Question

how many no of coulombs of charge required for the oxidation of mole of water to oxygen

Answer 1

2×96500C= 32g.m of oxygen
1 mole of water require =16 g.m oxygen or ½mole of oxygen
so charge require is 96500 C

Answer 2

Given:

Oxidation of mole of water to oxygen.

To find:

No. of coulombs of charge = ?

Formula to be used:

C = nF

where 'n' is difference of charge in ions, and 'F' is constant with standard value 96487.

Calculation:

When one mole of water is oxidized, it yields one oxygen atom as per the following chemical equation

$H_2O\rightarrow H_2 + \frac{1}{2}O_2$

Charge of O = -2

When this value is substitued in the following formula we get,

C=nF

$C = 2\times 96487$

$C = 192974$

Conclusion:

Thus it is inferred that for the oxidation of mole of water to oxygen 192974 C is required.

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