Draw an electrolytic cell in which Mn2+ is reduced to Mn and Sn is oxidized to Sn2+. Label the anode and cathode, indicate the direction of electron flow, and write an equation for the half- reaction occurring at each electrode. What minimum voltage is necessary to drive the reaction?

Video transcript

hi everyone for this problem. It reads which of the following images shows the correct electrolytic cell with labeled an Oid and Kathy Lloyd and indicated electron flow direction where cobalt is oxidized and nickel is reduced. What are the half reactions at the an Oid? And the catholic how much voltage is needed to run the reaction? Okay, so the question we want to answer here is which image shows the correct Okay, electrolytic cell. Okay. And then we also want to answer the question how much voltage is needed to run the reaction. Okay, so here let's start off by defining what is an electrolytic cell that is a cell that converts electrical energy into chemical energy. And we have some images here that we need to identify which one is correct. So the first thing that we need to look at is the an Oid and Kathy Oid. Okay, so the an Oid end is positive and the catholic end is negative. So let's take a look at our options. So far. We should have our an Oid on the positive side and our cath oid on the negative side of our voltage source. Okay. And so if we take a look so far at our answer choices, we can see that answer choice. D is not going to be correct because our catholic here is positive and are an Oid is negative. Okay, so let's just write a note here. So are an oid is positive in our catholic Oid is negative. Okay, so the next thing that we should look at is our flow. Okay, so we have a voltage source and with electrolytic cells, the redox reaction is not spontaneous. And so electrical energy has to be supplied, which is why we have this voltage source to supply and initiate the reaction. Okay, so our voltage source and we're dealing with electrons. So we need to see for our an Oid which is positive. Our electron flow is going to flow towards that positive charge. Okay, so let's take a look at a we have our electrode here, we have our electron here and our electron is negative. So it should be flowing towards our positive charge. So so far answer choice A shows that let's take a look at B. B. Also shows that answer choice C. We see that our electron flow is going in the opposite direction because it's a negative charge, it should be going towards that positive voltage source. So we can see here that we can eliminate answer choice C. Okay, because the flow is not correct. Alright, let's look at the next thing. So when we have an an Oid, this is dealing with oxidation. Okay. And so the question also asked us to write our half reactions. Okay, so because our an OID is oxidation are cobalt is going to be remember oxidation is loss. So let's go ahead and write out our an Oid half reaction. Okay. And we'll do that one in a different color. So we have our cobalt and we said that an Oid is positive and its oxidation. If we remember acronym oil rig oxidation is loss. Okay, so that means we're going to be losing an electron or we're going to be losing two electrons. Okay, so this is going to become this is going to become Our Cobalt two plus plus two electrons. So that's our half reaction for an android and then for a catholic oid are catholic, Oid is reduction. Okay. And from our acronym oil rig reduction is gained. So we're gaining two electrons. So we have our nickel two plus is gaining two electrons. So this becomes nickel salt. Okay, so these are two half reactions here. So let's take a look at our option choices to see which one is going to be correct. Okay, so if we take a look at at A and B are nickel is is being is going is undergoing reduction. Okay. And so that means our nickel two plus ion is gaining two electrons. And so if you can look at our image here, this image does not match that. Okay, because here we have our in our image the opposite. Okay, so that gives us answer choice. A as our correct answer here, this is going to be the correct image. Okay, so is the correct image we wrote the two half reactions the last question asks us how much voltage is needed to run the reaction. Okay, so let's go ahead and do that. So for our voltage we have a equation and our equation is the electrolysis of our catholic oid minus the electro license of our an Oid. Okay, so our value for a catholic oid is going to be negative 0.23 minus our value for an oid, which is negative 0.28. That gives us our voltage is going to equal 0.05. Okay, so that is going to be the voltage, the minimum voltage needed. So this is the end of this problem. So answer choice a is the correct answer. These are two half reactions based off of oxidation and reduction and then this is our our minimum voltage needed to run the reaction. Okay, so that is the end of this problem. I hope this was helpful.