Given the following reduction half-reactions: Fe 3+ (aq) + e - → Fe 2+ (aq) E° red = +0.77 V S 2 O 6 2- (aq) + 4 H + (aq) + 2 e - → 2 H 2 SO 3 (aq) E° red = +0.60 V N 2 O(g) + 2 H + (aq) + 2 e - → N 2 (g) + H 2 O(l) E° red = -1.77 V VO 2+ (aq) + 2 H + (aq) + e - → VO 2+ + H 2 O(l) E° red = +1.00 V (a) Write balanced chemical equations for the oxidation of Fe 2+ (aq) by S 2 O 6 2- (aq), by N 2 O(aq), and by VO 2+ (aq).

hi everyone. So I had to identify the balance redox reaction For the oxidation of under acidic conditions. A period day nitrate and permanent night. So for the first reaction we're gonna have s. n two plus Plus. i owe 4 -4 The shields. SN four plus I minus. I need to first separate the whole reaction into two half reactions. SN two plus Initials SN four plus. Yeah, I04 building. I'm minus now. Need to balance the non hydrogen and non oxygen elements first and this is already balanced. So now we can balance the oxygen by adding H 20 liquid, decided it needs oxygen. So in the second round we have four options on the right hand side, but none on the product side. You can add four H 20 to the outside. Now we can balance our hydrogen by adding H plus two hydrogen. We have no house on the right hand side but we have eight on the product side and H plus the react inside. Now we can balance the charges by adding electrons to the more positive side. If you look at our charges, you have a plus two here. Last four here Plus eight Here -1 -10. We need to add two electrons in the reacting side And then here at eight electrons. So they react inside. Now we need to balance our electrons on the half reaction. They have two electrons in the first reaction but eight in the second reaction you need to pop off for the first reaction. Get four SN 2 plus. And this shows four SN four plus plus eight electrons. You're gonna have four SN two plus Plus. i owe 4 -4 Plus. eight h plus plus eight electrons. This shows four s. World class plus eight electrons and minus Last four H 20. didn't cancel eight electrons on both sides. And now for the balanced reaction, The last four SN 2 plus plus, I owe four minus plus H plus Initials, four s and four plus and minus Plus four H 2 L. For the 2nd 1, S. N two plus, N 03 minus Initials s. n four plus plus N. O. I need to first separate the whole reaction into two half reactions, FSN two plus. And this goes to s. n four plus. We have N 03 minus, going to N. O. Now we need to balance the 900 non oxygen elements first and this is already balanced. Now we need to balance the oxygen by adding H 20, liquid oxygen. On the second reaction, you have three options on the raft inside. Put one on the product side, We can add to H20. To the park side. Now, we need to balance our hydrogen by adding H plus this, I've been to hydrogen. We have no hydrogen on the acting side. We have four on the product side. We can add four H plus the react inside. Now we need to bounce our charges by adding electrons to the more positive side. Have a look at our charges. We have a plus two here, last four here Plus four here -100. I need to add two electrons to the product side And they need to add three electrons here to the wrapping side. Now we need to balance our electrons on the two half reactions. There are two in the first equation and three in a second. They need to find at least common multiple which is six months, probably three in the top equation. Most part two. In the second equation We're gonna have three s. n two plus. Building three S N four plus plus six electrons. They have six electrons plus H plus Plus two n. 03 getting to N O Plus four H 20. Now we need to get the overall reaction by adding the two reactions. Any common species? We're gonna have three S. N. Plus just two. N. 03 minus plus. H plus plus six electrons. Building three s. n four plus plus six electrons Plus two and out Plus four H 20. The balanced reaction we get three S. N two plus Plus two and plus H plus Building three S and four Plus. That's two and two As for H 20. And then for our last reaction We're gonna have s. n two plus plus M N. Of four minus Building SN four plus M. N. 0. 2. We're gonna have SN two plus Building SN four Plus. We have M N. Afford MNO two. Now we need to balance the 900 non option elements First. This is already balanced. Now need to balance the oxygen by adding H 20 liquid beside that needs oxygen went for auction on the right side and the second reaction, but only two on the product side. Look at two H 20. The product side. Now we need to balance the hydrogen by adding H plus to the side. That needs hydrogen. Have no hydrogen on the raft inside, but we have four on the product side. Look at four H Plus. Now I need to balance the charges by adding electrons to the more positive side. If you look at our charges plus two, last four, Last four, 0000. I need to add two electrons to the product side And added four electrons over here to the reacting side. Actually, this is the negative side B -1 here. I need to add plus three electrons on the right hand side. Now I need to bounce our electrons on the two half reactions. I have to in the first reaction and three in the second. Almost three on the top two on the bottom, We get three s in two plus, Item three SN 4 plus I have six electrons. Then we have six electrons plus four H plus H plus Plus two MN 04 Leading to MN. or two Plus four H 20. We have three SN 2 plus plus two In a 4 -4 Plus eight h plus plus six electrons. Building three s and four plus six electrons Plus two MNO plus four H +20. Our balanced reaction, The three SN plus, that's two and a four minus plus H plus Building three SN plus Plus two MNO Plus four H 2 out. Thanks for tomorrow video and I hope it was helpful.

