Calculate ∆G° rxn and K for each reaction. a. The disproportionation of Mn 2+ (aq) to Mn(s) and MnO 2 (s) in acid solution at 25 °C. b. The disproportionation of MnO 2 (s) to Mn 2+ (aq) and MnO 4 – (aq) in acid solution at 25 °C.

Welcome back everyone at 25 °C. What are the equilibrium constant? And the gibbs free energy change of the reaction for the disproportionate reaction of liquid hydrogen peroxide to liquid water and oxygen gas. As the problem suggests, we are looking at a reaction of hydrogen peroxide h2o 2 that decomposes into water. Add oxygen ot we need to balance the equation. And essentially, we can notice that we have the same number of hydrogens, but we don't have the same number of auctions on the right side, we have two auctions and three total. So what we're going to do is just add two on the right hand side to add an even number of oxygens, which essentially allows us to get four oxygens on the left as well. So we have four oxygens on the left, four on the right and our hygen are also now balanced. So this is our overall equation. But now if we think about the individual half reactions, we essentially need to use two of them. One of them is going to be represented by age 202, combining with modes of hydro. We can obtain this half equation from the cell potentials or the standard reduction potentials stable. So now we're just adding an aqueous state as well as two electrons to produce two moles off of water. If we look at the table, we understand that the standard reduction potential would be equal to 1.78 volts. Now, for the other one, we want to also consider the reduction of oxygen. So if we identify our equation that we 02, yes, combining with 4h plus and for electrons to produce two moles of water, now the standard reduction potential is equal to 1.23 volts. What we want to do is just reverse one of these half equations. And we're going to, we're going to reverse the second one because it has oxygen on the left side and not the right side. So what we're going to do is just rewrite how first half equation just as it is right? We don't want to reverse it because it has peroxide on the left hand side. And now for the second one, we actually want to reverse it by reversing it. We are also going to add a negative sign or the potential, right? So let's add a negative sign sign have water on the left hand side producing 02 plus 4h plus plus four electrons. And now to cancel out the electrons, we need to multiply the first half equation by two. So we get two or, or electrons and or most of water. If we essentially combine them, we will notice that we get two modes of peroxide reducing oxygen, right? Our H plus would cancel each other out. And, and now, instead of having four moles of water, on the right hand side, we would only have two because they on the left. So we indeed get our target equation and it would have a cell potential which would be equal to the sum of the individual value. So 1.78 volts plus negative, 1.23 volts. What do we get? Well, that would be zero point 55 volts. So we have our value for the cell potential. And now what we're going to do from here is evaluate the equilibrium constant. So we want to get our equilibrium constant keq and we also want to get a word delta G value. So let's recall that we can calculate the change in the gibbs free energy as negative NFE how many moles of electrons do we have? Well, we have four total, right. So we're going to take negative for most of the electrons multiply by the far is constant and then multiply by the value that we got for eno we got 0.55 votes. That's our calculation for delta G. Let's evaluate the result specifically, we get negative 212.3 kilojoules. We can only use two significant figures. So it would be more accurate to say negative 2.1 multiplied by it, sense the power of second kilojoules. So we essentially have our first answer. That'd be negative 2.1 multiplied by sense, the power of second killer jus. And now we want to determine the equilibrium constant. How do we do that? Well, essentially we need to recall the relationship between the equilibrium constant and delta G. Let's recall that delta G not is equal to negative RT Ln of the equilibrium constant. So if we rearrange the equation, then the equilibrium constant is simply of negative delta G divided by RT, that's everything that we need to get. And we can substitute the values. So the equilibrium constant would be e to the power of negative negative, right? So that would be positive. We can essentially say it's the power of positive 2.1 multiplied by 10. The power of fifth joules. Why is it fifth? Well, essentially we need to multiply delta G by 1000 to get Juuls, right? And multiplying by 1000 would increase the exponent by three units. Now, what is the value of R eight point 314 Jules with Calvin promo, what is the temperature? 98 Calvin? Because as the problem suggests that's 25 °C. And now we want to get our answer. Now, before we get our answer, we can also introduce a small fix. Instead of using the rounded value, right, we are going to actually take our delta G value of negative 212.3 kg joules. So it would be more accurate to use two point 123. That's what we actually have to avoid the propagation of error. So that'd be 2.123 multiplied by 10, the power of fifth tools, right? And now if we evaluate the result from here, we simply end up with a value of 1.6 multiplied by sense, the power of negative 10 to the power of positive 37th. And that will be our final answer. So let's label it and conclude the video. Thank you for watching.

Ch.20 - Electrochemistry

Tro - Chemistry: A Molecular Approach 5th Edition

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Ch.20 - Electrochemistry

Problem 121

Chapter 20, Problem 121

Calculate ∆G°_rxn and K for each reaction.
a. The disproportionation of Mn²⁺(aq) to Mn(s) and MnO₂(s) in acid solution at 25 °C.
b. The disproportionation of MnO₂(s) to Mn²⁺(aq) and MnO₄^–(aq) in acid solution at 25 °C.

Verified step by step guidance

Step 1: Identify the half-reactions involved in the disproportionation of Mn²⁺(aq) to Mn(s) and MnO₂(s). The two half-reactions are: Mn²⁺(aq) → Mn(s) and Mn²⁺(aq) → MnO₂(s).

Step 2: Write the balanced half-reactions and determine the standard reduction potentials (E°) for each half-reaction from a standard reduction potential table.

Step 3: Calculate the standard cell potential (E°_cell) for the overall reaction using the formula: E°_cell = E°_cathode - E°_anode.

Step 4: Use the Nernst equation to calculate the standard Gibbs free energy change (∆G°_rxn) for the reaction: ∆G°_rxn = -nFE°_cell, where n is the number of moles of electrons transferred and F is the Faraday constant.

Step 5: Calculate the equilibrium constant (K) for the reaction using the relationship: ∆G°_rxn = -RTlnK, where R is the universal gas constant and T is the temperature in Kelvin.

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Key Concepts

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

Gibbs Free Energy (∆Gr°)

Gibbs Free Energy (∆Gr°) is a thermodynamic potential that measures the maximum reversible work obtainable from a thermodynamic process at constant temperature and pressure. It indicates the spontaneity of a reaction; a negative ∆Gr° suggests that the reaction can occur spontaneously, while a positive value indicates non-spontaneity. The standard Gibbs free energy change can be calculated using standard enthalpy and entropy values.

Equilibrium Constant (K)

The equilibrium constant (K) is a dimensionless number that expresses the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. It provides insight into the extent of a reaction; a large K value indicates that products are favored, while a small K suggests that reactants are favored. The relationship between K and Gibbs free energy change is given by the equation ∆Gr° = -RT ln(K).

Disproportionation Reaction

A disproportionation reaction is a specific type of redox reaction in which a single substance is simultaneously oxidized and reduced, resulting in two different products. In the case of Mn2+(aq), it is transformed into elemental manganese (Mn) and manganese dioxide (MnO2) under acidic conditions. Understanding the half-reactions involved is crucial for calculating the overall Gibbs free energy change and equilibrium constant for the reaction.