A metal is known to form fluoride \(M \mathrm{~F}_{2}\). When 10 ampere electricity is passed throwgh i multen salt for \(330 \mathrm{sec}, 1.95 \mathrm{~g}\) metal is deposited. Find out the atomic weight of metal. What will be the quantity of charge required to deposit the same mass of \(\mathrm{Cu}\) from \(\mathrm{CuSO}_{4}\) (aq.) (At. wt. of \(\mathrm{Cu}=63.6\) )

Understanding the electrochemical equivalent is crucial for solving electrolysis problems. Imagine it as the bridge that connects the electrical world with the chemical world. In technical terms, it is the amount of a substance that is deposited or dissolved at an electrode during electrolysis for every unit of electric charge that passes through the electrolyte.

How do we find it? You would compute it by dividing the mass of the substance deposited by the total charge used in the process, essentially using Faraday's First Law as your formula guide. In our problem, after finding out how much charge was used to deposit metal (calculated by multiplying the current by the time), we can determine the electrochemical equivalent of the metal simply by rearranging the formula from the solution steps. This value is pivotal for calculating the atomic weight of the metal, which is our next step.

The atomic weight of a metal in electrolysis isn't just a number; it's a scale for measuring how much 'stuff' makes up one mole of that metal. When dealing with electrolysis, we're not only interested in the mass of a substance that deposits but also in knowing the amount of that substance on a molecular level.

For our problem, the atomic weight is derived from the electrochemical equivalent, and the electrochemical equivalent is directly proportional to the atomic weight. Once we have the electrochemical equivalent and we know the valency (which is the charge the ion would have in an ionic compound), we can apply both to Faraday’s First Law to find the atomic weight. This figure is indispensable for chemists and physicists alike for calculations involving molecular amounts.

Facing electrolysis problems can sometimes leave students with more questions than answers. Understanding not just one concept but the interplay between physics and chemistry is the key to solving these problems. Firstly, identify what information is given and what needs to be calculated; in our exercise, we're after the atomic weight and the charge required to deposit a different metal.

Common issues often arise from misinterpreting the relationship between atomic weight and electrochemical equivalent or misunderstanding the valency involved. Precision in applying the correct formulas, as we've done in following Faraday's First Law, saves the day. Clear steps are the roadmap to finding solutions—know what you're solving for, use the right formulas, and pay attention to the units involved. With a well-designed strategy, you'll move from being puzzled by electrolysis to skillfully unraveling its mysteries.

In electrolysis, the quantity of electricity refers to the total electric charge that is transferred through the electrolyte, facilitating the chemical change at the electrodes. Recollect that charge is measured in coulombs, and one ampere of current passing for one second equals one coulomb. To frame it simply, it's like measuring the flow of water through a pipe—the quantity of electricity is the total flow of our 'electric water'.

In the given problem, the charge needed to deposit copper is found by using the electrochemical equivalent of copper, which in turn needs the copper's atomic weight and valency. By substituting these values in the appropriate formula, you quantify the electrical charge required to deposit a specific mass of copper. This concept isn't just useful for solving textbook problems, it is essential for any real-world application of electrolysis, such as electroplating or purification of metals.