A hydrogen-oxygen fuel cell operates on the simple reaction given below.
H2(g) + 1/2 O2(g) H2O(l)
If the cell is designed to produce 1.8 A of current, and if the hydrogen is contained in a 2.9 L tank at 200. atm pressure at 25°C, how long can the fuel cell operate before the hydrogen runs out? (Assume there is an unlimited supply of O2.)
so i first used pv=nrt and solved for n and got 23.71 moles but i don't know where to go from there
I have the same problem! I don't know how to solve it, so please if you found the solution can you send it to me?? I don't know how this webpage works so my email is nazarenoroggero @ gmail . com
Great job using the ideal gas law (PV = nRT) to find the number of moles of hydrogen gas in the tank. Now let's proceed further to calculate the time the fuel cell can operate before the hydrogen runs out.
Given:
Current (I) = 1.8 A
The balanced equation for the reaction shows that 2 moles of hydrogen gas are consumed to produce 2 moles of water (H2O). So, based on the stoichiometry of the reaction, 1 mole of hydrogen gas would produce 1 mole of water.
We can use Faraday's law of electrolysis to relate the current to the amount of substance consumed. According to Faraday's law, the amount of substance consumed (in moles) is directly proportional to the current and inversely proportional to the charge per mole of electrons. The charge per mole of electrons for this reaction is 2F, where F is the Faraday constant (approximately 96,485 C/mol).
Let's calculate the amount of hydrogen gas that will be consumed:
Amount of hydrogen gas consumed (in moles) = (Current (in A) × Time (in seconds)) / (2F)
To find the time the fuel cell can operate, we need to rearrange the equation:
Time (in seconds) = (Amount of hydrogen gas (in moles) × 2F) / Current (in A)
We already know the amount of hydrogen gas in moles, which is 23.71 moles.
Plugging in the values, we get:
Time (in seconds) = (23.71 moles × 2 × 96,485 C/mol) / 1.8 A
Now, calculate the time.