Animated Solution for Chemistry - Electrochemistry: How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane? (Atomic weight of B = 10.8 $\mu$)

The problem we are tackling is a beautiful bridge between two distinct chemical worlds: the fiery combustion of diborane and the steady, electrical decomposition of water. It requires us to think backwards—starting from the end goal and tracing our steps back to the source.

Our mission begins with a specific target: we need to completely burn $27.66\text{ g}$ of diborane ($B_2{H}_6$). In chemistry, mass is just a disguise; the true currency is moles.

To find the moles, we first calculate the molar mass of diborane. Using the given atomic weight of Boron ($10.8\text{ u}$) and Hydrogen ($1\text{ u}$): $M_{B_2{H}_6}=2\times 10.8+6\times 1=27.6\text{ g/mol}$

Now, we convert the given mass into moles: $n_{B_2{H}_6}=\frac{27.66}{27.6}\approx 1\text{ mol}$

We have exactly $1\text{ mol}$ of diborane to burn.

How much oxygen does this $1\text{ mol}$ of diborane need? To answer this, we must look at the balanced chemical equation for its combustion. When diborane burns, it reacts with oxygen to form boric oxide and water: $B_2{H}_6+3O_2⟶B_2{O}_3+3H_{2}O$

The stoichiometry is clear: $1\text{ mol}$ of $B_2{H}_6$ requires exactly $3\text{ mol}$ of $O_2$. This is our oxygen demand. We now know exactly what the electrolysis cell needs to produce.

To generate these $3\text{ mol}$ of oxygen, we turn to the electrolysis of water. At the anode, water molecules are oxidized to release oxygen gas. Let's examine the half-reaction: $2H_{2}O⟶O_2+4H^{+}+4e^{-}$

This equation tells us a crucial fact: producing $1\text{ mol}$ of $O_2$ requires the transfer of $4\text{ mol}$ of electrons.

Since our demand is $3\text{ mol}$ of $O_2$, the total moles of electrons required will be: $n_{e^{-}}=3\times 4=12\text{ mol}$

In electrochemistry, $1\text{ mol}$ of electrons corresponds to $1\text{ Faraday}$ ($F$) of charge. Therefore, the total charge $Q$ required is $12F$. Knowing that $1F=96500\text{ C}$, we have: $Q=12\times 96500\text{ C}$

We have the total charge, and we are given a steady current $I=100\text{ A}$. The relationship between charge, current, and time is one of the most fundamental in physics: $Q=I\times t$

Substituting our known values: $12\times 96500=100\times t$

Solving for $t$ gives us the time in seconds: $t=\frac{12\times 96500}{100}\text{ seconds}$

However, our options are in hours. To convert seconds to hours, we divide by $3600$: $t=\frac{12\times 965}{3600}\text{ hours}$

Simplifying the fraction: $t=\frac{11580}{3600}\approx 3.216\text{ hours}$

Rounding to the nearest given option, we get $3.2\text{ hours}$. The mission is complete!