Animated Solution for Chemistry - Electrochemistry: 108 g of silver (molar mass $108{\text{ g mol}}^{-1}$) is deposited at cathode from ${\text{AgNO}}_3(aq)$ solution by a certain quantity of electricity. The volume (in L) of oxygen gas produced at 273 K and 1 bar pressure from water by the same quantity of electricity is ………

Imagine an electrolytic cell humming with energy. We are passing a certain amount of electricity through an aqueous solution of silver nitrate. At the cathode, beautiful silver metal begins to deposit, while at the anode, water oxidizes to release bubbles of oxygen gas. This is the physical reality of our problem.

But how do we connect the amount of silver deposited to the volume of oxygen gas evolved? The bridge between these two seemingly different events is Faraday's Second Law of Electrolysis.

Equivalents

Faraday's Second Law states that when the same quantity of electricity is passed through different substances, the number of equivalents produced at the electrodes is exactly the same. This is a powerful tool because it allows us to bypass calculating the exact charge in Coulombs.

We can write our master equation as:

Let's find out how many equivalents of silver we have. The mass of silver deposited is given as $108\text{ g}$. The molar mass of silver is also $108{\text{ g mol}}^{-1}$. Since the silver ion is ${\text{Ag}}^{+}$, its n-factor is $1$, making its equivalent mass equal to its molar mass.

So, we have exactly $1$ equivalent of silver. This immediately tells us that we also have $1$ equivalent of oxygen gas.

Now, we need to convert the equivalents of oxygen into moles. Let's look at the oxidation reaction of water at the anode:

Notice that producing one molecule of ${\text{O}}_2$ requires the transfer of $4$ electrons. This means the n-factor for oxygen gas is 4.

Knowing the n-factor, we can easily find the moles of oxygen. Moles are simply equivalents divided by the n-factor:

Volume

Finally, we need the volume of this oxygen gas at $273\text{ K}$ and $1\text{ bar}$ pressure. Remember, at these standard conditions (STP defined by IUPAC since 1982), one mole of an ideal gas occupies $22.7\text{ L}$.

So, we multiply our moles by the molar volume:

Rounding this to two decimal places as per the given answer key, we get $5.66\text{ L}$.

Always pay close attention to the standard conditions mentioned in the question! If the pressure was $1\text{ atm}$ instead of $1\text{ bar}$, the molar volume would be $22.4\text{ L}$, and the answer would be exactly $5.6\text{ L}$.