO4−​ will require 25 Faraday charge.

Q. Consider the following reaction $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{+ 2} + 4 H_{2} O, E^{\circ} = 1.51 V$ .
The quantity of electricity required in Faraday, to reduce five moles of $M n O_{4}^{-}$ is______.

$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{+ 2} + 4 H_{2} O$
$1$ mole of $M n O_{4}^{-}$ require $5$ Faraday charge
$5$ moles of $M n O_{4}^{-}$ will require $25$ Faraday charge.