Use tabulated electrode potentials to calculate ∆G° rxn for each reaction at 25 °C. b. Br 2 (l) + 2 Cl – (aq) → 2 Br – (aq) + Cl 2 (g) c. MnO 2 (s) + 4 H + (aq) + Cu(s) → Mn 2+ (aq) + 2 H 2 O(l) + Cu 2+ (aq)

hey everyone in this example, we need to determine our change in gibbs free energy for our reaction at 25 degrees Celsius. Using the tabulated electrode potentials. So we're given our electrode cell below and we want to recognize that we need to write out two half reactions are first half reaction is going to show the reduction of our lengthen iem caddy on which forms solid length any um by gaining three electrons. Our second half reaction is to show the oxidation of nitrogen monoxide gas, which forms the following ions where we have the nitrate, an ion n 03 minus which gains three electrons. And we also should recognize that in our cell we have water, which is why we will use water to balance out our oxygen's in this second half reaction. So we should recognize that we have only one atom of oxygen on the reactant side here and three atoms of oxygen here and so because we recall that we use water to balance out our oxygen's, we're going to add to our reactant side, two moles of water. So we would say plus two moles of H +20. And we'll go ahead and just write this in a different color. And by adding water into our reaction, we have also added hydrogen protons. And so we need to balance out our hydrogen protons by adding on the product side for hydrogen protons. So now all of our atoms in the second half reaction as well as our first half reaction are balanced here because we only had one atom of Lentini um for both sides of our equation. And our next step is to make sure that our electrons are balanced as well as our charges. And so we would say that in both half reactions, we do have three electrons. However, looking at the net charges, we have a net charge of zero here because we have a plus three net charge that got canceled by the addition of three electrons. So we have a net charge of zero on both sides of our equation. In the first half reaction and in the second half reaction we have a net charge of plus four minus one which gives us plus three. But because we added three electrons, we actually have a net charge of zero which matches our net charge of zero on both sides here. And so we don't need to manipulate our charges or electrons. So we can go ahead and actually just add up these 22 half reactions together to form one half reaction. And we should recognize that because we have electrons on the reaction side in the first half reaction. And on the product side in the second half reaction we can cancel the electrons out so we can cancel out both three electrons. And now we will just add up everything else. So we would have our lengthen ian caddy on Plus nitrogen monoxide gas plus two moles of water produces our product where we have solid length any um plus the nitrate an ion plus four moles of hydrogen proton. And so this would be our full reaction here. Where now we want to make note of our first half reaction, which we should recognize occurred as a reduction because we had electrons added on the reactant side. And recall that our reduction occurs at the cathode Half cell. Now for our standard electoral potential table in our textbooks or online, we would see that this reduction of lengthening in Catalan has a cell potential equal to .96V. Looking at our second half reaction, we should recognize that we have electrons added on the product side of this reaction, which means that this reaction occurs as an oxidation. And recall that our oxidation reaction occurs at the half cell. Now, when we look up in our textbooks or online for our standard reduction potential table, we would see that for the oxidation of nitrogen monoxide gas, we have a cell potential value equal to negative two 0.38 volts. And just to be clear, I just realized that these two values are mixed up. So we actually have for the cell potential of the reduction of Anthony in carry on the value of negative 2.38 volts. And for the oxidation of nitrogen monoxide gas, we actually have a cell potential of 0.96 volts. So these are the orders for the reduction and oxidation cell potential values here. And so now that we have these values noted, were able to calculate our standard cell potential degree cell, which we should recall is found by taking the cell potential of our cathode subtracted from the cell potential of our anodes. And so this would allow us to say that our cell potential of our cathode which above is negative 2.38V can be subtracted from our cell potential of our anodes, which above we stated as 0.96V for oxidation half reaction. And this difference gives us our standard cell potential equal to a value of negative 3.34 volts. Now because we need to solve for the change and gives free energy for reaction. We should recall that. That utilizes units of jewels. And we would ultimately want our final answer to be in units of kilo jewels. And so we want to recall that for one volt that is equivalent to one Juul per column. And so we want to interpret our standard cell potential here as a value of -3. jewels per column. And so our next step is to recall our formula to find gibbs free energy. And so we would say that the change and gives free energy of our reaction is equal to negative one times are electrons transferred. N multiplied by Faraday's constant, which is then multiplied by the standard cell potential E degree cell. And so we want to make note of our electrons transferred. So looking at our half reactions, we canceled out three electrons from both half reactions. And that means that our electrons transferred is equal to a value of three, sorry, not four. And so we're going to plug that down below. For our calculation for the change in Gibbs free energy of our reaction. So we would say that delta g degree of our reaction is equal to negative one times the three moles of electrons that are transferred. Which we will then multiply by Faraday's constant. Which we should recall is equal to a value of 96,485 columns per moles of electrons. This allows us to cancel out moles of electrons with moles of electrons here. And now we want to move from units of columns into jewels. So we're going to plug in our standard cell potential above which we interpreted in jewels per columns here because we use this conversion factor. So we're gonna multiply By that factor next. So we have for one column a value of negative 3.34 jewels. And this represents our standard cell potential. And now we're able to cancel out units of columns and we're left with units of jewels for our change in gibbs free energy. So we did say above that. We want our change in gibbs free energy expressed in units of kilo jewels. So we're going to convert now from jewels to kill a jules. And so we should recall that our prefix kilo tells us that we have 10 to the third power jewels for one kg jule. And now we're able to cancel out jewels, which actually leaves us with kilo jewels as our final unit for the change in gibbs free energy. And this will give us a value equal to 967 kg jewels as our change in Gibbs. Free energy for our reaction. So this would be our final answer here. I hope that everything I viewed was clear, but if you have any questions, just 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 65b,c

