Question

How many Coulombs of electricity are required for complete oxidation of 90 gm of H$_2$O ?

Hint:

Always start by writing the balanced half-reaction to determine 'n' (number of electrons transferred per mole).

Answer 1

The oxidation reaction of water is:
H$_2$O $\to$ 2H$^+$ + $\frac{1}{2}$O$_2$ + 2e$^-$
1. Moles of Water:
Molar mass of H$_2$O = 18 g/mol.
Moles of H$_2$O = $\frac{90}{18} = 5$ moles.
2. Electrons required:
From the equation, oxidation of 1 mole of H$_2$O releases 2 moles of electrons.
So, 5 moles of H$_2$O will release $5 \times 2 = 10$ moles of electrons.
3. Charge Calculation:
Charge of 1 mole of electrons (1 Faraday) $\approx$ 96487 C.
Total Charge = $10 \times 96487 = 964870$ C.