Ch.20 - Electrochemistry

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Brown 14th Edition

Ch.20 - Electrochemistry

Problem 51a

Chapter 20, Problem 51a

Given the following reduction half-reactions:
Fe³⁺(aq) + e^- → Fe²⁺(aq) E°_red = +0.77 V
S₂O₆^2-(aq) + 4 H⁺(aq) + 2 e^- → 2 H₂SO₃(aq) E°_red = +0.60 V
N₂O(g) + 2 H⁺(aq) + 2 e^- → N₂(g) + H₂O(l) E°_red = -1.77 V
VO²⁺(aq) + 2 H⁺(aq) + e^- → VO²⁺ + H₂O(l) E°_red = +1.00 V
(a) Write balanced chemical equations for the oxidation of Fe²⁺(aq) by S₂O₆^2-(aq), by N₂O(aq), and by VO²⁺(aq).

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Identify the reduction half-reactions given in the problem and their respective standard reduction potentials (E°).

Determine the oxidation half-reaction for Fe^{2+} by reversing the given reduction half-reaction: Fe^{2+} \rightarrow Fe^{3+} + e^{-}.

For each oxidizing agent (S_2O_6^{2-}, N_2O, VO_2^{+}), write the corresponding reduction half-reaction as given in the problem.

Combine the oxidation half-reaction of Fe^{2+} with each reduction half-reaction to form a balanced redox equation. Ensure that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction.

Balance the overall chemical equation for each reaction by ensuring that the number of atoms and the charge are balanced on both sides of the equation.

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Key Concepts

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

Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions involve the transfer of electrons between species, where oxidation refers to the loss of electrons and reduction refers to the gain of electrons. In these reactions, the species that loses electrons is oxidized, while the one that gains electrons is reduced. Understanding the roles of oxidizing and reducing agents is crucial for balancing redox equations.

Standard Electrode Potentials (E°)

Standard electrode potentials (E°) are measured under standard conditions and indicate the tendency of a species to be reduced. A higher E° value means a greater likelihood of reduction occurring. In redox reactions, comparing the E° values of the half-reactions helps determine which species will oxidize and which will reduce, guiding the balancing of the overall reaction.

Balancing Redox Reactions

Balancing redox reactions involves ensuring that both mass and charge are conserved. This typically requires separating the reaction into half-reactions for oxidation and reduction, balancing each for atoms and charge, and then combining them. The use of coefficients and the adjustment of electrons transferred are essential steps in achieving a balanced equation.

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Key Concepts

Oxidation-Reduction Reactions

Standard Electrode Potentials (E°)

Balancing Redox Reactions