Question

Three faradays of electricity is passed through molten \(\mathrm{Al}_{2} \mathrm{O}_{3}\), aqueous solutions of \(\mathrm{CuSO}_{4}\) and molten \(\mathrm{NaCl}\). The amounts of \(\mathrm{Al}, \mathrm{Cu}\) and \(\mathrm{Na}\) deposited at the cathodes will be in the molar ratio of (a) \(1: 2: 3\) (b) \(3: 2: 1\) (c) \(1: 1.5: 3\) (d) \(6: 3: 2\)

Short answer

The amounts of \(\mathrm{Al}, \mathrm{Cu}\) and \(\mathrm{Na}\) deposited at the cathodes will be in the molar ratio of \(1: 1.5: 3\).

Step-by-step solution

Understanding the electrochemical reaction

First, we need to understand that in electrolysis, metals are deposited at the cathode equivalent to the number of faradays of charge passed through the solution, according to Faraday's laws of electrolysis. This is determined by the charge of the metal ion and the amount of electricity passed. For example, to deposit 1 mole of a monovalent ion (like Na+), 1 faraday is needed. For a divalent ion like Cu2+, 1 faraday deposits 0.5 moles, and for a trivalent ion like Al3+, 1 faraday deposits 1/3 moles.

Calculating moles of Aluminium (Al) deposited

For aluminium, which has a valency of 3 (from \(\mathrm{Al_{2}O_{3}}\)), it takes 3 faradays to deposit 1 mole of \(\mathrm{Al}\). Since we pass 3 faradays through molten \(\mathrm{Al_{2}O_{3}}\), we get \(\frac{3}{3} = 1\) mole of \(\mathrm{Al}\) deposited.

Calculating moles of Copper (Cu) deposited

For copper, which has a valency of 2 (from \(\mathrm{CuSO}_{4}\)), it takes 2 faradays to deposit 1 mole of \(\mathrm{Cu}\). Since we pass 3 faradays through an aqueous solution of \(\mathrm{CuSO}_{4}\), we get \(\frac{3}{2} = 1.5\) moles of \(\mathrm{Cu}\) deposited.

Calculating moles of Sodium (Na) deposited

For sodium, which has a valency of 1 (from \(\mathrm{NaCl}\)), it takes 1 faraday to deposit 1 mole of \(\mathrm{Na}\). Since we pass 3 faradays through molten \(\mathrm{NaCl}\), we get \(3 \times 1 = 3\) moles of \(\mathrm{Na}\) deposited.

Determining the molar ratio of the deposited metals

Now we have 1 mole of \(\mathrm{Al}\), 1.5 moles of \(\mathrm{Cu}\), and 3 moles of \(\mathrm{Na}\) deposited. Thus, the molar ratio of \(\mathrm{Al} : \mathrm{Cu} : \mathrm{Na}\) is \(1 : 1.5 : 3\).

Key concepts

Faraday's Laws of Electrolysis

When we talk about electrolysis, we're exploring a process where electrical energy causes a chemical change, particularly the movement of ions to produce a substance at the electrodes. To make sense of this, Faraday's laws come into play. The first law states that the mass of the substance deposited at the electrode is directly proportional to the quantity of electricity that passes through the electrolyte. This means more electricity, more substance!
Faraday's second law tells us that when the same amount of electricity is passed through different electrolytes, the masses of substances deposited are proportional to their chemical equivalent weights.
To apply these concepts, imagine you're charging your phone. The electricity used is directly associated with how much your phone charges—the more you charge, the more battery you have. Faraday's laws relate to electrolysis in a similar way—the more electricity you use, the more product you get at the electrode. This is crucial for solving electrolysis problems because it links electricity (which we can measure) to chemical changes (which we want to cause).

Molar Ratio Calculation

Understanding molar ratios is like being a master chef who knows exactly how much of each ingredient to use for the perfect recipe. In chemistry, the recipe is our reaction, and the ingredients are the reactants. The molar ratio provides the relationship between the amounts in moles of any two substances involved in a chemical reaction.
So, how do we calculate it? Take the coefficients from the balanced chemical equation and convert them into moles, if they aren’t already. If you’re baking cookies and the recipe calls for two eggs for every cup of flour, your egg to flour ratio is 2:1. Similarly, in chemistry, if the reaction requires two moles of hydrogen for every mole of oxygen to make water, the molar ratio is 2:1.
Using molar ratios, we can predict how much of each substance will be needed or produced in a reaction. When performing electrolysis, understanding the molar ratio helps determine the amount of substance deposited based on the electricity used, ensuring that not a single electron of energy goes to waste.

Electrochemical Reaction

Electrochemical reactions are the heart of the process in an electrolytic cell, where electricity becomes the catalyst for chemical changes. These reactions involve the transfer of electrons between the electrode and the electrolyte.
Here’s how it works—during electrolysis, positive ions (cations) are attracted to the negative electrode (cathode), where they gain electrons and are reduced. Conversely, negative ions (anions) head over to the positive electrode (anode), lose electrons, and are oxidized. This exchange brings about the deposition of substances at the electrodes.
Think of it as a dance floor where electrons are the dancers. At the cathode, it’s like adding more dancers to the party, while at the anode, it’s like some dancers are leaving. The overall effect is that the 'dance' chemistry of the substance is changing. Understanding the steps of these reactions is fundamental for solving electrolysis problems because it allows you to predict what will form at each electrode and how much you'll need to put into the system to get what you want out of it.