What current in amperes is required to produce `50.0` ml of `O_2` gas measured at STP by electrolysis of water for a period of 3 hrs ?

Explanation

To find the current required to produce 50.0 ml of O2 gas through the electrolysis of water for a period of 3 hours, we can follow these steps:

### Step 1: Calculate the number of moles of O2 produced
At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 liters. We need to convert 50.0 ml of O2 to liters:

\[ 50.0 \, \text{ml} = 50.0 \times 10^{-3} \, \text{L} = 0.0500 \, \text{L} \]

Now, we can calculate the number of moles of O2 produced:

\[ \text{Moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Molar Volume at STP}} = \frac{0.0500 \, \text{L}}{22.4 \, \text{L/mol}} \approx 0.00224 \, \text{mol} \]

### Step 2: Determine the charge required to produce the O2
From the electrolysis of water, the reaction is:

\[ 2H_2O \rightarrow 2H_2 + O_2 \]

This means that 1 mole of O2 requires 4 moles of electrons (since 4 electrons are needed to produce 1 mole of O2). The charge (Q) required can be calculated using Faraday's constant (F = 96500 C/mol):

\[ \text{Charge (Q)} = \text{Moles of } O_2 \times 4 \times F \]
\[ Q = 0.00224 \, \text{mol} \times 4 \times 96500 \, \text{C/mol} \approx 861.16 \, \text{C} \]

### Step 3: Calculate the current required
The current (I) can be calculated using the formula:

\[ Q = I \times t \]

Where:
- \( Q \) is the charge in coulombs
- \( I \) is the current in amperes
- \( t \) is the time in seconds

First, convert the time from hours to seconds:

\[ t = 3 \, \text{hours} = 3 \times 60 \times 60 \, \text{seconds} = 10800 \, \text{s} \]

Now, rearranging the formula gives us:

\[ I = \frac{Q}{t} = \frac{861.16 \, \text{C}}{10800 \, \text{s}} \approx 0.0798 \, \text{A} \]

### Final Answer
The current required to produce 50.0 ml of O2 gas by electrolysis of water for a period of 3 hours is approximately **0.0798 A**.

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