The quantity of charge (in Faraday) required to electrolyse 54 g H $_{2}$ O is :

Open in App

Solution

$2 H_{2} O (l) + 2 e^{-} \to H_{2} (g) + 2 O H^{-} (a q)$
Thus, the electrolysis of 1 mole of water requires 2 moles of electrons or 2 faraday of charge.
54 g of water corresponds to $\frac{54}{18} = 3$ moles which will require $3 \times 2 = 6$ faraday of charge for electrolysis.

Suggest Corrections

Similar questions

Q. The quantity of charge (in Faraday) required to reduce 96 g Mg from molten solution of MgCl

_{2}

is :

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

Q. The quantity of charge (in Faraday) required to liberate 33.6 litre Cl

_{2}

from molten NaCl:

Q. How many Faraday charge is required for

{K M n O}_{4} \to {M n}^{2 +}

Q. Assertion :The complete electrolysis of 1 mole of water requires 2 Faradays. Reason: Faraday is the quantity of electric charge carried by 1 mole of electrons.

Join BYJU'S Learning Program

The quantity of charge (in Faraday) required to electrolyse 54 g H2O is :

2H2O(l)+2e−→H2(g)+2OH−(aq)Thus, the electrolysis of 1 mole of water requires 2 moles of electrons or 2 faraday of charge.54 g of water corresponds to 5418=3 moles which will require 3×2=6 faraday of charge for electrolysis.

The quantity of charge (in Faraday) required to electrolyse 54 g H $_{2}$ O is :

$2 H_{2} O (l) + 2 e^{-} \to H_{2} (g) + 2 O H^{-} (a q)$
Thus, the electrolysis of 1 mole of water requires 2 moles of electrons or 2 faraday of charge.
54 g of water corresponds to $\frac{54}{18} = 3$ moles which will require $3 \times 2 = 6$ faraday of charge for electrolysis.