The cell potential of this electrochemical cell depends on the pH of the solution in the anode half-cell. Pt(s) | H 2 (g, 1 atm) | H + (aq, ? M) || Cu 2+ (aq, 1.0 M) | Cu(s) What is the pH of the solution if E cell is 355 mV?

Welcome back everyone. To another video. The cell potential of this electrochemical cell depends on the ph of the solution. In the A node half cell, we're given our cell notation and we're asked to calculate the ph of the solution. If E cell is 355 millivolts, there are four answer choices. A is two B 0.3 C one ND 0.9. So let's begin solving this problem. Remembering that E cell according to the nurse equation is equal to E knot cell, which is the standard potential at basically at standard conditions, right? Or basically the the standard potential of our cell minus. We can use many forms here. We can use RT divided by NF but for simplicity, we can just use another form of the nearest equation and say that it reduces to 0.05916 volts there N which is the number of modes of electrons transferred multiplied by the logarithm of Q. This assumes the standard temperature or basically it would be more accurate to say a room temperature of 298.15 Kelvin. So we're going to use this equation. But to use it, we first of all want to identify E knot cell and to do that, we need to, first of all identify our salt bridge in the cell notation. On the left side, we have our, a note which is a process of oxidation. And on the right, we have our cathode, it represents the process of reduction. If we read it from left to right. First of all, we understand that we have hydrogen gas producing H plus or protons, we can immediately balance this by adding two and adding two electrons on the right hand side. According to the table, the standard reduction potential would be zero in this case. And similarly add the cathode, we have copper two plus producing copper solid. Let's add two electrons to balance it out. Let's use the table to identify the standard reduction potential according to the table. That's zero point 34 volts. And now what we can do here is just combine them together to understand the net equation taking place. We have copper to CS reacting with hydrogen gas and producing two H plus and copper solid. We're interested in e eno cell, which is just the sum of these two because we're adding them together. So that's essentially 0.34 volts. OK. Now, what else do we need? Well, we need to know the number of modes of electrons transferred the least common multiple between two and two is two. So we have two moles of electrons. And now finally, what about the reaction quotient? Well, let's look at the product side and divide concentrations by the concentrations on the reacting side. So first of all, we have H plus, we need to square it due to the coefficient we have a solid, we can neglect it. It doesn't belong to Q. On the left hand side, we will include the concentration of the copper to cations as well as the partial pressure of hydrogen gas because it's a gas, right. So we're not using a concentration. And now we can essentially substitute our expression into the original formula. We notice that a cell becomes equal to E not C minus 0.05916 volts divided by M bloggers of H plus word divided by the concentration of copper two CS multiplied by the partial pressure of hygiene. What we understand here is that according to the context of the problem, the concentration of copper two plus is one molar. We have our standard conditions, the partial pressure of hygiene is one atmosphere. So we actually can get rid of our denominator, right? And this helps us tremendously because the math becomes much easier. So what can we do here? Well, essentially we can, first of all remember the rules of logarithms. If we have a logarithm of a specific value squared, we can actually bring this to in front of the logarithm. So it becomes a coefficient, meaning I am going to move that to in front. I'm just going to put it in front. And what I notice is that negative log of H plus is actually my ph isn't it? So the equation becomes E cell equals E not C. Right now, I have two in my numerator and I have to my denominator because I have two modes of electrons. They can essentially say that I'm subtracting 0.05916 votes. And now remember negative log of H plus is my ph. So I'm going to turn this into positive. And I will write P instead taking into account that I can combine this negative sign with the logarithm. And for that purpose, if I want to put Ph, I need to flip the sign. So let's talk for bh essentially looking at this equation P would be equal to. Now E cell minus E not thou divided by the coefficient 0.05916. Let's plug in the values we have 355 mills which is essentially the 0.355 walls. E cell is 0.34 volts. And we are dividing this by 0.05916. Now, if we evaluate the result, we end up with a ph value of 0.3 which corresponds to the answer choice B in the multiple choice. Thank you for watching.

Ch.20 - Electrochemistry

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Tro 6th Edition

Ch.20 - Electrochemistry

Problem 115

Chapter 20, Problem 115

The cell potential of this electrochemical cell depends on the pH of the solution in the anode half-cell. Pt(s) | H₂(g, 1 atm) | H⁺(aq, ? M) || Cu²⁺(aq, 1.0 M) | Cu(s) What is the pH of the solution if E_cell is 355 mV?

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Identify the half-reactions for the electrochemical cell. The anode half-reaction involves hydrogen: \( \text{H}_2(g) \rightarrow 2\text{H}^+(aq) + 2e^- \). The cathode half-reaction involves copper: \( \text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s) \).

Write the Nernst equation for the cell potential: \( E_{cell} = E^\circ_{cell} - \frac{RT}{nF} \ln Q \), where \( Q \) is the reaction quotient.

Calculate the standard cell potential \( E^\circ_{cell} \) using standard reduction potentials: \( E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \). Use standard reduction potentials for \( \text{Cu}^{2+}/\text{Cu} \) and \( \text{H}^+/\text{H}_2 \).

Express the reaction quotient \( Q \) in terms of concentrations: \( Q = \frac{[\text{H}^+]^2}{[\text{Cu}^{2+}]} \). Given \([\text{Cu}^{2+}] = 1.0 \text{ M}\), simplify \( Q \) to \( [\text{H}^+]^2 \).

Rearrange the Nernst equation to solve for \([\text{H}^+]\), and then calculate the pH using \( \text{pH} = -\log[\text{H}^+] \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Nernst Equation

The Nernst Equation relates the cell potential (E) of an electrochemical cell to the concentrations of the reactants and products. It is expressed as E = E° - (RT/nF)ln(Q), where E° is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred, F is Faraday's constant, and Q is the reaction quotient. This equation allows us to calculate the effect of concentration changes on the cell potential.

pH and Hydrogen Ion Concentration

pH is a measure of the acidity or basicity of a solution, defined as the negative logarithm of the hydrogen ion concentration: pH = -log[H+]. In electrochemical cells, the pH can influence the concentration of H+ ions, which in turn affects the cell potential. A lower pH indicates a higher concentration of H+ ions, which is crucial for determining the conditions in the anode half-cell.

Electrochemical Cell Components

An electrochemical cell consists of two half-cells: the anode, where oxidation occurs, and the cathode, where reduction takes place. In the given cell, the anode involves hydrogen gas and protons, while the cathode involves copper ions and solid copper. Understanding the roles of these components and their reactions is essential for analyzing how changes in conditions, such as pH, affect the overall cell potential.