An MnO 2 (s)/Mn 2+ (aq) electrode in which the pH si 10.24 is prepared. Find the [Mn 2+ ] necessary to lower the potential of the half-cell to 0.00 V (at 25°C)

hey everyone in this example, we have that. The ph of a prepared lead oxide lead two plus electrode is 12.38 at degrees Celsius. We need to figure out the concentration of lead two plus that is needed to decrease the potential of the half cell to zero volts. So what we should recognize is that we have a redox reaction where lead oxide produces lead two plus as a product. So when we have a redox reaction, we need to determine whether the atoms are balanced. Right now. We have one atom of lead on both sides of our equation. However, we have two atoms of oxygen on the reactant side and we should recall that to balance oxygen, we use um water. So because we have two moles of oxygen on the reacting side, we're going to go ahead and add two moles of water to the product side here. And so right now, in adding two moles of water, we've also added formals of hydrogen to the product side. And so we should next recall that we balance hydrogen with hydride or proton. And so this means that we would add to our react inside here, four hydride atoms. Our next step is to recognize the charges and whether they're balanced on both sides of the equation. So right now we have a net charge of plus four on the reacting side and we have a net charge of plus two on the product side. So we want to cancel out that or rather reduce that net charge of plus four on the reactive side by adding two electrons to the reacting side. Now because we recognize that we added electrons to the reacting side. We can say that this is therefore a reduction. And so we want to look up on our standard electrode potential table. The electrode potential for the reduction of lead oxide. And we would see that on this table. We have a value of 1.46V for the reduction of lead oxide. We also want to make note of the fact that on this part of our cell, which is our reaction side, we we've added these four moles of hydride. So we need to figure out our concentration of hydride. And we should recall that we can calculate the concentration of hydride by taking 10 to the negative value of our ph And so this means that we would take 10 to the negative 12.38 being our ph given in the prompt. This is going to give us our concentration concentration of hydride equal to 4.1687 times 10 to the negative 13th power. Next we want to recall the following formula where we have the electrode potential of our half cell is equal to the standard electrode potential of our half cell subtracted from Our constant value 0.0592 divided by N. Where we recall that N is going to be our electrons transferred in our reaction for the reduction half. And this is then going to be multiplied by the log of Q. Where we recall that Q is going to be our concentration of hydride which is raised to the coefficient four as a power and divided by the concentration of lead two plus. So that would look like Q equals concentration of lead two plus, divided by our concentration of hydride. Where we see that we have a coefficient of four here. So we're going to raise this to the fourth power. So right now we're going to go ahead and fill in what we know for our formula here. So we would have that according to the prompt, the energy of our half cell should be 0.0V. This is set equal to 1.46V which is subtracted from Our constant value here. 0.0592 divided by N. Where we see above that we have a transfer of two electrons on the reduction side of our reaction. So we have two for n. multiplied by the log of our concentration of lead two plus. Which is what we need to solve for Divided by our concentration of hydride which above we stated is 4.1687 times 10 to the negative 13th power. And this is raised to the fourth power here. So we're going to further simplify this so that now we just take this log term and we want to cancel it out by using a power of And we're going to place as an exponents, our negative value of our electrode potential 1.46 divided by negative 0.0592 which is divided by two. So we're going to place this in parentheses and complete this in brackets. This is then going to be set equal to our simplification where we now have the concentration of lead two plus divided by our concentration of hydride being 4.1687 times 10 to the negative 13th power raised to the fourth power. And to completely enter this question we want to isolate for the concentration of lead two plus. So this is going to be set equal to on both sides. We're gonna multiply by the denominator so that it cancels out on the right hand side. So this is set equal to our right hand side where we now have 4.1687 times 10 to the negative 13th power raised to the fourth power, multiplied by 10 to the brackets where we have negative 1.46 divided by negative 0.0592 divided by two. So simplifying the entire right side in our calculators, we're going to get that the concentration of lead two plus is equal to 0.6373. And this will be our final answer here to complete this example as our concentration of lead two plus needed to decrease our half cell potential to zero volts. So what's highlighted in yellow is our final answer. If you have any questions, leave them down below, and I will see everyone in the next practice video.

Ch.19 - Electrochemistry

Tro - Chemistry: A Molecular Approach 4th Edition

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Ch.19 - Electrochemistry

Problem 129

Chapter 19, Problem 129

An MnO₂(s)/Mn²⁺(aq) electrode in which the pH si 10.24 is prepared. Find the [Mn²⁺] necessary to lower the potential of the half-cell to 0.00 V (at 25°C)

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Identify the half-reaction for the MnO2(s)/Mn2+(aq) electrode: \( \text{MnO}_2(s) + 4\text{H}^+(aq) + 2\text{e}^- \rightarrow \text{Mn}^{2+}(aq) + 2\text{H}_2\text{O}(l) \).

Use the Nernst equation to relate the cell potential to the concentrations and pH: \( E = E^0 - \frac{RT}{nF} \ln Q \), where \( Q \) is the reaction quotient.

Determine the standard reduction potential \( E^0 \) for the half-reaction from a standard reduction potential table.

Calculate the reaction quotient \( Q \) using \( Q = \frac{[\text{Mn}^{2+}]}{[\text{H}^+]^4} \). Given \( \text{pH} = 10.24 \), find \( [\text{H}^+] = 10^{-10.24} \).

Set \( E = 0.00 \) V and solve the Nernst equation for \( [\text{Mn}^{2+}] \) to find the concentration necessary to achieve the desired potential.

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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 reduction potential 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 cell potential, E° is the standard 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 is essential for calculating the potential of the half-cell under non-standard conditions.

pH and its Effect on Electrochemical Reactions

pH is a measure of the hydrogen ion concentration in a solution, which can significantly influence the behavior of electrochemical reactions. In this case, a pH of 10.24 indicates a basic environment, affecting the solubility and speciation of manganese ions. Understanding how pH alters the equilibrium of the MnO2/Mn2+ system is crucial for determining the necessary concentration of Mn2+ to achieve the desired potential.

Equilibrium and Reaction Quotient (Q)

The reaction quotient (Q) is a measure of the relative concentrations of products and reactants at any point in a reaction. For the MnO2/Mn2+ half-cell, Q can be expressed as [Mn2+]/[MnO2]. At equilibrium, Q equals the equilibrium constant (K), and changes in concentration will shift the equilibrium position according to Le Chatelier's principle. This concept is vital for understanding how to manipulate [Mn2+] to achieve a specific cell potential.