Calculating PH After Electrolysis Of AgNO3 Solution

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Hey guys! Let's dive into a chemistry problem that involves electrolysis. We're going to figure out the pH of a silver nitrate (AgNO3) solution after we run electricity through it. Specifically, we'll analyze what happens when 9,650 Coulombs of current are passed through 1 liter of a 1 M AgNO3 solution using platinum (Pt) electrodes. This is a classic example of how electrolysis can change the concentration of ions in a solution and, consequently, its pH. So, buckle up; we're about to explore the fascinating world of electrochemistry! This kind of problem often appears in chemistry exams, so understanding the steps is crucial. We'll break it down into manageable parts to make sure we get it right.

Understanding the Basics of Electrolysis

First off, let's refresh our memories on what electrolysis actually is. Electrolysis is the process of using electricity to drive a non-spontaneous chemical reaction. In our case, we're using electricity to decompose the AgNO3 solution. When we apply a voltage to the solution with the Pt electrodes, ions move towards the electrodes. The key reactions happen at the electrodes: oxidation at the anode (where electrons are lost) and reduction at the cathode (where electrons are gained). In our AgNO3 solution, the ions present are Ag+ (silver ions), NO3- (nitrate ions), and, of course, H+ and OH- from the water itself since it's an aqueous solution. At the cathode, Ag+ ions will gain electrons and form solid silver (Ag). At the anode, water (H2O) will be oxidized, producing oxygen gas (O2), protons (H+), and electrons. These reactions are what change the pH of the solution. To determine the pH, we need to figure out how many moles of H+ ions are produced during the electrolysis. This involves using Faraday's laws of electrolysis, which link the amount of electricity passed through the solution to the amount of substance produced or consumed at the electrodes. Therefore, it's not just about memorizing formulas; it's about understanding the reactions and how they affect the concentration of ions in the solution. We will use Faraday's laws, which connect the amount of electricity to the number of moles of electrons transferred. Then we will relate this to the number of moles of H+ ions produced, which will help us calculate the concentration and, finally, the pH of the solution.

The Electrolysis Process

The electrolysis of the AgNO3 solution using Pt electrodes involves a series of redox reactions. Here's a breakdown of what happens at each electrode:

  • Cathode (Reduction): Silver ions (Ag+) are reduced to solid silver (Ag). The reaction is: Ag+ + e- -> Ag.
  • Anode (Oxidation): Water (H2O) is oxidized to produce oxygen gas (O2), hydrogen ions (H+), and electrons. The reaction is: 2H2O -> O2 + 4H+ + 4e-.

Faraday's Laws

Faraday's laws of electrolysis are fundamental to understanding the quantitative aspects of electrolysis. We'll use the following formula:

Q = I * t, and Q = n * F

Where:

  • Q is the total charge in Coulombs.
  • I is the current in Amperes.
  • t is the time in seconds.
  • n is the number of moles of electrons.
  • F is Faraday's constant (96,500 Coulombs/mol).

In our problem, we are given Q = 9650 Coulombs. We'll use Faraday's laws to find the number of moles of electrons that have passed through the solution. Once we know this, we can determine the moles of H+ ions produced. From there, we'll calculate the concentration of H+ and, finally, the pH of the solution.

Step-by-Step Calculation of the pH

Now, let's get into the nitty-gritty and calculate the pH of the solution step by step. This is where we put our knowledge into action. We will use the information given and some basic chemistry principles to solve for pH. Remember, the goal is to find the concentration of H+ ions in the solution and then calculate the pH using the formula: pH = -log[H+].

Step 1: Calculate the Moles of Electrons Transferred

We'll use Faraday's law to find the number of moles of electrons (n) that have passed through the solution.

Q = n * F

9650 C = n * 96500 C/mol

n = 9650 C / 96500 C/mol

n = 0.1 mol of electrons

Step 2: Determine the Moles of H+ Produced

From the anode reaction (2H2O -> O2 + 4H+ + 4e-), we know that 4 moles of electrons are produced for every 4 moles of H+ ions. Thus, the mole ratio is 4:4 or 1:1. Since 0.1 moles of electrons are involved, we can calculate the moles of H+ formed.

Moles of H+ = (Moles of electrons) * (Mole ratio of H+ to electrons) Moles of H+ = 0.1 mol * (4/4) Moles of H+ = 0.1 mol

Step 3: Calculate the Concentration of H+

We have 0.1 mol of H+ ions in 1 liter of solution, so the concentration is:

[H+] = moles of H+ / volume of solution

[H+] = 0.1 mol / 1 L

[H+] = 0.1 M

Step 4: Calculate the pH

Finally, we can calculate the pH using the formula:

pH = -log[H+]

pH = -log(0.1)

pH = 1

So, the pH of the solution after electrolysis is 1.

Conclusion: The Answer Revealed!

Alright, folks, we've successfully navigated the chemistry waters and found our answer. Through a careful application of Faraday's laws and an understanding of the redox reactions at the electrodes, we've determined that the pH of the AgNO3 solution after electrolysis is 1. The correct answer from the multiple-choice options is, therefore, D. 1. This entire process demonstrates how electrochemical principles can be used to predict and explain changes in the pH of solutions during electrolysis.

To recap:

  • We calculated the moles of electrons using Faraday's law.
  • We used the stoichiometry of the anode reaction to find the moles of H+ ions produced.
  • We calculated the concentration of H+ ions.
  • We used the concentration of H+ ions to calculate the pH.

This kind of problem helps us understand the relationship between electricity, chemical reactions, and the resulting changes in solution properties. Keep practicing, and you'll become a pro at these problems! Always remember the importance of balancing the chemical equations and applying the correct stoichiometric ratios. Keep up the great work! If you have any questions, feel free to ask. And remember, chemistry is all about understanding the how and the why. Great job, everyone! Keep up the awesome work!