During electrolysis of acidified water, \(2.8 L\) of oxygen gas is released (at STP). The charge passed through the anode is:

48250 C

The correct answer is option 1. 48250C.

To find the charge passed through the anode during the electrolysis of water, we can use Faraday's law of electrolysis, which states that the amount of substance produced or consumed in an electrolytic reaction is directly proportional to the quantity of electricity passed through it.

First, let's calculate the number of moles of oxygen gas produced:

Given that 1 mole of any gas at STP occupies 22.4 L, and we have 2.8 L of oxygen gas produced, we can calculate the number of moles of oxygen gas:

\( \text{Volume of oxygen gas (at STP)} = 2.8 L \)

\( \text{Number of moles of oxygen gas} = \frac{2.8\, \text{L}}{22.4\, \text{L/mol}} = 0.125\, \text{mol} \)

Now, each mole of oxygen gas (\(O_2\)) requires 4 moles of electrons (according to the balanced equation for the electrolysis of water), which means to produce 1 mole of oxygen gas, we need \(4 \times 96500\, \text{C}\) of charge (since 1 Faraday (\(F\)) is equivalent to \(96500\, \text{C}\)).

So, for \(0.125\) moles of oxygen gas, the charge passed would be:

\( \text{Charge passed} = 0.125 \times 4 \times 96500\, \text{C} \)

\( \text{Charge passed} = 4825\, \text{C} \)

Therefore, the correct answer is option 1: \(48250\, \text{C}\).