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Iron II Ammonium Sulfate & Nickel II Solutionwhat is the Half – Reaction inanode:what is the Half- Reaction incathode:what is the overall Balanced cellre...

Question

Iron II Ammonium Sulfate & Nickel II Solutionwhat is the Half – Reaction inanode:what is the Half- Reaction incathode:what is the overall Balanced cellreaction:

Iron II Ammonium Sulfate & Nickel II Solution what is the Half – Reaction in anode: what is the Half- Reaction in cathode: what is the overall Balanced cell reaction:



Answers

In the half-cell $\mathrm{Zn}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Zn}(s),$ what is the electrode and is this half-reaction an anodic reaction or a cathodic reaction?

So in this case we have a specific reaction. Iron reacts with copper and it's two plus form to you and iron and it's two plus form and copper. So we can separate this into half reactions where we can see the iron is going to be oxidized since iron goes from zero oxidation state to a plus two. So is it lower to higher oxidation state? This will occur at the end note. And then we have the reduction at the cathode or copper two plus is reduced to copper since it goes from plus to higher oxidation state to zero lower oxidation state. And essentially for this, this reaction will be spontaneous given that copper likes to essentially retain the lower oxidation state while iron likes to be in the higher oxidation state. So this is spontaneous reactions over spontaneous reactions. We would need essentially a voltaic so we would need a salt bridge that essentially allows for the exchange of essentially ions. So salt bridge and ions move towards the anote, which is negatively charged in this case and cat ions move to the capital, which is positively charged for will take so a galvanic saw actually, and essentially we observe for a voltaic cell like this, the mass of the android would decrease since we're losing essentially a solid form to go into solution. So the mass of the iron, an ode would decrease and essentially there would be essentially reduction and the Catherine mass technically would increase. But by a small amount and we would see basically that electrons flow in the wire and electrons move from the an ode to the cathode. And these are the principles mainly for this type of cell and that's it.

Let's consider the Redox reaction. We've got MG solid. Add copper two plus Sequus to produce N G two plus Equus and copper solid. So the oxidation here for the an ode half reaction will be, uh, magnesium solid to produce magnesium two plus and two electrons. And the reduction or the cathode half reaction will be the copper two plus plus two electrons producing copper solid. Let's draw a call batting sell here. So if we were to sketch this out here, two beakers uh, let's say decide. It's my an ode, uh, said It's my cathode and Theodore. Half reaction is the magnesium. Sorry. Magnesium solid going to magnesium. Two plus two electrons half the road copper here. We're gonna need a magnesium solid bar and in the solution were going toe have magnesium treats. One bowler on over here. We need a copper solid bar copper nitrate solution. We'll need assault bridge, which will be filled with one Moeller K. C. L would be our salt. We have a volt meter connecting the two electrodes together and there would be a simple, galvanic self. Electrons would move through the wire from the annals through to the cathode, and we

In this problem, we're gonna be looking at the following Redox reaction and practicing, making a galvanic cell out of it. And so to start with, I'm just gonna do some basic on. Ah, analyzing of what's going on here, Let's break it down to the reaction, reduction and oxidation. So let's start with Let's do some numbering. We have zero for our solid plus two for our eye on and then the same on the right. And to identify our reduction, we look for the reduction in oxidation States of this fell from plus 2 +20 to reduction, and this increased from zero to plus two. That's an oxidation. And so to remember our handy pneumonic we have an ox Katt read, which means that an out is the site of oxidation and the cathode is the site of reduction. So when I'm drawing my galvanic, so I'm going to remember that and that's a great thing to remember. You'll never forget it, if you can remember that. And so here we have our basic galvanic cell set up, and I'm gonna call this side my an ode in this side my cathode. And that means I'm going to have this side, my oxidation side be zinc and my reduction side you lead because lead is being played it onto here a solid and then, uh, zinc is also being, um it's being oxidized, which means it is going from its solid state to its a qui estate in solution and then on this side, we also have lead ions floating around because they are being, um, added to the plate and electrons air flowing from left to right from an 02 cathode. And we're generating a natural current here because this is a spontaneous process. And so what's happening here in terms of half reactions, we can analyze it, too. So are oxidation. Half reaction is gonna be zinc solid to zinc ion, and we'll see that that's two electrons are being lost and that our reduction is actually lead acquiesced to lead solid, and it's gonna involve two electron addition. And as you concede, these guys balance out nicely. We don't need toe change anything about the coefficients. And so this oxidation half reaction is occurring at the an ode. We are losing those electrons through this cop through this wire, and it's going all the way to this lead plate on then and conversely, our lead ions are coming out of this solution and going right up to our lead plate. And of course, we have a salt bridge going on here to to neutralize thes charges, and I hope that this video is helpful.

In this problem, we're gonna be going over a basic galvanic cell set up based on the following redox reaction on. So to start with, I will identify my oxidation and reduction going on. Let's just get our terms established here. So identifying the oxidation numbers it zero for this aluminum solid. Plus two for this Nikolai on plus three for this aluminum ion and zero for the nickel solid. And so I'm seeing that my reduction is taking place with my nickel because we felt from plus 2 to 0. And my oxidation is taking place with the aluminum because we increased from zero to plus three. And so I'm gonna remember my handy, uh, pneumonic for, uh, galvanic cells, which is an notice. The site of oxidation Katha to the side of reduction with an ox cat red. And so I'm gonna set up my basic out. Van Exel. Just pretend this is a lot prettier than it is. We can have a volt meter in here. Great. And so I'm going to say that this is my an ode, and this is my cat foad. Great. And so since we know that an artist side of oxidation we're gonna need to place the easily oxidized metal on the left. And so that is gonna be aluminum solid. And we're gonna make sure that we're swimming in a nice little pool of aluminum I owns. And then on the right hand side, we're gonna have our nickel, which is being plated as a solid and again it's swimming in a nice pool of nickel ions. And so we are going to then observe Ah, what is happening with, um, each side? So Anote is the site of oxidation electrons air flowing from the an ode to the cathode and being plated onto this nickel over here. And just to break this down in terms of half reactions as well. So we have a l. L three plus and then we have Nikolai on going to nickel solid and will notice that the charges are a bit different here, So there's probably some sort of balancing that needs to happen. So we have three electrons that need to be added to this side and then to that need to be added to this side. And so we'll see that, um, we'll need to do some strike yama tree to resolve this, um, and then put together that's gonna lead us to our final product out here. But essentially, these half reactions, uh, are gonna end up being to aluminum to l three plus in six electrons and then we have six electrons plus three Nikolai ons, goes to three nickels solids, and so that is essentially are half reaction for our cathode and are an ode, and I hope that was helpful.


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