Electrolysis Of NaI: Mass Deposited At Cathode Calculation

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Hey guys! Today, we're diving into a classic electrochemistry problem: figuring out how much stuff gets deposited on the cathode during the electrolysis of a sodium iodide (NaI) solution. This is a super important concept in chemistry, and it's something you'll definitely encounter in your studies. We'll break it down step by step, so don't worry if it seems a bit intimidating at first. Let's get started!

Understanding Electrolysis and Cathode Reactions

First things first, let's make sure we're all on the same page about electrolysis. Electrolysis is basically using electricity to drive a non-spontaneous chemical reaction. Think of it like forcing a reaction to happen that wouldn't normally occur on its own. In our case, we're passing an electric current through a solution of NaI, and this current is going to cause some interesting chemical changes.

Now, the cathode is the negatively charged electrode in an electrochemical cell. Remember, opposites attract! So, positive ions (cations) in the solution are drawn towards the cathode. At the cathode, these ions gain electrons in a process called reduction. This is where the magic happens, and where we'll see the deposition of a substance.

In a NaI solution, we have Na+ cations and I- anions. Water (H2O) is also present, and it can participate in the electrolysis process too. So, the million-dollar question is: what gets reduced at the cathode? To figure that out, we need to consider the reduction potentials of the possible species. Reduction potential is a measure of how likely a species is to be reduced – the higher the reduction potential, the more likely it is to be reduced.

Here's the breakdown of possible cathode reactions and their standard reduction potentials:

  1. Reduction of Na+: Na+(aq) + e- → Na(s) E° = -2.71 V
  2. Reduction of H2O: 2H2O(l) + 2e- → H2(g) + 2OH-(aq) E° = -0.83 V

Notice that the reduction potential for water is much higher (less negative) than that for sodium ions. This means that water is much more likely to be reduced at the cathode than Na+. So, the primary reaction occurring at the cathode will be the reduction of water, producing hydrogen gas (H2) and hydroxide ions (OH-).

Key takeaway: The cathode reaction in the electrolysis of aqueous NaI is the reduction of water, not the reduction of sodium ions.

Calculating the Mass of Substance Released at the Cathode

Okay, now that we know what's happening at the cathode (water is being reduced), we can figure out how much substance is being released. The question asks for the mass of substance released when a current of 5 Amperes is passed through the solution. Since the primary cathode reaction is the reduction of water, the substance being released is hydrogen gas (H2).

To calculate the mass of H2 produced, we'll need to use Faraday's laws of electrolysis. These laws relate the amount of substance produced at an electrode to the amount of electric charge passed through the electrolytic cell. Here's the key equation we'll use:

m = (M * I * t) / (n * F)

Where:

  • m = mass of substance produced (in grams)
  • M = molar mass of the substance (in g/mol)
  • I = current (in Amperes)
  • t = time (in seconds)
  • n = number of moles of electrons transferred per mole of substance
  • F = Faraday's constant (96485 C/mol)

Let's break down each of these variables for our specific problem:

  • M (Molar mass of H2): The molar mass of H2 is approximately 2 g/mol.
  • I (Current): The problem states that the current is 5 Amperes.
  • t (Time): This is where things get a little tricky. The problem doesn't explicitly give us a time. However, it does tell us that the solution contains 1 mole of NaI. We'll need to use this information to figure out how long the current is applied. We will revisit this shortly.
  • n (Moles of electrons transferred): From the balanced half-reaction for the reduction of water (2H2O(l) + 2e- → H2(g) + 2OH-(aq)), we can see that 2 moles of electrons are transferred for every 1 mole of H2 produced. So, n = 2.
  • F (Faraday's constant): Faraday's constant is a fundamental constant in electrochemistry, equal to 96485 C/mol.

Now, let's figure out the time (t). We know that 1 mole of NaI is present in the solution. During electrolysis, the iodide ions (I-) will be oxidized at the anode. The half-reaction for this is:

2I-(aq) → I2(s) + 2e-

This tells us that 2 moles of electrons are released for every 2 moles of iodide ions oxidized, or 1 mole of electrons per mole of iodide ions. Since we have 1 mole of NaI, we effectively have 1 mole of I- ions. This oxidation reaction at the anode will continue until all iodide ions are consumed. The cathode reaction (reduction of water) occurs simultaneously. So, the total charge passed through the solution will correspond to the oxidation of 1 mole of iodide ions.

The total charge (Q) can be calculated using the following equation:

Q = n * F

Where:

  • Q = total charge (in Coulombs)
  • n = number of moles of electrons (in this case, 1 mole)
  • F = Faraday's constant (96485 C/mol)

So, Q = 1 mol * 96485 C/mol = 96485 Coulombs

We also know that current (I) is the rate of flow of charge, so:

I = Q / t

Therefore, we can rearrange this to solve for time (t):

t = Q / I

t = 96485 C / 5 A = 19297 seconds

Now we have all the pieces of the puzzle! We can plug everything into Faraday's law equation:

m = (M * I * t) / (n * F)

m = (2 g/mol * 5 A * 19297 s) / (2 * 96485 C/mol)

m ≈ 1 gram

So, the mass of hydrogen gas produced at the cathode is approximately 1 gram.

Therefore, the correct answer is C) 1.

Key Considerations and Common Mistakes

Electrolysis problems can be tricky, so let's go over some key considerations and common mistakes to help you ace these questions:

  • Identifying the Cathode and Anode Reactions: Always start by identifying the possible reactions that can occur at the cathode (reduction) and anode (oxidation). Consider the reduction potentials to determine which reaction is most likely to occur.
  • Water Electrolysis: Remember that water can be electrolyzed in aqueous solutions. In many cases, water will be reduced at the cathode instead of the metal cation, especially for alkali metals and alkaline earth metals.
  • Faraday's Laws: Make sure you understand and can apply Faraday's laws of electrolysis. Pay close attention to the units and make sure they are consistent.
  • Stoichiometry: Don't forget the stoichiometry of the half-reactions. The number of moles of electrons transferred per mole of substance is crucial for the calculation.
  • Units: Always double-check your units! Convert time to seconds, and make sure you're using the correct units for Faraday's constant.

Practice Problems

To really solidify your understanding, let's try a couple of practice problems:

  1. What mass of copper will be deposited at the cathode when a current of 2.5 A is passed through a solution of copper(II) sulfate (CuSO4) for 30 minutes?
  2. A solution of silver nitrate (AgNO3) is electrolyzed using a current of 1.0 A. If 1.0 gram of silver is deposited at the cathode, how long did the electrolysis last?

Working through these problems will help you get comfortable with the calculations and the concepts involved.

Final Thoughts

Electrolysis is a fascinating and important topic in chemistry. By understanding the principles behind it and practicing problem-solving, you'll be well-equipped to tackle any electrolysis question that comes your way. Remember to break the problem down into smaller steps, identify the key reactions, and apply Faraday's laws carefully. Keep practicing, and you'll become an electrolysis expert in no time! Good luck, and happy calculating! This guide should provide a comprehensive breakdown for students and anyone interested in understanding electrolysis calculations, especially regarding mass deposition at the cathode. Remember, practice makes perfect, so keep working on those electrochemistry problems! If you have any questions, feel free to ask!