Question Video: Calculating the Mass of Hydrogen Produced in a Cell | Nagwa Question Video: Calculating the Mass of Hydrogen Produced in a Cell | Nagwa

# Question Video: Calculating the Mass of Hydrogen Produced in a Cell Chemistry • Third Year of Secondary School

A current of 2.345 A passes through the cell shown in the figure for 45 minutes. What is the mass of the hydrogen produced, in units of grams, to 1 decimal place and in scientific notation? [H = 1 g/mol]

04:10

### Video Transcript

A current of 2.345 amps passes through the cell shown in the figure for 45 minutes. What is the mass of the hydrogen produced, in units of grams, to one decimal place and in scientific notation? The molar mass of hydrogen is one gram per mole.

The apparatus in the diagram is known as a Hoffman voltameter. In the cathode of the apparatus, hydrogen ions are reduced to form hydrogen gas. At the anode, water is oxidized to form oxygen gas and hydrogen ions.

Since this question asks us about the mass of hydrogen produced, we’re interested in the reaction occurring at the cathode. Our first step will be to determine the total amount of charge, or quantity of electrons, that were used to reduce the hydrogen ions to hydrogen gas. We can do that with this equation, which tells us the total charge 𝑄 is equal to the current 𝐼 multiplied by the time.

The current given in the problem is 2.345 amps, which we can go ahead and plug into our equation. The time given in the problem is 45 minutes. But we need to express this time in seconds to plug it into our equation. There are 60 seconds in one minute, so we can convert our time from minutes to seconds if we multiply by 60. This gives us 2700 seconds, which we can plug into our equation now. One amp is equivalent to one coulomb per second. We can substitute those units in. Now, the seconds will cancel, giving us 6331.5 coulombs.

But to determine the amount of hydrogen gas produced, we need to know how many electrons, in moles, this amount of charge is equivalent to. We can figure that out using this equation, which tells us the total charge is equal to the amount of charged particles, in moles, multiplied by Faraday’s constant. We need to isolate the amount of electrons so we can solve the problem, which we can do if we divide both sides of the equation by Faraday’s constant, which gives us this equation. Let’s just flip that around so that the amount of charged particles is on the left-hand side. And then we’ll solve.

We solved for the amount of charge previously; it’s 6331.5 coulombs. Faraday’s constant has a value of 96500 coulombs per mole. Performing the calculation gives us 0.06561 moles of electrons. The reaction equation tells us that every two moles of electrons produces one mole of hydrogen gas. In other words, the amount of hydrogen gas produced will be half the amount of electrons that reacted, or 0.032805 moles.

We’ve almost solved the problem now. We just need to convert this amount of hydrogen gas to a mass. We can do that by multiplying the amount of hydrogen gas by the molar mass of hydrogen gas. We can plug in the amount of hydrogen gas we just calculated. And we can use the molar mass data provided in the problem to determine the molar mass of hydrogen gas. One hydrogen atom has a molar mass of one gram per mole and there are two hydrogen atoms in one molecule of hydrogen gas. So the molar mass of hydrogen gas is two grams per mole. The units of moles cancel, which gives us 0.06561 grams.

The problem told us to express our answer in scientific notation, which is a number between one and 10 multiplied by a power of 10. Expressing the answer in scientific notation would be 6.561 times 10 to the negative two grams. Finally, the problem said to report our answer to one decimal place, so we can round to get our final answer. So the mass of hydrogen produced when a current of 2.345 amps passes through the Hoffman voltameter for 45 minutes is 6.6 times 10 to the negative two grams.

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