Chapter 19, Problem 65b,c

Use tabulated electrode potentials to calculate ∆G°_rxn for each reaction at 25 °C. b. Br₂(l) + 2 Cl^–(aq) → 2 Br^–(aq) + Cl₂(g) c. MnO₂(s) + 4 H⁺(aq) + Cu(s) → Mn²⁺(aq) + 2 H₂O(l) + Cu²⁺(aq)

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Identify the half-reactions involved in the overall reaction. For the reduction half-reaction: MnO2(s) + 4 H+(aq) + 2 e- → Mn2+(aq) + 2 H2O(l). For the oxidation half-reaction: Cu(s) → Cu2+(aq) + 2 e-.

Look up the standard electrode potentials (E°) for each half-reaction from the standard electrode potential table. You will find E° for MnO2/Mn2+ and Cu/Cu2+.

Calculate the standard cell potential (E°cell) for the overall reaction using the formula: E°cell = E°cathode - E°anode, where MnO2/Mn2+ is the cathode (reduction) and Cu/Cu2+ is the anode (oxidation).

Use the relationship between the Gibbs free energy change (∆Gr°) and the cell potential to calculate ∆Gr°xn. The formula to use is: ∆Gr° = -nFE°cell, where n is the number of moles of electrons transferred in the balanced equation (which is 2 in this case), and F is the Faraday constant (approximately 96485 C/mol).

Ensure all units are consistent, particularly that temperature is in Kelvin if needed, and calculate ∆Gr°xn for the reaction at 25 °C (298 K).

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Key Concepts

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

Electrode Potentials

Electrode potentials, measured in volts, indicate the tendency of a chemical species to be reduced. Standard electrode potentials (E°) are measured under standard conditions (1 M concentration, 1 atm pressure, and 25 °C). These values are crucial for predicting the direction of redox reactions and calculating the Gibbs free energy change (∆Gr°) for reactions.

Gibbs Free Energy (∆Gr°)

Gibbs free energy change (∆Gr°) is a thermodynamic quantity that indicates the spontaneity of a reaction at standard conditions. It can be calculated using the equation ∆Gr° = -nFE°, where n is the number of moles of electrons transferred, F is Faraday's constant, and E° is the cell potential. A negative ∆Gr° indicates a spontaneous reaction.

Redox Reactions

Redox reactions involve the transfer of electrons between species, where one species is oxidized (loses electrons) and another is reduced (gains electrons). In the given reaction, manganese dioxide (MnO2) is reduced to manganese ions (Mn2+), while copper (Cu) is oxidized to copper ions (Cu2+). Understanding the oxidation states and half-reactions is essential for calculating overall cell potentials.