What is the coefficient on Sn 2+ when the following reaction is balanced in acidic solution? (a) 2 (b) 4 (c) 5 (d) 7

Hello. Everyone in this video, we're trying to balance out this reaction over here. So what I'm gonna do first is break this down into two half reactions and balance them out separately. So We do. So we have our c. 0. 3 plus to give us C. 02 plus And then the other half reaction is RMN two plus To give us MN. -. Alright, so the first step is to go ahead to balance out our oxygen atoms. So we see for the first half reaction this doesn't need any oxygen's because we only purely have the cobalt. But on the second reaction we definitely do need to add this source of hydrogen or oxygen. How we add a source of oxygen is by adding H20. So on the right side we have four atoms of auction but on the right or on the left we have none on the right, we have four. So how can balance out the auctions is by adding four moles of H20 to the left side. So kind of break this up. So this is our half reactions. And now we're going to do in this step is balance our oxygen's. Okay. So again, the first reaction will just stay as is that they need to change yet in terms of balancing out our oxygen atoms by the left side, we'll add those waters. So again we have our Mn two plus and then we said we're adding four moles of H 20. All right now let's go ahead and balance are hydrogen atoms. Again on the first half reaction. We don't need to do anything because we don't need any oxygen or hydrogen sources but we definitely need to balance out our high jeans for the second reaction. Take a look at the second reaction. Then on the left side we have eight moles or eight hydrogen atoms, and then on the right side we have none. So what we'll do is purely add hydrogen sources and we can do that by adding our H plus ions. So let's first start off with the MN two plus. Then the four moles of H 20. To give us M N 04 minus. So because we have eight atoms of hydrogen on the left side, we need eight moles of H plus on the right side. Alright, so, so we can see here now that our oxygen's and our high regions are balanced. Next step to take is to go ahead and balance out the charges. Alright, I'm just gonna go ahead and scroll down to give us a little bit more space to work with. Alright, so we can see the first reaction here definitely needs to have a um charge balance. So we see here that we have a three plus chart on the left and then we have a two plus charge on the right, we can balance this out is by adding a negative source, so one electron to the left side to bounce it out. So for the first reaction we have C 03 plus, we're giving one electron here And that gives us CO two plus. Alright, let's take a look at the second half reaction then. So on the left side we have a two plus charge on right side. Let's see, we have minus one and then eight plus charges. So we have a seven plus charge. We need a two plus charged on the right side to kind of balance it out. How we can do this is by adding five sources of a negative source which is gonna be an electron. So we add five moles of electrons. So then we have MN two plus Writing with four moles of H 20. And that gives us what we have M and 04 minus Reacting with eight moles of hydrogen can ions. And lastly to bounce our charges will add five moles of electrons. Alright, so lastly the last thing to do is to balance out our in terms of balancing at least we can balance out our electrons. So we see on the second half reaction we have five electrons but on the first one we only have one. So what we need to do is Multiply this whole first reaction by five and then if we do so We'll get five moles of CO three plus Plus five electrons to give us five CO 2 plus. And then the second reaction will just remain as is again just scrolling down to give us a little bit more space. Alright, so now we can combine these two half reactions to give us the overall net reaction. So we see here that the five electrons will cancel out. All right, if we have that then our final reaction will be five C 03 plus, Reacting with MN two plus And four moles of H 20. And on product side we have C O or five moles of C plus, Reacting with MN 04 And eight moles of H. Plus. So the question was asking was the coefficient of CO3 plus. In our net reaction, I will see here that five is going to be that coefficient. So formally then our answer IS five. And right here is going to be my final answer for this problem. Thank you all so much for watching

Ch.19 - Electrochemistry

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McMurry 8th Edition

Ch.19 - Electrochemistry

Problem 2

Chapter 19, Problem 2

What is the coefficient on Sn²⁺ when the following reaction is balanced in acidic solution?
(a) 2 (b) 4 (c) 5 (d) 7

Verified step by step guidance

insert step 1> Identify the oxidation and reduction half-reactions in the given unbalanced reaction.

insert step 2> Balance the atoms other than O and H in each half-reaction.

insert step 3> Balance the oxygen atoms by adding H2O molecules to the side that needs oxygen.

insert step 4> Balance the hydrogen atoms by adding H+ ions to the side that needs hydrogen.

insert step 5> Balance the charge by adding electrons to one side of each half-reaction, then combine the half-reactions ensuring the electrons cancel out.

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Key Concepts

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

Balancing Redox Reactions

Balancing redox reactions involves ensuring that both mass and charge are conserved. In acidic solutions, this typically requires the use of half-reactions, where oxidation and reduction processes are balanced separately before combining them. The coefficients in front of each species indicate the number of moles involved in the reaction, which must be adjusted to achieve balance.

Oxidation States

Oxidation states are a way to keep track of electrons in chemical reactions. Each element in a compound is assigned a specific oxidation state based on its electron configuration and bonding. Understanding the changes in oxidation states during the reaction is crucial for identifying which species are oxidized and reduced, which directly influences the coefficients used in balancing the equation.

Acidic Solution Conditions

In an acidic solution, the presence of hydrogen ions (H⁺) affects the balancing of redox reactions. When balancing in acidic conditions, additional H⁺ ions and water molecules may be added to the half-reactions to ensure that both mass and charge are balanced. This is particularly important when determining the coefficients for species like Sn²⁺ in the overall balanced equation.